MAS439/MAS6320 COMMUTATIVE ALGEBRA
AND ALGEBRAIC GEOMETRY
MOTY KATZMAN
Contents
1. Introduction 3
2. Commutative rings– definitions and examples 5
3. Subrings 7
4. Homomorphisms of rings 9
5. Ideals 11
6. Prime ideals 15
7. Radicals and nilradicals 18
8. Sneak preview: Algebraic sets 19
8.1. The Zariski Topology 22
8.2. The Nullstellensatz 23
8.3. Affine varieties 26
8.4. Coordinate rings 28
8.5. Polynomial maps 30
9. Modules 32
10. Homomorphisms of R-modules 33
11. Quotients of modules 34
12. The first Isomorphism Theorem 35
13. Generators of modules 35
14. Direct sums and products 37
15. Free modules 37
16. Localization of rings 38
17. Localization of modules 41
18. Exact sequences 42
19. The exactness of localization 43
20. Local properties 44
21. Further properties of localization 45
22. Noetherian rings 46
23. Hilbert’s Basis Theorem 48
24. Primary decomposition 49
1
MAS439/MAS6320 COMMUTATIVE ALGEBRA AND ALGEBRAIC GEOMETRY 2
25. Artinian rings 53
26. The height on an ideal 54
27. Matrix algebra over commutative rings 56
28. Algebras 57
29. Integral extensions 58
30. Noether’s Normalization Theorem 60
31. Hilbert’s Nullstellensatz 62
32. The Going Up Theorem 63
33. Dimensions of rings 64
34. Algebraic sets, again 65
35. Hilbert’s Nullstellensatz– strong form 66
36. Irreducible algebraic sets, again 68
37. The dimension of an algebraic set 69
MAS439/MAS6320 COMMUTATIVE ALGEBRA AND ALGEBRAIC GEOMETRY 3
1. Introduction
Commutative algebra is the study of comuutative rings. These are sets R having
two composition laws (called addition and multiplication) which behave in the same
way as addition and multiplication of integers. Commutative rings come up all over
pure mathematics. Here are two examples:
• Number theory. The Gaussian integers
Z[i] = {a+ bi : a, b ∈ Z} ⊂ C
form a commutative ring under the usual operations of addition and mul-
tiplication of complex numbers. Similarly, if we consider the third root of
unity
ω = exp(2pii/3) = (−1 +
√
3 i)/2
then the subset
Z[ω] = {a+ bω : a, b ∈ Z} ⊂ C
form a commutative ring. To see this note the relations
ω3 = 1, ω2 + ω + 1 = 0.
Rings of algebraic integers like these are useful for studying Diohantine equa-
tions. For example, to prove the first case n = 3 of Fermat’s last theorem one
writes
z3 = x3 + y3 = (x+ y)(x+ ωy)(x+ ω2y)
and considers prime factorizations of both sides in the ring Z[ω].1
• Geometry. We can study a topological space X (for example a subset X ⊂ Rn)
via its ring of functions. This is the set of all continuous functions f : X → R.
These can be added and multiplied pointwise:
(f + g)(x) = f(x) + g(x), (f · g)(x) = f(x) · g(x).
Similarly we can consider rings of differentiable or analytic functions.
Algebraic geometry is the study of systems of polynomial equations. The set of
solutions to such a system is called an affine variety. We first choose a field K we wish
to work over (e.g. K = Q, R or C). Given a collection of polynomials f1, f2, · · · , fr
in n variables x1, · · · , xn with coefficients in K, the corresponding affine variety is the
subset
V (f1, · · · , fr) ⊆ Kn
consisting of those points (a1, · · · , an) ∈ Kn satisfying
f1(a1, · · · , an) = f2(a1, · · · , an) = · · · = fr(a1, · · · , an) = 0.
1See Hardy and Wright, An Introduction to the Theory of Numbers, Chapters 12–13.
MAS439/MAS6320 COMMUTATIVE ALGEBRA AND ALGEBRAIC GEOMETRY 4
Perhaps surprisingly, algebraic geometry occupies a central place in modern pure
mathematics. The basic reason is that the study of polynomial equations has many
different aspects, which relate to lots of other areas of pure mathematics. We give a
few examples of this:
• Number theory. If we take the field k = Q then polynomial equations become
Diophantine equations. For example Fermat’s last theorem is the statement
that if n > 2 the variety
xn + yn = 1
has no points over the field k = Q with x, y nonzero.
• Geometry and topology. Working over C we can study the geometry and
topology of varieties. For example, consider the variety
y2 = x3 − x.
This is an example of an elliptic curve. Topologically it is a torus (dough-
nut) with 3 points removed. To get the full torus we must work with the
corresponding projective variety2.
• Physics. In string theory the world is supposed to have 10 dimensions. 4 of
these dimensions form spacetime, the other 6 are supposed to be curled up very
small, and account for the properties of the fundamental forces. For string
theory to work properly these curled up dimensions must have a special shape:
they should be Calabi-Yau threefolds3. Examples of Calabi-Yau threefolds are
most easily described using polynomial equations. For example, again working
over C, the variety
x51 + x
5
2 + x
5
3 + x
5
4 + x1x2x3x4 = 0
is a Calabi-Yau threefold known as the quintic threefold. Note that it has
4− 1 = 3 complex dimensions, and hence 6 real dimensions.
• Algebra. The basic calculational tool in algebraic geometry is commutative
algebra. Every affine variety X has an associated co-ordinate ring
K[V ] = K[x1, · · · , xn]/(f1, · · · , fr),
which is a commutative ring. It is obtained by quotienting the polynomial
ring K[x1, · · · , xn] by the ideal generated by the polynomials defining V .
This course will focus on the relationship between commutative algebra and alge-
braic geometry. We aim to introduce the basic commutative algebra needed to study
2See Frances Kirwan, ‘Complex algebraic curves’, Chapter 5.
3See Brian Greene, ‘The Elegant Universe’.
MAS439/MAS6320 COMMUTATIVE ALGEBRA AND ALGEBRAIC GEOMETRY 5
affine varieties, and to build up an intuition for the relationship between the geomet-
rical properties of an affine variety V and the algebraic properties of the associated
co-ordinate ring K[V ].
2. Commutative rings– definitions and examples
Definition 2.1. A ring is a set R together with two binary operations + : R×R→ R
(“addition”) and · : R×R→ R (“multiplication”) such that
(a) (R,+) is an Abelian group (we call the additive identity 0R or just 0),
(b) multiplication is associative: (a · b) · c = a · (b · c) for all a, b, c ∈ R, and there
exists a multiplicative identity 1R: a1R = 1Ra = a for all a ∈ R, and
(c) for all a, b, c ∈ R, a · (b+ c) = a · b+ a · c and (a+ b) · c = a · c+ b · c.
If a ring R satisfies the additional condition that a · b = b · a for all a, b ∈ R then we
call R a commutative ring.
Remark. Notice that for any r ∈ R, −1R · r = −r and 0R · r = (1R + (−1R))r =
r + (−r) = 0R.
Remark. Notice that a field is just a commutative ring R in which every element other
that 0R has a multiplicative inverse.
Example. The integers Z with its usual addition and multiplication is a commutative
ring. Also, Z/nZ the integers mod n form a commutative ring.
Example. Let C[0, 1] be the set of continuous real functions defined on [0, 1]. For
f, g ∈ C[0, 1] define (f + g)(x) = f(x) + g(x) and (f · g)(x) = f(x)g(x). This turns
C[0, 1] into a commutative ring.
Example (Polynomial rings). Let R be a commutative ring. The set of polynomials
in one variable with coefficients in R forms a commutative ring R[x]. The elements
are finite sums of the form
f(x) =
d∑
i=0
aix
i = a0 + a1x+ a2x
2 + · · ·+ adxd
with a0, a1, · · · ad ∈ R. If ad 6= 0 we call the number d ≥ 0 the degree of the polynomial
f(x) (we consider the zero polynomial to have degree 0). Polynomials of degree 0 are
called constant polynomials. The addition and multiplication laws are( n∑
i=0
aix
i
)
+
( n∑
i=0
bix
i
)
=
n∑
i=0
(ai + bi)x
i.
( m∑
i=0
aix
i
) · ( n∑
i=0
bjx
j
)
=
mn∑
k=0
ckx
k where ck =
k∑
i=0
ai · bk−i.
MAS439/MAS6320 COMMUTATIVE ALGEBRA AND ALGEBRAIC GEOMETRY 6
The zero element is the zero polynomial 0R. The identity is the constant polynomial
1R.
Now we can define R[x1, . . . , xn], the polynomial ring in x1, . . . , xn with coefficients
in R, recursively as R[x1, . . . , xn] = R[x1, . . . , xn−1][xn].
Example (Power series rings). Let R be a commutative ring and define R[[x]] to be
the set of all power series in x with coefficients in R, i.e., all expressions a0 + a1x +
· · ·+ anxn + . . . where a0, . . . , an, · · · ∈ R with the following operations( ∞∑
i=0
aix
i
)
+
( ∞∑
j=0
bjx
j
)
=
∞∑
i=0
(ai + bi)x
i,
( ∞∑
i=0
aix
i
)
·
( ∞∑
j=0
bjx
j
)
=
∞∑
d=0
d∑
k=1
(ak · bd−k)xd.
Its not hard to see that R[[x]] is a commutative ring with additive identity 0R + 0Rx+
. . . 0Rx
n + . . . and multiplicative identity 1R + 0Rx+ 0Rx
2 . . . 0Rx
n + . . . .
Now we can define R[[x1, . . . , xn]], the power series ring in x1, . . . , xn with coefficients
in R, recursively as R[[x1, . . . , xn]] = R[[x1, . . . , xn−1]][[xn]].
Example (Direct Products). Let R and S be commutative rings. Define the direct
product of R and S to be the Cartesian product R×S with addition (r1, s1)+(r2, s2) =
(r1 + r2, s1 + s2) and multiplication (r1, s1) · (r2, s2) = (r1 · r2, s1 · s2). This is a
new commutative ring, which we denote R × S, with additive identity (0R, 0S) and
multiplicative identity (1R, 1S).
Definition 2.2. Let R be a commutative ring. An element a ∈ R is a zero-divisor if
there exists a b ∈ R \ {0R} such that ab = 0R.
We say that R is an integral domain (or just a domain, for short) if 0R is its only
zero-divisor.
Example. Z is a domain and so is any field; Z/nZ is a domain if and only if n is
prime.
Example. Let R and S be commutative rings; R × S is never a domain because
(1R, 0S) · (0R, 1S) = (0R, 0S) = 0R×S.
Proposition 2.3. If R is a domain, so are R[x] and R[[x]].
MAS439/MAS6320 COMMUTATIVE ALGEBRA AND ALGEBRAIC GEOMETRY 7
Proof. Suppose that for non-zero f, g ∈ R[[x]] we have fg = 0. Since f, g are not
zero we can write f = amx
m + am+1x
m+1 + . . . and g = bnx
n + bn+1x
n+1 + . . . with
am, bn 6= 0R. Now
0R[[x]] = f · g = ambnxm+n + (am+1bn + ambn+1)xm+n+1 + . . .
so ambn = 0R. But R is an integral domain and am, bn 6= 0R so these are not zero-
divisors and we obtain a contradiction.
The proof that R[x] is an integral domain is similar.
Remark. From now on we will refer to the additive and multiplicative identities of a
ring R as 0 and 1 rather that 0R and 1R if no confusion is likely.
3. Subrings
Definition 3.1. Let S be a commutative ring. We say that R ⊆ S is a subring of S
if
(a) the addition and multiplication of S restricted to R make R into a commuta-
tive ring, and
(b) 1S ∈ R.
Proposition 3.2 (The subring criterion). Let S be a commutative ring and let R ⊆ S
be a subset. Then R is a subring of S if and only if
(a) 1S ∈ R,
(b) R is closed under addition and multiplication,
(c) for all a ∈ R, −a ∈ R.
Example. The following is a chain of subrings
Z ⊆ Q ⊆ R ⊆ C
and so is
R ⊆ R[[x1]] ⊆ R[[x1, x2]] ⊆ R[[x1, x2, x3]] ⊆ . . . .
Example (Polynomial rings). Let R be a commutative ring; let R[x] be the subset
of R[[x]] consisting of finite sums, i.e., all elements
∑∞
i=0 aix
i ∈ R[[x]] for which there
exists an n ≥ 0 such that ai = 0R for all i ≥ n.
Lemma 3.3. Let R be a commutative ring, let Λ be a set, and for each element
λ ∈ Λ, let Sλ be a subring of R. Then the intersection S =
⋂
λ∈Λ Sλ ⊂ R is a subring.
MAS439/MAS6320 COMMUTATIVE ALGEBRA AND ALGEBRAIC GEOMETRY 8
Proof. This is immediate from the definitions. Let us check for example that the
intersection S is closed under addition. Suppose given elements a, b ∈ S. This means
precisely that a, b ∈ Sj for all j ∈ J . Then since the subsets Sj ⊂ R are subrings we
have a+ b ∈ Sj for all j ∈ J . This then implies that a+ b ∈ S.
Definition 3.4. Let R be a ring and T ⊂ R a subset. The subring of R generated
by T is the intersection of all subrings S ⊂ R containing T . It is denoted 〈T 〉 ⊂ R.
Note that 〈T 〉 ⊂ R is indeed a subring by Lemma 3.3. In fact it is the smallest
subring of R containing T : if S ⊂ R is any other subring containing T then by
definition we have 〈T 〉 ⊂ S. To describe 〈T 〉 ⊂ R explicitly we first define the subset
(1) Tˆ = {t1 · t2 · · · tn : n ≥ 0, ti ∈ T} ⊂ R,
consisting of all finite (possibly empty) products of elements of T . Note that 1R ∈ Tˆ
by definition, since we interpret the empty product as meaning 1R.
Lemma 3.5. The subring 〈T 〉 ⊂ R consists of those elements of R which can be
written as finite (possibly empty) sums of the form
(2) r = ±p1 ± p2 ± · · · ± pk with k ≥ 0 and pi ∈ Tˆ .
Here, when k = 0, we interpret the empty sum as meaning 0R.
Proof. Let X ⊂ R be the set of all elements of the form (2). It is easy to see that any
subring S ⊂ R containing T also contains X, since by definition S is closed under
addition, multiplication and additive inverses. In particular X ⊂ 〈T 〉. On the other
hand, we claim that the subset X ⊂ R is itself a subring. Since T ⊂ X it then follows
from Definition 3.4 that 〈T 〉 ⊂ X, and hence that 〈T 〉 = X.
To prove the claim note that X is clearly closed under addition and additive inverses
and contains 1R. To see that it is closed under multiplication we use
(±p1 ± · · · ± pk) · (±q1 ± · · · ± ql) =
∑
i,j
±(pi · qj).
Since pi · qj ∈ Tˆ for all i, j, the sum on the right also lies in X.
Definition 3.6. We say that a ring R is generated by a subset T ⊂ R if R = 〈T 〉.
We say that a ring R is finitely-generated if it is generated by a finite subset.
Example.
(a) The residue class ring Z/n and the integers Z are generated by the empty
subset ∅.
(b) The polynomial ring Z[x] is generated by the single element set {x} ⊂ Z[x].
(c) Take R = Z[x] and consider the one-element subset T = {x2} ⊂ R. The
subring S = 〈T 〉 ⊂ R consists of all polynomials in x2 with integer coefficients.
MAS439/MAS6320 COMMUTATIVE ALGEBRA AND ALGEBRAIC GEOMETRY 9
Example. The ring Q is not finitely-generated. Indeed, given any finite subset T ⊂ Q
there is a finite set of prime numbers Π such that every element of T can be written
as a fraction a/b, with the integer b being a product of primes from Π. The subset
S ⊂ Q consisting of all rational numbers which can be written in this form is easily
seen to be a subring of Q. Hence 〈T 〉 ⊂ S and since S ( Q is a proper subring, it
follows that Q is not generated by T .
4. Homomorphisms of rings
Definition 4.1. Let R and S be rings. A homomorphism from R to S is a function
φ : R→ S with the following properties:
(a) for all a, b ∈ R, φ(a+ b) = φ(a) + φ(b) and φ(ab) = φ(a)φ(b), and
(b) φ(1R) = 1S.
A homomorphism is an isomorphism if it is a bijection. We say that R and S are
isomorphic if there is an isomorphism from R to S, and we denote this R ∼= S.
Remark. Notice that condition (a) above implies that that any homomorphism φ :
R → S, is an homomorphism of the additive groups of R and S so φ(0R) = 0S and
for all r ∈ R, φ(−r) = −φ(r).
Example. Let S be a commutative ring and let R be a subset of S. Then R is a
subring of S if and only if the inclusion map R ⊆ S is a homomorphism.
Example (The evaluation map). Let R be a commutative ring, let r ∈ R. The
function φ : R[x]→ R defined as φ(a0 + a1x + · · · + adxd) = a0 + a1r + · · · + adrd is
an homomorphism.
Can a similar construction produce an homomorphism φ : R[[x]]→ R?
Proposition 4.2. For any ring R there exists a unique ring homomorphism ψ : Z→
R.
Proof. For all n ∈ Z define
ψ(n) =

1R + · · ·+ 1R︸ ︷︷ ︸
n times
if n > 0
0R if n = 0
−
1R + · · ·+ 1R︸ ︷︷ ︸
−n times
 if n < 0

MAS439/MAS6320 COMMUTATIVE ALGEBRA AND ALGEBRAIC GEOMETRY 10
Remark. The ring homomorphism in Proposition 4.2 may or may not be injective;
in the former case Z is a subring of R but in any case we can make sense of an
expression nr with n ∈ Z and r ∈ R: this is interpreted to be ψ(n) · r, e.g. nr =1R + · · ·+ 1R︸ ︷︷ ︸
n times
 r = r + · · ·+ r︸ ︷︷ ︸
n times
for n > 0 and (−1)r = −r for n = −1.
Example. The map φ : R[x] → C defined as φ(a0 + a1x + · · · + adxd) = a0 + a1i +
· · ·+ adid is a ring homomorphism.
Definition 4.3. Let R, S be commutative rings and let φ : R → S be a ring homo-
morphism. The kernel of φ, denoted kerφ, is {r ∈ R |φ(r) = 0} and the image of φ,
denoted Imφ, is {φ(r) | r ∈ R}
Proposition 4.4. Let R, S be commutative rings and let φ : R → S be a ring
homomorphism. The image of φ is a subring of S.
Proof. We apply the subring criterion:
(a) 1S = φ(1R) ∈ Imφ,
(b) if s1, s2 ∈ Imφ, there exist r1, r2 ∈ R such that s1 = φ(r1), s2 = φ(r2); now
s1 + s2 = φ(r1) + φ(r2) = φ(r1 + r2) ∈ Imφ,
s1 · s2 = φ(r1) · φ(r2) = φ(r1 · r2) ∈ Imφ,
and
(c) for any s ∈ Imφ, there exists a r ∈ R such that s = φ(r); now −s = −φ(r) =
φ(−r) ∈ Imφ.

Remark. Why isn’t the kernel of a ring homomorphism φ : R→ S a subring of R?
Proposition 4.5.
(a) If f : R→ S and g : S → T are ring homomorphisms then so is the composite
map g ◦ f : R→ T .
(b) If f : R→ S is a bijective ring homomorphism then the inverse map f−1 : S →
R is also a ring homomorphism. (So if R is isomorphic to S, then S is also
isomorphic to R.)
MAS439/MAS6320 COMMUTATIVE ALGEBRA AND ALGEBRAIC GEOMETRY 11
Proof. Part (a) is immediate from the definitions and is left to the reader.
For part (b) note first that the assumption f(1R) = 1S ensures that f
−1(1S) = 1R.
To prove that f−1 preserves addition we must show that
(3) f−1(s1 + s2) = f−1(s1) + f−1(s2)
for all s1, s2 ∈ S. Since f is a bijection we can take r1, r2 ∈ R such that f(ri) = si.
Thus f−1(si) = ri. Since f preserves addition we have f(r1 + r2) = s1 + s2. Hence
f−1(s1 + s2) = r1 + r2. This proves (3). A similar argument works for multiplication.

5. Ideals
Definition 5.1. Let R be a commutative ring. A non-empty subset I ⊆ R is an ideal
of R if
(a) for all a, b ∈ I, a+ b ∈ I, and
(b) for all a ∈ I and r ∈ R, ra ∈ I.
Notice that (a) just says that I is a subgroup of the additive group of R.
Example. For any commutative ring, {0R} and R are ideals of R.
Example. For any integer n, nZ is an ideal of Z.
Example. Let R = C[x1, . . . , xn] and let (a1, . . . , an) ∈ Cn. The set
{f ∈ R | f(a1, . . . , an) = 0}
is an ideal. More generally, for any S ⊆ Cn,
{f ∈ R | f(s) = 0 for all s ∈ S}
is an ideal.
Proposition 5.2. The kernel of a ring homomorphism φ : R→ S is an ideal of R.
Proof. Pick any a, b ∈ kerφ and r ∈ R. Now φ(a + b) = φ(a) + φ(b) = 0 + 0 = 0 so
a+ b ∈ kerφ and φ(ra) = φ(r)φ(a) = φ(r)0 = 0 so ra ∈ kerφ.
Proposition 5.3. A ring homomorphism φ : R → S is injective if and only if
kerφ = {0R}.
Proof. This is true for homomorphisms of groups.
MAS439/MAS6320 COMMUTATIVE ALGEBRA AND ALGEBRAIC GEOMETRY 12
Proposition 5.4. Let φ be a ring homomorphism φ : R→ S and let I be an an ideal
of S. Then φ−1(I) = {a ∈ R |φ(a) ∈ I} is an ideal of R.
Proof. If a, b ∈ φ−1(I), i.e., if φ(a), φ(b) ∈ I, then φ(a + b) = φ(a) + φ(b) ∈ I, so
a+ b ∈ φ−1(I). If now r ∈ R, φ(ra) = φ(r)φ(a) ∈ I, so ra ∈ φ−1(I).
With the notation above, we call φ−1(I) the contraction of I to R.
Proposition 5.5. Let R be a commutative ring, let Λ 6= ∅ be a set and let {Iλ}λ∈Λ
be a set of ideals of R. Then ∩λ∈ΛIλ is an ideal.
Proof. Pick any a, b ∈ ∩λ∈ΛIλ and r ∈ R. Now for all λ ∈ Λ, a, b ∈ Iλ, so a + b ∈ Iλ
for all λ ∈ Λ, i.e., a+ b ∈ ∩λ∈ΛIλ. Also for all λ ∈ Λ, ra ∈ Iλ, so ra ∈ ∩λ∈ΛIλ.
Definition 5.6. Let R be a commutative ring and let A ⊂ R be any subset. We
define the ideal generated by A, denoted 〈A〉, to be the set of all elements of R of the
form r1a1 + · · ·+ rnan where n ≥ 1, r1, . . . , rn ∈ R and a1, . . . , an ∈ A.
If A is a finite set, say A = {a1, . . . , an} we also write a1R + · · ·+ anR for 〈A〉.
Example. Every ideal in a PID R has the form aR for some a ∈ R.
Proposition 5.7. Let R be a commutative ring and let A ⊂ R be any subset. Then
〈A〉 is an ideal. Furthermore, if I ⊆ R is any ideal containing A, 〈A〉 ⊆ I and 〈A〉 is
the intersection of all ideals which contain A.
Proof. The fact that 〈A〉 is an ideal follows easily from its definition.
Write J for the intersection of all ideals which contain A.
If I is an ideal which contains A, it must also contain all elements r1a1 + · · ·+ rnan
where n ≥ 1, r1, . . . , rn ∈ R and a1, . . . , an ∈ A, i.e., it must contain 〈A〉.
Any ideal containing A must contain 〈A〉 so J ⊇ 〈A〉. But 〈A〉 is one of these ideals
in the intersection, so J ⊆ 〈A〉, and we deduce that the intersection is 〈A〉.
Definition 5.8. With the notation as above, if J is an ideal of R, φ(J) need not be
an ideal of S, e.g., let φ be the inclusion map Z ⊂ Q, clearly φ(Z) is not an ideal
of Q. However, we can talk about the expansion of J to S, which is the ideal of S
generated by φ(J) and which we will often denote JS if no confusion can occur.
MAS439/MAS6320 COMMUTATIVE ALGEBRA AND ALGEBRAIC GEOMETRY 13
Definition 5.9. Let R be a commutative ring and let {Iλ}λ∈Λ be a set of ideals of
R. We define the sum of ideals
∑
λ∈Λ Iλ to be the ideal 〈∪λ∈ΛIλ〉.
Definition 5.10. Let R be a commutative ring and let I1, . . . , In be a finite set of
ideals of R. We define the product of ideals
∏n
j=1 Ij to be the ideal generated by all
products {a1 · · · · · an | a1 ∈ I1, . . . , an ∈ In}. For any ideal I ⊆ I and any n ≥ 1 we
also write In = I · . . . · I︸ ︷︷ ︸
n times
.
Example. Let R = K[x1, . . . , xn], I = x1R + · · ·+ xnR. What is Im?
Notice that x21 + x
2
2 ∈ I2 cannot be written as a product of two elements in R.
Ideals play a role similar to that of normal subgroups of groups: they provide a
way to construct a quotient.
Definition 5.11. Let R be a commutative ring and let I ⊆ R be an ideal. We
can forget for the moment the multiplicative structure of R and consider it as an
(additive) Abelian group; now I is a (normal) subgroup of R and we can form the
quotient R/I to obtain a new Abelian group. We turn the Abelian group R/I into a
commutative ring by defining multiplication as follows: for all r, s ∈ I let
(r + I) · (s+ I) = r · s+ I.
This is well defined because, if r+I = r′+I and s+I = s′+I, i.e., if r−r′, s−s′ ∈ I,
(r+ I) · (s+ I) = r ·s+ I = r ·s− r · (s−s′)+ I = rs′+ I = r′s′+ I = (r′+ I) · (s′+ I).
It is straightforward to verify that R/I with this multiplication satisfies the axioms
defining commutative rings. For example, for any r + I, s + I, t + I ∈ R/I, (r +
I) ((s+ I)(t+ I)) = rst+ I = ((r + I)(s+ I)) (t+ I). Also notice that 1R/I = 1 + I
and 0R/I = 0 + I = I.
Example. Let n ∈ Z; we observed that nZ, set of integers divisible by n, is an ideal
of Z so we can form the quotient Z/nZ which is nothing but the ring of integers mod
n.
Example. Let R = R[x] and let I be the ideal of R generated by the one element
x2 + 1 ∈ R, i.e., I is the set of polynomials in R which are divisible by x2 + 1. The
commutative ring R/I isomorphic to C.
Define φ : R/I → C as follows: for any coset f(x) + I let φ(f(x) + I) = f(i). This
is well defined because, if f(x) + I = f ′(x) + I then f ′(x) = f(x) + r(x2 + 1) for some
r ∈ R and then
φ(f ′(x) + I) = φ(f(x) + r(x2 + 1) + I) = f(i) + r(i2 + 1) = f(i) = φ(f(x) + I).
Since any complex number a+ ib equals φ(a+ bx), φ is surjective.
MAS439/MAS6320 COMMUTATIVE ALGEBRA AND ALGEBRAIC GEOMETRY 14
To verify that φ is injective, we show that kerφ = 0. Suppose that φ(f(x)+I) = 0,
i.e., that f(i) = 0, i.e., that i is a root of f . Now since f has real coefficients, −i is a
root of f , too. Thinking of f as an element of C[x] we deduce that f is divisible by
(x− i)(x+ i) = x2 +1, and this holds in R. We now deduce that f ∈ I, i.e., f+I = 0.
Proposition 5.12. Let R be a commutative ring, I ⊆ R an ideal. Define a function
φ : R→ R/I with φ(r) = r + I. Then φ is a ring homomorphism which is surjective
and whose kernel is I.
Proof. This is an easy exercise.
Remark. We refer to the homomorphism φ in the proposition above as the quotient
homomorphism.
Proposition 5.13. Let R be a commutative ring, I ⊆ R an ideal and let φ : R→ R/I
be the quotient map. There is a bijection between the ideals of R/I and the ideals of
R which contain I given by
J ⊆ R/I −→ φ−1(J) = {a ∈ R |φ(a) ∈ J}
K ⊆ R,K ⊇ I −→ φ(K) = {φ(a) ∈ R/I | a ∈ K}.
Proof. We first show that for K ⊆ R,K ⊇ I , φ(K) is indeed an ideal of R/I (this
is false for a general homomorphism φ). If a + I, b + I ∈ φ(K), i.e., if a, b ∈ K then
(a+ I) + (b+ I) = (a+ b+ I) ∈ φ(K). If now also r+ I ∈ R/I then (r+ I)(a+ I) =
ra+ I = φ(ra) ∈ φ(K).
Now notice that for any ideal J ⊆ R/I, φ−1(J) ⊇ φ−1(0) = kerφ = I.
Finally, since φ is surjective, for any ideal J ⊆ R/I, φ (φ−1(J)) = J and, since
kerφ = I ⊆ K, φ−1 (φ(K)) = K.
Theorem 5.14 (The first isomorphism theorem). Let R and S be commutative
rings and let φ : R → S be an homomorphism. Then φ induces a isomorphism
φ : R/ kerφ→ Imφ given by φ(r + kerφ) = φ(r).
MAS439/MAS6320 COMMUTATIVE ALGEBRA AND ALGEBRAIC GEOMETRY 15
Proof. We first show that φ is well defined: if r + kerφ = r′ + kerφ then r′ = r + a
with a ∈ kerφ. Now
φ(r′ + kerφ) = φ(r′) = φ(r + a) = φ(r) + φ(a) = φ(r) + 0S = φ(r) = φ(r + kerφ).
φ is clearly surjective, and we now show that it is injective:
φ(r + kerφ) = φ(r) = 0⇔ r ∈ kerφ⇔ r + kerφ = 0R/ kerφ.

Example.
C ∼= R[x]/(x2 + 1)R[x].
To see this let φ : R[x] → C be the evaluation maps φ(p(x)) = p(i); φ is surjective
and we just need to show that kerφ = (x2 + 1)R[x]. If p(x) ∈ kerφ then we can write
p(x)+(x2 +1)R[x] = (ax+b)+(x2 +1)R[x] for some a, b ∈ R, because we can subtract
from the coset representative p(x) multiples of x2 +1, and now 0 = φ(ax+b) = ai+b,
so a = b = 0 and p(x) ∈ (x2 + 1)R[x].
6. Prime ideals
Definition 6.1. An ideal I of a commutative ring R is a prime ideal if I 6= R and
whenever ab ∈ I for some a, b ∈ R, then a ∈ I or b ∈ I.
Example. The ideal nZ of Z is prime if and only if n is a prime integer.
Example. The ideal 〈f〉 of R[x] is prime if and only if f is irreducible, i.e., cannot
be written as a product of two polynomials of positive degree.
Proposition 6.2. Let P be a prime ideal in a commutative ring R. If r1r2 . . . rn ∈ P
for some r1, . . . , rn, then of one the factors is in P .
Proof. Easy induction on n.
Proposition 6.3. Let R and S be commutative rings and let φ : R → S be a ring
homomorphism. If P ⊂ S is a prime ideal, so is φ−1(P ) ⊂ R.
MAS439/MAS6320 COMMUTATIVE ALGEBRA AND ALGEBRAIC GEOMETRY 16
Proof. Let a, b ∈ R be such that, ab ∈ φ−1(P ), i.e., φ(ab) ∈ P . Now φ(ab) =
φ(a)φ(b) ∈ P and since P is prime, φ(a) ∈ P or φ(b) ∈ P , i.e., a ∈ φ−1(P ) or
b ∈ φ−1(P ).
Proposition 6.4. Let R be a commutative ring and let I ⊂ R be an ideal. If P is a
prime ideal of R which contains I then the image of P in R/I is prime. Consequently,
the bijection described in 5.13 between the ideals of R/I and the ideals of R which
contain I restricts to a bijection between the prime ideals of R/I and the prime ideals
of R which contain I.
Proof. Write P for the image of P in R/I. If a + I, b + I ∈ R/I are such that
(a+ I)(b+ I) ∈ P , i.e., if ab+ I = p+ I with p ∈ P , then ab ∈ P + I = P . Since P
is prime, a ∈ P or b ∈ P , i.e., a+ I ∈ P or b+ I ∈ P .
Proposition 6.5. Let R be a commutative ring and let I ( R be an ideal. The ideal
I is prime if and only if R/I is a domain.
Proof. Assume first that I is prime. If a+ I and b+ I are non-zero elements of R/I,
i.e., if a, b /∈ I, then ab /∈ I and(a+ I)(b+ I) = ab+ I 6= 0R/I .
Conversely, if R/I is a domain and for some a, b ∈ R, ab ∈ I, then (a+ I)(b+ I) =
0R/I and so a+ I = 0R/I or b+ I = 0R/I , i.e., a ∈ I or b ∈ I.
Proposition 6.6. Let R be a commutative ring and let I1, . . . , In ⊂ R be ideals and
let P ⊂ R be a prime ideal. The following are equivalent:
(1) P ⊇ Ij for some 1 ≤ j ≤ n,
(2) P ⊇ I1 ∩ · · · ∩ In, and
(3) P ⊇ I1 . . . In.
Proof. The implications (a) ⇒ (b) ⇒ (c) are trivial; we prove (c) ⇒ (a). Assume
that P ⊇ I1 . . . In but for all 1 ≤ j ≤ n we can pick an aj ∈ Ij \ P . Now a1 . . . an ∈
I1 . . . In \ P , a contradiction.
Definition 6.7. An ideal I of a commutative ring R is a maximal ideal if, I 6= R
and the only ideal properly containing I is R.
MAS439/MAS6320 COMMUTATIVE ALGEBRA AND ALGEBRAIC GEOMETRY 17
Proposition 6.8. Maximal ideals of a commutative ring R are prime.
Proof. Let I be a maximal ideal and suppose that for some a, b ∈ R we have ab ∈ I.
Now if neither a ∈ I nor b ∈ I, the ideals I+aR and I+bR are strictly bigger that I,
but as I is maximal, we have I+aR = R and I+bR = R and also (I+aR)(I+bR) = R.
But (I + aR)(I + bR) = I2 + aI + bI + abR and since each of these summands is in
I so is the sum. We deduce that R ⊆ I, a contradiction.
Proposition 6.9. The bijection described in 5.13 between the ideals of R/I and the
ideals of R which contain I restricts to a bijection between the maximal ideals of R/I
and the maximal ideals of R which contain I.
Proof. Easy exercise.
Remark. Let R and S be commutative rings and let φ : R → S be a ring homo-
morphism. We noticed that if P ⊂ S is prime, so is φ−1(P ). However, note that if
P ⊂ S is maximal, φ−1(P ) need not be maximal. E.g., let φ be the inclusion map
Z ⊂ Q; we’ll soon see that {0Q} ⊂ Q is a maximal ideal, but φ−1({0Q}) = {0Z} is
not maximal.
Proposition 6.10. Let R be a commutative ring.
(a) R is a field if and only if R are {0R} are the only ideals of R.
(b) An ideal I ⊂ R is maximal if and only if R/I is a field.
Proof. (a) Assume first R is a field. If I is an ideal of R which contains a non-zero
element r, it must also contain r−1r = 1R, and so must contain R1R = R. Assume
now that R are {0R} are the only ideals of R. Pick any non-zero r; the ideal 〈r〉 must
be R so there exists an a ∈ R such that ar = 1R, so any non-zero element in R has a
multiplicative inverse, i.e., R is a field.
(b) If I is maximal then Proposition 5.13 tells us that R/I cannot have an ideal other
than {0R/I} and R/I so R/I is a field. Conversely, if R/I is a field, any ideal of R
which contains I must be I or R, so I is maximal.
MAS439/MAS6320 COMMUTATIVE ALGEBRA AND ALGEBRAIC GEOMETRY 18
Theorem 6.11. Any proper ideal I in a commutative ring R is contained in a max-
imal ideal.
Proof. Consider the set S of all proper ideals of R which contain a given ideal I. Given
any totally ordered subset A ⊆ S with respect to inclusion (i.e., for any I, J ∈ A,
I ⊆ J or J ⊆ I) denote with IA the union of all the ideals in A. We now show that
IA is also a proper ideal. If r ∈ R and a ∈ IA, then a ∈ Iα for some α ∈ A and so
ra ∈ Iα ⊆ IA. If now we have another b ∈ IA, say b ∈ Iβ for some β ∈ A then either
Iα ⊆ Iβ, in which case a + b ∈ Iβ ⊆ IA or Iβ ⊆ Iα, in which case a + b ∈ Iα ⊆ IA
(or both). So IA is an ideal, and it is proper because if 1R ∈ IA, there must exist an
α ∈ A for which 1R ∈ Iα, contradicting the fact that Iα is a proper ideal.
We now invoke Zorn’s Lemma, one of the equivalent forms of the Axiom of Choice
which says that if a partially ordered set is such that every totally ordered subset
has an upper bound, the partially ordered set has a maximal element. This maximal
element in S is a proper ideal maximal with respect to inclusion, i.e., a maximal ideal.

7. Radicals and nilradicals
Definition 7.1. Let R be a commutative ring. An element r ∈ R is nilpotent if
there exists an m ≥ 1 such that rm = 0R. The nilradical of R the set of all nilpotent
elements in R.
Proposition 7.2. Nil(R) is an ideal and it equals the intersection of all prime ideals
of R.
Proof. If r, s ∈ Nil(R) pick positive integers α and β such that rα = sβ = 0. Now
(r + s)α+β =
α+β∑
i=0
(
α + β
i
)
risα+β−i
and for all 0 ≤ i ≤ α + β, i ≥ α or α + β − i ≥ β, which implies that ri = 0 or
sα+β−i = 0, hence (r + s)α+β = 0.
Also for any t ∈ R, (tr)α = tαrα = 0 and we deduce that Nil(R) is an ideal.
For any r ∈ Nil(R) such that rα = 0 and any prime ideal P ⊂ R, the fact that
rα ∈ P implies that r ∈ P hence Nil(R) is contained in all prime ideals of R.
MAS439/MAS6320 COMMUTATIVE ALGEBRA AND ALGEBRAIC GEOMETRY 19
It remains to show that if r is in all prime ideals, it is nilpotent. By way of
contradiction we assume that r is not nilpotent, in which case 0 /∈ {ri}i≥1 and the set
S = {I ⊂ R | I is an ideal and ri /∈ I for all i ≥ 1}
is not empty because it contains {0}. Obviously, any totally ordered subset of S
has an upper bound, namely its union, so Zorn’s Lemma implies that there exists a
maximal element J ∈ S. I claim that J is prime: if ab ∈ J for some a, b ∈ R \ J
consider the ideals K = J +Ra and L = J +Rb; these are strictly larger than J and
the maximality of J implies that there exist α, β ≥ 1 such that rα ∈ K and rβ ∈ L,
say rα = j1 + u1a and r
β = j2 + u2b with j1, j2 ∈ J and u1, u2 ∈ R. Multiply these to
obtain
rα+β = (j1 + u1a)(j2 + u2b) = j1j2 + j1u2b+ j2u1a+ u1u2ab ∈ J
contradicting the fact that J ∈ S.
Definition 7.3. For any ideal I of a commutative ring R we define the radical of I
to be √
I = {r ∈ R | rα ∈ I for some α ≥ 1}.
Proposition 7.4.
√
I is an ideal and is the intersection of all prime ideals in S which
contain I.
Proof. Let f : R → R/I be the quotient map and notice that r ∈ √I if and only if
f(r) ∈ Nil(R/I).
The first statement follows from the fact that
√
I = f−1 Nil(R/I).
Also
f−1 (Nil(R/I)) = f−1
 ⋂
P⊂R/I prime
P
 = ⋂
P⊂R/I prime
f−1
(
P
)
=
⋂
I⊆P⊂R prime
P.

8. Sneak preview: Algebraic sets
In this section we introduce the application of commutative algebra to algebraic
geometry, i.e., to the study of algebraic sets. We will revisit this topic at the end
of the course when we have developed the tools to define the dimension of algebraic
sets.
MAS439/MAS6320 COMMUTATIVE ALGEBRA AND ALGEBRAIC GEOMETRY 20
Fix a field K, for example K = Q,R or C. Define the affine space
AnK = {(a1, · · · , an) ∈ Kn}.
Note that this is just the set Kn; the funny notation is traditional in algebraic geom-
etry.
Given a polynomial f = f(x1, · · · , xn) ∈ K[x1, · · · , xn] and a point
p = (a1, · · · , an) ∈ AnK
we have an element f(p) ∈ K obtained by substituting xi = ai in the polynomial f .
In other words, we view elements of the ring K[x1, · · · , xn] as functions f : AnK → K,
and f(p) is the evaluation of the function f at the point p.
Consider the maps
(4) Ideals I ⊂ K[x1, · · · , xn]
V
--
Subsets V ⊂ AnK
I
nn
defined by
V (I) = {p ∈ AnK : f(p) = 0 for all f ∈ I},
I(V ) = {f ∈ K[x1, · · · , xn] : f(p) = 0 for all p ∈ V }.
Note that these maps are order-reversing:
I1 ⊆ I2 =⇒ V (I1) ⊇ V (I2); V1 ⊆ V2 =⇒ I(V1) ⊇ I(V2).
We now make the very important
Definition 8.1. A subset X ⊂ AnK is said to be algebraic if it is of the form X = V (I)
for some ideal I ⊂ K[x1, · · · , xn].
We will see later (Theorem 23.1) that any ideal I ⊂ K[x1, · · · , xn] is finitely-
generated and, therefore, of the form I = (f1, · · · , fr) for some finite set of polymonials
f1, · · · , fr. Then
V (I) = V (f1, · · · , fr) = {(a1, · · · , an) ∈ AnK : fi(a1, · · · , an) = 0 for 1 ≤ i ≤ r}.
Thus, a subset X ⊂ AnK is algebraic precisely if it is the vanishing locus of a finite set
of polynomials.
Example.
(a) The empty-set ∅ ⊂ AnK is an algebraic subset; it is the vanishing locus of the
unit ideal K[x1, · · · , xn].
(b) The set AnK itself is an algebraic subset; it is the vanishing locus of the zero
ideal (0) ⊂ K[x1, · · · , xn].
MAS439/MAS6320 COMMUTATIVE ALGEBRA AND ALGEBRAIC GEOMETRY 21
(c) The circle
V = {(a, b) ∈ A2R : a2 + b2 = 1} ⊂ A2R
is an algebraic subset. Indeed, V = V (I) where I ⊂ R[x, y] is the principal
ideal generated by the polynomial
f = x2 + y2 − 1 ∈ R[x, y].
(d) The union of the two co-ordinate axes
V = {(a, b) ∈ A2R : ab = 0} ⊂ A2R
is an algebraic subset; being equal to V (I) where
I = (xy) ⊂ R[x, y].
(e) Any point (a, b) ∈ A2R is an algebraic subset. It is the vanishing locus of the
non-principal ideal
I = (x− a, y − b) ⊂ R[x, y].
(f) The subset
V = {a ∈ R : a ≥ 0} ⊂ A1R
is not algebraic. Indeed, any nonzero polynomial in R[x] has only finitely
many roots, so any proper algebraic subset of A1R consists of at most finitely
many points.
Finite intersections and arbitrary unions of algebraic subsets are also algebraic:
Proposition 8.2.
(a) Given an arbitrary collection of ideals Iλ ⊂ R indexed by a set Λ we have
V
(∑
λ∈Λ
Iλ
)
=
⋂
λ∈Λ
V
(
Iλ
)
.
(b) Given two ideals I1, I2 ⊂ R we have
V (I1 ∩ I2) = V (I1) ∪ V (I2).
Proof. For part (a) recall that the ideal
∑
λ∈Λ Iλ consists of all finite sums of elements
of
⋃
λ∈Λ Iλ. The statement P ∈ V (Iλ) for all λ ∈ Λ is equivalent to the statement that
for any f ∈ ⋃λ∈Λ Iλ we have f(P ) = 0. But then the same is true for any element
f ∈∑λ∈Λ Iλ.
For part (b) note that there are obvious inclusions V (Ij) ⊆ V (I1 ∩ I2) which gives
an inclusion
V (I1 ∩ I2) ⊇ V (I1) ∪ V (I2).
MAS439/MAS6320 COMMUTATIVE ALGEBRA AND ALGEBRAIC GEOMETRY 22
For the reverse inclusion suppose that p /∈ V (I1) ∪ V (I2). Then we can find f1 ∈ I1
and f2 ∈ I2 with f1(p) 6= 0 and f2(p) 6= 0. But then f1 · f2 ∈ I1 ∩ I2 and f1 · f2(p) 6= 0
so that p /∈ V (I1 ∩ I2).
Example. Consider the ideals I1 = (x, y) ⊂ C[x, y, z] and I2 = (z) ⊂ C[x, y, z]. Thus
V (I1) is the z-axis, and V (I2) is the (x, y)-plane. We claim that
I1 + I2 = (x, y, z), I1 ∩ I2 = (xz, yz).
For the second statement suppose that f ∈ I1∩I2. Then since f ∈ I1, all monomials
appearing with nonzero coefficient in f(x, y, z) must contain either an x or a y. On
the other hand, since f ∈ I2, any monomial appearing with nonzero coefficient in
f(x, y, z) must contain a z. Hence all monomials appearing in f(x, y, z) contain
either xz or yz, and it follows that f ∈ (xz, yz).
Finally note that V (I1 + I2) = {(0, 0, 0)} and V (I1 ∩ I2) is the union of the z-axis
with the (x, y)-plane.
8.1. The Zariski Topology.
Remark. Digression on the definition of a topological space. Recall that Rn has a
distance function on it: given two points x, y ∈ Rn we define
d(x, y) =
( n∑
i=1
(yi − xi)2
) 1
2
.
A subset U ⊂ Rn is open if for any x ∈ U there exists an r > 0 such that
x ∈ Br(x) = {y ∈ Rn : d(x, y) < r} ⊂ U.
A subset F ⊂ Rn is called closed if it is the complement of an open subset. One can
make the same definitions in any metric space X.
Open subsets of a mteric space X (e.g. X = Rn) have the following properties:
(a) The empty set ∅ ⊂ X and the set X itself are both open subsets;
(b) If U1, U2 ⊂ X are open subsets then so is the intersection U1 ∩ U2 ⊂ X;
(c) If Ui ⊂ X is an arbitrary collection of open subsets of X indexed by some set
I then the (possibly infinite) union
⋃
i∈I Ui ⊂ X is also an open subset.
A set X with a collection of subsets (called open subsets) satisfying these conditions
is called a topological space. There are some rather extreme examples: you could take
just ∅, X as your open sets, or you could take all subsets to be open. A metric
space always gives rise to a topological space by defining open subsets as above. But
different metrics (equivalent ones) can give rise to the same topological space. And
some topological spaces do not come from a metric space at all.
MAS439/MAS6320 COMMUTATIVE ALGEBRA AND ALGEBRAIC GEOMETRY 23
Remark. A topological space X is called Hausdorff if given any two distinct points
x 6= y ∈ X, there exist open subsets U, V ⊂ X such that
x ∈ U, y ∈ V and U ∩ V = ∅.
The topological space arising from a metric space (X, d) is always Hausdorff: given
x, y ∈ X we can take U and V to be open balls at x and y with radius r < d(x, y)/2.
But for example, if a set X has more than one element, then the topology whose only
open subsets are ∅ and X is never Hausdorff.
Proposition 8.2 shows that the algebraic subsets form the closed subsets of a topol-
ogy on AnK . This is called the Zariski topology. It’s a rather weird topology, and
almost never comes from a metric structure.
Example. Consider the case n = 1. Any polynomial f ∈ K[x] of degree d has at most
d roots. Hence the algebraic subsets of A1K = K are just the finite sets of points.
Note that if K is infinite then any two non-empty open subsets intersect. Thus this
topology is certainly not Hausdorff in general.
8.2. The Nullstellensatz. Consider again the correspondences (4). We would like
to say that V and I define inverse bijections, but this is of course not true as they
are currently defined.
Example.
(a) Let I = (x2) ⊂ C[x] and J = (x) ⊂ C[x]. Then V (I) = V (J) = {0} ⊂ A1C.
(b) Let I = (x2 + 1) ⊂ R[x] and J = R[x]. Then V (I) = V (J) = ∅.
(c) Let k = Z/p for some prime number p > 0 and take I = xp − x. Then
V (I) = V (0) = A1k because by Fermat’s little theorem, any element a ∈ Z
satisfies ap ≡ a modulo p.
The problem with Example (a) is that the ideal I is not radical. Recall the definition
of the radical of an ideal
√
I = {r ∈ R : rn ∈ I for some n ≥ 1}.
Note that V (I) = V (
√
I) because fn(p) = 0 =⇒ f(p) = 0. Moreover, for the same
reason, any ideal of the form I(V ) is radical. Thus we can restrict our maps V and
I to correspondences
Radical ideals I ⊂ R
V
--
Algebraic subsets V ⊂ AnK
I
mm
The basic problem in Example (b) is that we considered a polynomial x2 + 1 which
had no real roots. Of course this would not have worked over C, where the polynomial
MAS439/MAS6320 COMMUTATIVE ALGEBRA AND ALGEBRAIC GEOMETRY 24
x2 +1 has the two roots ±i. Example (c) shows a different problem when the field k is
finite: there are nonzero polynomials which vanish at every point of A1k. To eliminate
both of these problems we restrict to a special class of fields.
Definition 8.3. A field K is said to be algebraically closed if every non-constant
polynomial f ∈ K[x] has a root in K.
Using polynomial division it follows that if k is algebraically closed then any poly-
nomial f ∈ K[x] can be written in the form
f(x) = c · (x− a1) · · · (x− ak),
with c ∈ k and ai ∈ k.
Example.
(a) The fields Q and R are not algebraically closed because f(x) = x2 + 1 has no
roots.
(b) The field C is algebraically closed: this is called the fundamental theorem of
algebra.
(c) A finite field K is never algebraically closed because the polynomial
f(x) = 1 +
∏
a∈K
(x− a)
has no roots. Thus algebraically closed fields have infinitely many elements.
Remark. It can be proved that any field K can be embedded as a subfield K ⊆ K¯ of
a minimal algebraically closed field known as the algebraic closure of k. It is defined
by formally adjoining roots of all polynomials in K[t]. Thus for example C is the
algebraic closure of R, and the algebraic closure of Q is the field Q¯ ⊂ C of algebraic
numbers.
The following important result is called Hilbert’s Nullstellensatz (German for ‘Ze-
roes theorem’).
Theorem 8.4. Assume that K is algebraically closed. Then
I(V (I)) =
√
I.
In particular, if I is a radical ideal then I(V (I)) = I.
We defer the proof to later in the course (in section 31.)
MAS439/MAS6320 COMMUTATIVE ALGEBRA AND ALGEBRAIC GEOMETRY 25
Example. Consider the ideal
I = (x2 + y2 − 1, y − 1) ⊂ C[x, y].
The algebraic set V = V (I) is the intersection of the circle of unit radius centered
at the origin with the line y = 1. It therefore consists of a single point (0, 1). The
element x ∈ C[x, y] lies in the ideal I(V ) since it vanishes on the set V . Thus by the
Nullstellensatz we must have x ∈ √I. In fact
x2 = x2 + y2 − 1− (y + 1) · (y − 1) ∈ I.
But we claim that x /∈ I. Indeed, if x ∈ I we can write
x = p(x, y) · (x2 + y2 − 1) + q(x, y) · (y − 1)
for some polynomials p, q ∈ C[x, y]. Applying the homomorphism C[x, y] → C[x]
obtained by setting y = 1 gives x = p(x, 1) · x2 which gives a contradiction.
The following important corollary is absolutely vital for algebraic geometry.
Corollary 8.5. Assume that K is algebraically closed. The two maps I, V give
mutually-inverse order-reversing bijections
Radical ideals I ⊂ R
V
--
Algebraic subsets V ⊂ AnK
I
mm
Proof. The Theorem shows that if I a radical ideal then I(V (I)) = I. Now suppose
that V ⊂ AnK is an algebraic subset. By definition we can write V = V (I) for some
ideal, which we can take to be radical since V (I) = V (
√
I). But now
V (I(V )) = V (I(V (I)) = V (I) = V
so the result is proved.
It follows that we get a bijection between maximal ideals of R and mimimal al-
gebraic subsets of AnK . Note that any point p = (a1, · · · , an) ∈ AnK is an algebraic
subset since it is the vanishing locus of the ideal
I(p) = (x1 − a1, · · · , xn − an).
Hence we get a bijection
Maximal ideals I ⊂ R
V
--
Points p ∈ AnK
I
mm
.
MAS439/MAS6320 COMMUTATIVE ALGEBRA AND ALGEBRAIC GEOMETRY 26
8.3. Affine varieties. We have given geometric interpretations of radical and max-
imal ideals in polynomial rings above. The following definition allows a similar inter-
pretation of prime ideals.
Definition 8.6. We say that an algebraic set V ⊂ AnK is irreducible if it cannot be
written in the form V = V1 ∪ V2 with Vi ( V being proper algebraic subsets.
Example.
(a) The algebraic set V = V (xz, yz) ⊂ A3C is not irreducible since it is the union
V = V (z) ∪ V (x, y) which are both proper algebraic subsets.
(b) The algebraic set V = V (xy) ⊂ A2C is not irreducible since it is the union
V = V (x) ∪ V (y) which are both proper algebraic subsets.
Lemma 8.7. An algebraic set X ⊂ AnK is irreducible iff the ideal I(X) ⊂ R is prime.
Proof. Write R = K[x1, . . . , xn].
Suppose first that X is irreducible and suppose that fg ∈ I(X) for some f, g ∈ R.
Now X ⊆ V (fg) = V (f) ∪ V (g) and so
X = (V (f) ∩X) ∪ (V (g) ∩X).
Both sets in the union are algebraic, and since X is irreducible we deduce that either
X = V (f) ∩X or X = V (g) ∩X, i.e., X ⊆ V (f) or X ⊆ V (g), and we deduce that
f ∈ I(X) or g ∈ I(X).
Assume now that I(X) is prime and that X = X1 ∪ X2 where X1 and X2 are
algebraic sets. We assume further that X 6= X1 and show that X = X2. Since
X % X1, I(X) $ I(X1) and we can pick a f ∈ I(X1) \ I(X). Now for all g ∈ I(X2)
we have fg ∈ I(X), and since I(X) is prime, we deduce that g ∈ I(X). We conclude
that I(X2) ⊆ I(X), and since the reverse inclusion also holds, we conclude that
I(X2) = I(X) and hence that X = X2.
Thus we get a final bijection
Prime ideals I ⊂ R
V
..
Irreducible algebraic subsets V ⊂ AnK
I
mm
It is fairly obvious that we can break any given algebraic set into irreducible pieces.
More precisely we have
Lemma 8.8. Every algebraic subset X ⊂ AnK has a unique (up to reordering) decom-
position
X = X1 ∪X2 ∪ · · · ∪Xr
with each Xi ⊆ AnK an irreducible subset, and Xi 6⊆ Xj for i 6= j.
MAS439/MAS6320 COMMUTATIVE ALGEBRA AND ALGEBRAIC GEOMETRY 27
Proof. This is true regardless of whether K is algebraically closed or not (see theorem
36.2. In the case where K is algebraically closed we proceed as follows.
We assume Hilbert’s Basis Theorem (Theorem 23.1 which we will prove later in
the course) and which states that polynomial rings over a field do not have infinite
strictly ascending chains of ideals.
Let us call an algebraic subset X ⊂ AnK good if it can be written as a finite union
of irreducible algebraic subsets. Suppose X0 ⊂ AnK is bad. Then in particular X0 is
not irreducible, so we can write it as a union X0 = X1 ∪ Y with X1, Y ( X. If X1
and Y are both good then obviously X0 would also be good. Hence without loss of
generality X1 must be bad. We can repeat the argument and obtain a bad subset
X2 ( X1. In this way we obtain an infinite descending chain
· · · ( X3 ( X2 ( X1 ( X0
of bad subsets of X0. Now the order-reversing bijection of Corollary 8.5 implies
the existence of an infinite ascending chain of radical ideals, contradicting Hilbert’s
Basis Theorem. We have therefore shown that any algebraic subset X is good, and
hence has a decomposition as in the statement. We can easily ensure the minimality
condition Xi 6⊆ Xj for i 6= j by simply discarding any unnecessary factors.
For uniqueness, suppose we have
X = X1 ∪ · · · ∪Xn = Y1 ∪ · · · ∪ Ym
both satisfying the minimality condition. Then we have
Y1 = (Y1 ∩X1) ∪ · · · ∪ (Y1 ∩Xn).
Since Y1 is irreducible we must have Y1∩Xi = Y1 for some i. Thus Y1 ⊂ Xi. The same
argument shows that we must also have Xi ⊂ Yj for some j. Since this implies that
Y1 ⊂ Yj the minimality condition implies that j = 1 and hence Xi = Y1. Reordering
the Xis we have proved that X1 = Y1. Repeating this argument gives the result.
Example. Consider I = (x2− yz, xz−x) ⊂ C[x, y, z] and the corresponding algebraic
subset V = V (I) ⊂ A3C. Suppose that (a, b, c) ∈ V . Then the second equation shows
that either a = 0 or c = 1. In the first case we must have either b = 0 or c = 0. In
the second case we must have b = a2. It is then easy to see that
V = V (x, y) ∪ V (x, z) ∪ V (y − x2, z − 1).
Let us label the subsets appearing on the right hand side V1, V2 and V3 respectively.
Note that the minimality condition Vi 6⊆ Vj for i 6= j is satisfied because the points
(0, 0, 2), (0, 1, 0) and (1, 1, 1) are contained only in V1, V2 and V3 respectively.
To prove that the subsets Vi are irreducible it is enough to show that the defining
ideals are prime. For the first two this follows as in the last example. For the last we
MAS439/MAS6320 COMMUTATIVE ALGEBRA AND ALGEBRAIC GEOMETRY 28
note that
C[x, y, z]/(y − x2, z − 1) ∼= C[x]
via the unique C-algebra homomorphism sending (x, y, z) 7→ (x, x2, 1). We have now
proved that the irreducible components of V are the subsets V1, V2 and V3.
Example. Consider I = (xz−y2, x3−yz) ⊂ C[x, y, z] and the corresponding algebraic
subset V = V (I) ⊂ A3C. Suppose that (a, b, c) ∈ V . Multiplying the first equation
by z and the second by y and subtracting gives xz2 = x3y. Hence either a = 0 or
c2 = a2b. In the first case the first equation then implies that b = 0. Conversely, if
a = b = 0 then both defining equations are satisfied. Thus we can write
V = V (x, y) ∪ V (xz − y2, x3 − yz, z2 − x2y).
Let us call the two subsets appearing on the right hand side V1 and V2 respectively.
Note that we don’t have V1 ⊂ V2 or V2 ⊂ V1. Indeed (0, 0, 1) ∈ V1 \ V2 and (1, 1, 1) ∈
V2 \ V1.
The subset V1 is the line x = y = 0. This is easily seen to be irreducible geometri-
cally, because it is a copy of the affine line A1C, so the only algebraic subsets are finite
sets of points. Algebraically we note that C[x, y, z]/(x, y) ∼= C[z]. This shows firstly
that the ideal (x, y) is radical, which ensures that I(V1) = (x, y), and secondly that
(x, y) = I(V1) is prime, and hence that the subset V1 is irreducible.
We will show later that the other subset V2 ⊂ V is also irreducible (see Example
8.5). We will then have proved that the irreducible components of V are V1 and V2.
One last definition
Definition 8.9. An affine variety is an irreducible algebraic subset X ⊂ AnK .
8.4. Coordinate rings. Let X ⊂ AnK be an algebraic set. The coordinate ring of X
is defined to be the quotient ring
K[X] = K[x1, · · · , xn]/I(X).
Note that K[X] is always reduced, since the ideal I(X) is radical.
Remark. An algebraic subset X ⊂ AnK is a variety iff the co-ordinate ring K[X] is an
integral domain.
We can think of the elements of K[X] as functions on the algebraic set X.
Definition 8.10. Let X ⊂ AnK be an algebraic set. A function f : X → K is
called polynomial if there is a polynomial g ∈ K[x1, · · · , xn] such that for all points
p = (a1, · · · , an) ∈ X we have f(p) = g(a1, · · · , an).
MAS439/MAS6320 COMMUTATIVE ALGEBRA AND ALGEBRAIC GEOMETRY 29
Lemma 8.11. The set of polynomial functions on an algebraic subset X ⊂ AnK forms
a K-algebra under pointwise operations. This algebra is isomorphic to the co-ordinate
ring K[X].
Proof. Let R denote the ring of polynomial functions X → K with pointwise oper-
ations and give it the obvious K-algebra structure, where the structure map sends
an element λ ∈ K to the corresponding constant function X → K. There is then
a K-algebra homomorphism K[x1, · · · xn] → R sending a polynomial to the induced
polynomial function on X. This homomorphism is surjective, and by definition its
kernel is I(X). The result therfore follows from the isomorphism theorem.
For any algebraic set X ⊂ AnK we have a correspondence
Ideals I ⊂ K[X]
V
--
Algebraic subsets Y ⊂ X
I
mm
.
Indeed, the two sides can be identified with
Ideals I(X) ⊂ I ⊂ K[x1, · · · , xn]
V
..
Algebraic subsets Y ⊂ X ⊂ AnK
I
nn
.
Applying the Nullstellensatz we get
Theorem 8.12. Let K be an algebraically closed field, and X ⊂ AnK an algebraic
subset. Then the two maps I, V give mutually-inverse order-reversing bijections
Radical ideals I ⊂ K[X]
V
--
Algebraic subsets V ⊂ X
I
nn
In particular, the points of X (which are minimal algebraic subsets) correspond to
maximal ideals of K[X]. Similarly, irreducible algebraic subsets V ⊂ X correspond
to prime ideals in the ring K[X].
Example. Consider the algebra R = C[x, y]/(xy2−x). To understand it geometrically
we would like to view it as a co-ordinate ring of some algebraic subset. Write I =
(xy2 − x) ⊂ C[x, y]. The obvious thing to do is take V = V (I) ⊂ A2C. Then by the
Nullstellensatz I(V ) =
√
I and the co-ordinate ring C[V ] is C[x, y]/
√
I. So the first
thing is to check that the ideal I is radical, so that I =
√
I.
MAS439/MAS6320 COMMUTATIVE ALGEBRA AND ALGEBRAIC GEOMETRY 30
The easiest way to do that is to use the fact that all polynomial rings over fields
are unique factorisation domains. For any f ∈ C[x, y] of positive degree we can write
f in the form
f = g1 · g2 · · · gk,
with the polynomials gi ∈ C[x, y] being irreducible: that is, not the product of lower-
order polynomials. The important fact is that this decomposition of f is basically
unique: the factors gi appearing are unique up to reordering and multiplication by
scalar factors. It is immediate from this that f ∈ I precisely if it is divisible by all
three of the polynomials x, y − 1 and y + 1. It is then clear that fn ∈ I =⇒ f ∈ I
and hence I is radical.
The maximal ideals of R are in bijection with the points of the set V which consists
of the y-axis and the lines y = ±1. They are therefore of the form
(x, y − b), (x− a, y + 1), (x− a, y − 1) with a, b ∈ C.
The prime ideals which are not maximal correspond to irreducible subsets of V which
are not points. There are three such, namely
(x), (y − 1), (y + 1).
These correspond to the y-axis, and the two lines y = ±1.
We equip an algebraic subset X ⊂ AnK with the topology induced from the Zariski
topology on AnK . This means that the closed subsets Y ⊂ X are precisely the algebraic
subsets Y ⊂ AnK which are contained in X.
8.5. Polynomial maps. We now start to consider maps between algebraic subsets.
Definition 8.13. Suppose X ⊂ Amk and Y ⊂ Ank are algebraic subsets. A map
φ : X → Y is called polynomial if there are polynomials
φ1 ∈ K[x1, · · · , xm], φn ∈ K[x1, · · · , xm],
such that
φ(a1, · · · , am) =
(
φ1(a1, · · · , am), · · · , φn(a1, · · · , am)
)
.
Note that we could equivalently view the φi as being polynomial functions on X
or as elements of the co-ordinate ring K[X].
Example.
(a) Cuspidal cubic. Consider the polynomial map φ : A1C → A2C given by t 7→
(t2, t3). Its image is contained in the algebraic subset V = V (y2 − x3) ⊂ A2C.
We thus get an induced map
φ : A1C → V.
MAS439/MAS6320 COMMUTATIVE ALGEBRA AND ALGEBRAIC GEOMETRY 31
This polynomial map is a bijection: indeed, given any point (x, y) ∈ A2C
satisfying y2 = x3 we can take t = ±√x and choose the sign uniquely so that
t3 = y.
(b) Consider the polynomial map φ : A2C → A2C given by (s, t) 7→ (s2, st, t2). The
image is the algebraic subset V = (y2−xz) ⊂ A3C. Indeed, given any (x, y, z) ∈
V we can take s = ±√x and t = ±√z and then, changing one of the signs if
necessary, we have st = y. We thus get an induced polynomial map
φ : A2C → V
in which the inverse image of every point except the origin consists of 2 points
±(s, t). The inverse image of the origin (0, 0) is just (0, 0).
Note that if f : Y → K is a polynomial function on Y then the composition f ◦φ is
a polynomial function on X. This defines a K-algebra homomorphism φ∗ : K[Y ] →
K[X].
Proposition 8.14. Suppose X ⊂ AmK and Y ⊂ AnK are algebraic subsets. Sending a
regular map φ : X → Y to the algebra homomorphism φ∗ : K[Y ] → K[X] defines a
bijection{
Polynomial maps X → Y }→ {K-algebra homomorphisms K[Y ]→ K[X]}.
Proof. First of all, a polynomial map φ : X → Y is determined by the correspond-
ing algebra homomorphism φ∗ : K[Y ] → K[X]. Indeed, if we take the co-ordinate
functions yi ∈ K[Y ] then φ∗(yi) = φi ∈ K[X] is the ith component of the map φ.
Knowing all these clearly determines the map φ. This shows that the map φ 7→ φ∗ is
injective.
To prove that φ 7→ φ∗ is surjective, suppose we have an algebra map h : K[Y ] →
K[X]. Set φi = h(yi) and consider the resulting regular map φ : X → Y given by
φ(a1, · · · , am) =
(
φ1(a1, · · · , am), · · · , φn(a1, · · · , am)
)
.
Then φ∗(yi) = φi = h(yi). But K[Y ] is generated as an K-algebra by the elements
yi. Since both h and φ
∗ are K-algebra maps it follows that they are equal.
Example. Consider again the algebraic subset of Example 8.3
X = V (xz − y2, x3 − yz, z2 − x2y) ⊂ A3K .
MAS439/MAS6320 COMMUTATIVE ALGEBRA AND ALGEBRAIC GEOMETRY 32
We claim that X is the image of the morphism
φ : A1K → A3K , t 7→ (t3, t4, t5).
Indeed, given a point (x, y, z) ∈ X note that x = 0 =⇒ y = z = 0. On the other
hand, if x 6= 0, set t = y/x 6= 0 and u = z/x 6= 0. The defining relations become
u = t2, x = t · u, u2 = xt.
These reduce to u = t2 and x = t3 and hence (x, y, z) = (t3, t4, t5).
We now show that X = V (xz − y2, x3 − yz, z2 − x2y) ⊂ A3K is irreducible.
Since φ in surjective, φ∗ is injective.
Now φ∗ : K[X]→ K[t] is injective, K[t] is a domain, hence K[X] is also a domain.
Definition 8.15. Let X ⊂ AnK and Y ⊂ Amk be algebraic subsets. A polynomial map
X → Y is said to be an isomorphism if there is a polynomial map ψ : Y → X such
that ψ ◦ φ = idX and φ ◦ ψ = idY .
Note that φ : X → Y being an isomorphism implies that φ∗ : K[Y ] → K[X] is
an isomorphism of K-algebras with inverse ψ∗. Thus isomorphic algebraic sets have
isomorphic co-ordinate rings.
Example. The map φ : A1C → V of Example 8.5(a) corresponds to the K-algebra
homomorphism
φ∗ : C[V ] = C[x, y]/(y2 − x3)→ K[A1C] = C[t], (x, y) 7→ (t2, t3).
Note that this is not an isomorphism: its image is C[t2, t3] ⊂ C[t], which does not
contain the element t. Thus φ is also not an isomorphism: there is no inverse polyno-
mial map ψ : V → A1C. Of course there is an inverse to φ, as we discussed in Example
8.5(a), but it is not a polynomial map since it involves taking square-roots.
9. Modules
Definition 9.1. Let R be a commutative ring. An R-module (or a module over R)
is an abelian group (M,+) together with a function R ×M → M whose values for
r ∈ R and m ∈ M are denoted with (r,m) 7→ r ·m (or just (r,m) 7→ rm) which has
the following properties:
(a) for all m ∈M , 1Rm = m,
(b) for all r, s ∈ R and m ∈M , (rs) ·m = r · (s ·m),
(c) for all r, s ∈ R and m ∈M , (r + s) ·m = r ·m+ s ·m, and
(d) for all r ∈ R and m,n ∈M , r · (m+ n) = r ·m+ r · n.
Example. The Abelian group with one element 0 is a module over any commutative
ring; we call this the zero module and denote it with 0.
Any commutative ring R is an R-module.
MAS439/MAS6320 COMMUTATIVE ALGEBRA AND ALGEBRAIC GEOMETRY 33
Example. Let K be a field; K-modules are precisely K-vector spaces.
Example. Let R be a commutative ring. Any ideal I ⊆ R is an R-module if for all
r ∈ R and m ∈ I we define (r,m) 7→ r ·m where “·” is the multiplication in R.
Example. Let R be a commutative ring and let I ⊆ R be an ideal. We can endow
R/I a natural structure of an R-module by defining r · (s+ I) to be rs+ I. We can
also endow R/I the structure on an R/I-module as an ideal of R/I.
Example. Let R be a commutative ring, let Λ be a set and let Mλ be an R-module for
every λ ∈ Λ. The direct sum of {Mλ}λ∈Λ, denoted ⊕λ∈ΛMλ consists of all sequences
(mλ |λ ∈ Λ,mλ ∈Mλ and mλ = 0 for all but finitely many λ ∈ Λ)
with addition
(mλ)λ∈Λ + (nλ)λ∈Λ = (mλ + nλ)λ∈Λ
and for any r ∈ R,
r(mλ)λ∈Λ = (rmλ)λ∈Λ.
In the special case when Mλ ∼= R for all λ ∈ Λ, and for all λ ∈ Λ, Mλ = Rmλ, we
obtain the free module with basis {mλ}λ∈Λ. If Λ is a finite set, we call the free module
above a free module of finite rank.
Proposition 9.2. Let R be a commutative ring, and let F be a free R-module with
free basis e1, . . . , eα. Let M be any R-module, and let m1, . . . ,mα ∈M . There exists
a unique homomorphism of R-modules φ : F → M such that φ(ei) = mi for all
1 ≤ i ≤ α.
10. Homomorphisms of R-modules
Definition 10.1. Let R be a commutative ring and let M and N be R-modules. A
function φ : M → N is a homomorphism of R-modules if
(a) for all m1,m2 ∈M , φ(m1 +m2) = φ(m1) + φ(m2),
(b) for all r ∈ R and m ∈M , φ(rm) = rφ(m).
An homomorphism of R-modules φ : M → N is a isomorphism of R-modules if it
is injective and surjective.
Example. Let R be a commutative ring and let I ⊆ R be an ideal. The quotient
map R→ R/I is an R-module homomorphism.
MAS439/MAS6320 COMMUTATIVE ALGEBRA AND ALGEBRAIC GEOMETRY 34
Example. Let R be a commutative ring and let M be a R-module. For any r ∈
R the function µr : M → M defined by µr(m) = rm for all m ∈ M is an R-
module homomorphism. In particular, the identity function µ1R is an R-module
homomorphism.
Example. Let K be a field; K-module homomorphisms are nothing but linear maps
between vector spaces.
Definition 10.2. Let R be a commutative ring and let M be an R-module. A
subset N ⊆ M is an R-submodule of M if N is is an additive subgroup and closed
under the action of R. If N ⊆ M is a R-submodule of M , the inclusion map is an
homomorphism of R-modules.
Example. Let R be a commutative ring, let I ⊆ R be an ideal and let M be an
R-module.
IM = {a1m1 + · · ·+ asms|s ≥ 0, a1, . . . , as ∈ I, m1, . . . ,ms ∈M}
is a submodule of M .
Example. Let R be a commutative ring, let M be an R-module, let Λ be a non-
empty set and let Mλ ⊆M be an R-submodule for every λ ∈ Λ. Then ∩λ∈ΛMλ is an
R-submodule of M .
Proposition 10.3. Let R be a commutative ring, let M and N be R-modules and let
φ : M → N be an homomorphism of R-modules. The kernel of φ, denoted kerφ and
defined to be {m ∈M |φ(m) = 0} is an R-submodule of M . The image of φ, denoted
Imφ and defined to be {n ∈ N |n = φ(m) for some m ∈ M} is an R-submodule of
N .
11. Quotients of modules
Definition 11.1. Let R be a commutative ring, let M be an R-module and let
N ⊆M be a submodule. We define a new R module, the the quotient of M by N as
follows: as an Abelian group it is the quotient group of the Abelian group M by the
(normal) subgroup N and for each r ∈ R and each coset m+N we define r · (m+N)
to be the coset rm+N .
The function φ : M → M/N given by φ(m) = m + N is an homomorphism of
R-modules and we refer to it as the quotient homomorphism.
MAS439/MAS6320 COMMUTATIVE ALGEBRA AND ALGEBRAIC GEOMETRY 35
Proposition 11.2. Let R be a commutative ring, let M be an R-module and let
N ⊆ M be a submodule. There is a bijection between the submodules of M/N and
the submodules of M which contain N given by K → φ−1(K) and K 7→ φ(K) where
φ : M →M/N is the quotient homomorphism.
Proof. A straightforward modification of the proof of Proposition 5.13.
Example. Let M be an R-module and define AnnRM = {r ∈ R | rM = 0}. Then
M is an R/AnnRM -module where (r + AnnRM)m is defined to be rm for all (r +
AnnRM) ∈ R/AnnRM and all m ∈M .
So for any R-module M and ideal I ⊆ R, M/IM is an R/I-module.
12. The first Isomorphism Theorem
Theorem 12.1. Let φ : M → N be an homomorphism of R-modules. We have an
isomorphism of R-modules φ : M/ kerφ→ Imφ given by φ(m+ kerφ) = φ(m).
Proof. First we check that φ is well defined: if m1 + kerφ = m2 + kerφ then φ(m1) =
φ(m2). It easy to check that φ is a surjective homomorphism; If φ(m + kerφ) = 0,
then φ(m) = 0 so m+ kerφ = 0 and we deduce that φ is also injective.
13. Generators of modules
Definition 13.1. Let R be a commutative ring, let M be an R-module and let
G ⊆ M be any subset. The submodule of M generated by G, denoted 〈G〉, consists
of all
{r1g1 + . . . rsgs | s ≥ 1, g1, . . . , gs ∈ G, r1, . . . , rs ∈ R}.
The R-module M is finitely generated if there exists a finite G ⊆ M such that
M = 〈G〉.
Proposition 13.2. Let R be a commutative ring. For any R-module M there ex-
ists a free module F and a surjective homomorphism of R-modules φ : F → M .
Furthermore, if M is finitely generated, F can be chosen to have finite rank.
MAS439/MAS6320 COMMUTATIVE ALGEBRA AND ALGEBRAIC GEOMETRY 36
Proof. Any R-module M can be written as 〈G〉 for some G ⊆ M , e.g., take G = M .
Now let F be the free module with basis G define φ : F → M with φ ((rg)g∈G) =∑
g∈G rgg (it is crucial here to realize that the sum is finite.) It is easy to verify that
this is an homomorphism of R-modules and the fact that φ is surjective follows from
the fact that G generates M .
The second statement is now easy.
Example. Let R be a commutative ring and let I ⊆ R be an ideal. The R-module
R/I is finitely generated, in fact it is cyclic, e.g., generated by one element, namely
by 1 + I.
Definition 13.3. Let R be a commutative ring. The Jacobson radical of R is the
intersection of all its maximal ideals.
Proposition 13.4. Let R be a commutative ring and let J be its Jacobson radical.
For any element a ∈ J , 1− a is a unit.
Proof. Let a ∈ J . If 1 − a is not a unit, 1 /∈ (1 − a)R and so (1 − a)R is a proper
ideal, which must then be contained in a maximal ideal m ⊂ R (cf. Theorem 6.11.)
But a is also in m, hence so is (1− a) + a = 1, which is impossible.
Theorem 13.5 (Nakayama’s Lemma). Let R be a commutative ring, let M be a
finitely generated R-module and let I ⊂ R be an ideal contained in the Jacobson
radical of R. If IM = M then M = 0.
Proof. Assume that M 6= 0 and that the set {g1, . . . , gn} (n ≥ 1) is a minimal set
of generators of M . Since M = IM , we can write gn = a1g1 + · · · + angn where
a1, . . . , an ∈ I. We rearrange this to obtain (1 − a1)gn = a1g1 + · · · + an−1gn−1 and
we notice that 1 − a1 is a unit so gn ∈ 〈g1, . . . , gn−1〉 and so {g1, . . . , gn−1} is a set
of generators for M , contradicting the minimality of {g1, . . . , gn}. So our assumption
that n ≥ 1 is false, and so n = 0 hence M = 0.
Corollary 13.6. Let R be a commutative ring, let I ⊂ R be an ideal contained in
the Jacobson radical of R, let M be a finitely generated R-module and let N ⊆M be
a submodule. If M = N + IM then M = N .
Corollary 13.7. Let (R,m) be a local ring and let M be a finitely generated R-module.
MAS439/MAS6320 COMMUTATIVE ALGEBRA AND ALGEBRAIC GEOMETRY 37
(1) The R-module M/mM is a finitely dimensional vector space over the field
R/m.
(2) Let B ⊂ M be a set of elements whose images in M/mM form a basis for
M/mM as a R/m-vector-space. Then B is a minimal set of generators of M
as an R-module.
Proof. LetN ⊂M be theR-submodule ofM generated byB. We haveM = N+mM ,
hence N = M . If B′ ⊂ B and N ′ is the R-submodule of M generated by B′, then
N +mM 6= N ′ +mM .
14. Direct sums and products
Let R be a commutative ring, let Λ be a set and let Mλ be an R-module for every
λ ∈ Λ.
Definition 14.1. The direct product of {Mλ}λ∈Λ, denoted
∏
λ∈ΛMλ consists of all
sequences
(mλ |λ ∈ Λ,mλ ∈Mλ)
with addition
(mλ)λ∈Λ + (nλ)λ∈Λ = (mλ + nλ)λ∈Λ
and for any r ∈ R,
r(mλ)λ∈Λ = (rmλ)λ∈Λ.
Definition 14.2. The direct sum of {Mλ}λ∈Λ, denoted ⊕λ∈ΛMλ consists of all se-
quences
(mλ |λ ∈ Λ,mλ ∈Mλ and mλ = 0 for all but finitely many λ ∈ Λ)
with addition
(mλ)λ∈Λ + (nλ)λ∈Λ = (mλ + nλ)λ∈Λ
and for any r ∈ R,
r(mλ)λ∈Λ = (rmλ)λ∈Λ.
Note that, if Λ is finite,
∏
λ∈ΛMλ = ⊕λ∈ΛMλ.
15. Free modules
Definition 15.1. A free R-module is a module isomorphic to ⊕λ∈ΛR.
Proposition 15.2. An R-module M is free if and only if there exists a set of gener-
ators {mλ}λ∈Λ of M such that whenever r1mλ1 + · · ·+ rnmλn = 0 with λ1, . . . , λn ∈ Λ
and r1, . . . , rn ∈ R, we must have r1 = · · · = rn = 0.
MAS439/MAS6320 COMMUTATIVE ALGEBRA AND ALGEBRAIC GEOMETRY 38
Proof. The only if part is easy.
Assume that we have a set of generators as above and define a function φ : ⊕λ∈ΛR→
M by setting φ (rλ) =
∑
λ∈Λ rλmλ (note that the sum is finite.) It is not hard to check
that φ is an isomorphism of R-modules.

We call a set of generators as in the proposition a free basis.
The rank of a free module is the cardinality of a free basis.
Proposition 15.3. Let R be a commutative ring. For any R-module M there ex-
ists a free module F and a surjective homomorphism of R-modules φ : F → M .
Furthermore, if M is finitely generated, F can be chosen to have finite rank.
Proof. Any R-module M can be written as 〈G〉 for some G ⊆ M , e.g., take G = M .
Now let F be the free module with basis G define φ : F → M with φ ((rg)g∈G) =∑
g∈G rgg (it is crucial here to realize that the sum is finite.) It is easy to verify that
this is an homomorphism of R-modules and the fact that φ is surjective follows from
the fact that G generates M .
The second statement is now easy.
16. Localization of rings
Definition 16.1. Let R be a commutative ring. A subset U ⊂ R is called a multi-
plicative set if (a) 1R ∈ U and (b) for any a, b ∈ U , ab ∈ U .
Example. Let R be a commutative ring and a ∈ R \ {0}; {an}∞n=0 is a multiplicative
set.
Example. Let R be a commutative ring and P ⊂ R a prime ideal; R \ P is a
multiplicative set (because P is a proper ideal so 1 /∈ P and whenever a, b /∈ P ,
ab /∈ P .)
Example. Let R be a commutative ring; R is an integral domain if and only if R\{0}
is a multiplicative set.
Example. Let R be a commutative ring; the set of non zero-divisors of R is a mul-
tiplicative set.
Given a commutative ring R and a multiplicative set U ⊆ R construct a new
commutative ring as follows. We start by letting S be the set of all symbols r/u with
MAS439/MAS6320 COMMUTATIVE ALGEBRA AND ALGEBRAIC GEOMETRY 39
r ∈ R and u ∈ U . We define a relation “∼” on S by declaring a/u ∼ b/v if and only
if there exists a t ∈ U such that t(av − bu) = 0R. This relation is obviously reflexive
(take t = 1R ∈ U) and symmetric. To see that it is also transitive, assume that for
a, b, c ∈ R and u, v, w ∈ U we have t1(av − bu) = 0R and t2(bw − cv) = 0R for some
t1, t2 ∈ U ; multiply the first equality by t2w and the second by t1u and add them to
obtain t1t2v(aw − cu) = 0R and since t1t2v ∈ U , we see that a/u ∼ c/w.
Now “∼” is an equivalence relation, and we denote the equivalence class of a/u ∈ S
with [a/u]. Notice that for any a/u ∈ S and any t ∈ U , [a/u] = [(at)/(ut)].
We construct a new commutative ring U−1R, which we call the localization of R at
U as the set of equivalence classes [a/u] of elements a/u ∈ S with the addition[a
u
]
+
[
b
v
]
=
[
av + bu
uv
]
and multiplication [a
u
]
·
[
b
v
]
=
[
ab
uv
]
.
We need to show that the operations we have introduced are well defined, i.e., the
result does not depend on the choice of representative in the equivalence class. For
example, we need to show that if [a′/u′] = [a/u] and [b′/v′] = [b/v] then[
av + bu
uv
]
=
[
a′v′ + b′u′
u′v′
]
.
To see that this is indeed true, pick t1, t2 ∈ U such that t1(au′ − a′u) = 0 and
t2(bv
′ − b′v) = 0. Now[
a′v′ + b′u′
u′v′
]
=
[
t1t2uva
′v′ + t1t2uvb′u′
t1t2uvu′v′
]
=
[
t1t2u
′vav′ + t1t2uv′bu′
t1t2uvu′v′
]
=
[
av + bu
uv
]
and a similar calculation shows that multiplication in U−1R is also well defined. We
can now check that the additive identity of U−1R is [0/1], that the multiplicative
identity is [1/1] and that U−1R is indeed a commutative ring.
Theorem 16.2. Let R be a commutative ring and let U ⊆ R be a multiplicative set.
The function φ : R → U−1R defined by φ(r) = [r/1] is a ring homomorphism and it
has the following properties:
(a) for every u ∈ U , φ(u) is a unit, and
(b) if some other commutative ring S and ring homomorphism ψ : R → S have
the property that for every u ∈ U , ψ(u) is a unit then there exists a unique
MAS439/MAS6320 COMMUTATIVE ALGEBRA AND ALGEBRAIC GEOMETRY 40
ring homomorphism f : U−1R→ S such that ψ = f ◦ φ
(5) R
ψ
//
φ

S
U−1R
f
<<
(thus, in a sense, U−1R is the “smallest” way to extend R so that all elements
of U are units.)
Proof. To verify the first statement we check that φ(1) = [1/1] is the multiplicative
identity of U−1R, that φ(r+s) = [(r+s)/1] = [r/1]+[s/1] and that φ(rs) = [(rs)/1] =
[r/1][s/1].
We next notice that for all u ∈ U , φ(u)[1/u] = [u/1][1/u] = [u/u] = [1/1].
Now let S and ψ : R → S be as in (b) above. Define a function f : U−1R → S
with f([r/u]) = ψ(r)ψ(u)−1: this is well defined because if [r/u] = [r′/u′] we can find
a t ∈ U such that t(ru′ − r′u) = 0 and
f([r/u]) = ψ(r)ψ(u)−1
= ψ(t)−1ψ(t)ψ(r)ψ(u′)ψ(u′)−1ψ(u)−1
= ψ(t)−1ψ(t)ψ(r′)ψ(u)ψ(u′)−1ψ(u)−1
= ψ(r′)ψ(u′)−1 = f([r′/u′])
It is straightforward to verify that ψ is a ring homomorphism, e.g., for all [r/u], [s/v] ∈
U−1R,
f([r/u]) + f([s/v]) = ψ(r)ψ(u)−1 + ψ(s)ψ(v)−1
= ψ(r)ψ(v)ψ(v)−1ψ(u)−1 + ψ(s)ψ(u)ψ(u)−1ψ(v)−1
= ψ(rv + su)ψ(uv)−1
= f([(rv + su)/uv])
The uniqueness of f follows from the fact that the condition ψ = f ◦ φ leaves us no
choice in its definition: for all r ∈ R we must have f([r/1]) = ψ(r) and for all u ∈ U
we have 1S = f([1/1]) = f([u/u]) = f([u/1])f([1/u]) = ψ(u)f([1/u]) so f([1/u]) =
ψ(u)−1 and so f [r/u] = f([r/1][1/u]) = f([r/1])f([1/u]) = ψ(r)ψ(u)−1.
Example. The localization of Z at the multiplicative set Z \ {0} is Q. We generalize
this construction as follows: let R be a integral domain, the field of fractions of R is
the localization of R at the multiplicative set R \ {0}. Notice that this is indeed a
field since every non-zero element is a unit.
MAS439/MAS6320 COMMUTATIVE ALGEBRA AND ALGEBRAIC GEOMETRY 41
Example. Let K be a field. The ring of rational functions K(x) is the localization
of R = K[x] at R \ {0}.
Definition 16.3. Let R be a commutative ring and let P ⊂ R be a prime ideal. The
localization of R at P , denoted RP , is the localization of R at the multiplicative set
R \ P .
Definition 16.4. A commutative ring R is local if it has a unique maximal ideal.
Proposition 16.5. Let R be a commutative ring and let P ⊂ R be a prime ideal.
The ring RP is local with maximal ideal PRP = {r/u | r ∈ P, u /∈ P}, the expansion
of P to RP .
Proof. For any element s/v ∈ RP \ PRP we have s ∈ R \ P so v/s ∈ RP and so s/v
is a unit. So every ideal of PRP which is not contained in PRP is the whole of RP ,
and hence PRP is maximal.
17. Localization of modules
Suppose that we start with a commutative ring R, a multiplicative set U ⊂ R and
an R-module M . We may construct U−1R and we may now want to turn M into
a U−1R-module in a “natural” way, and we do this as follows: let M ′ be the set of
all symbols m/u with m ∈ M and u ∈ U and define a relation “∼” on M ′ with
m/u = n/v if and only if there exists a t ∈ U such that tvm − tun = 0. Just as
in the previous section this is an equivalence relation and we denote the equivalence
class of m/u with [m/u]. We now define the U−1R-module U−1M to be the U−1R-
module whose elements are the equivalence classes of elements in M ′ with addition
[m/u] + [n/v] = [(vm+un)/uv] and U−1R action [r/u][m/v] = [(rm)/(uv)]. One can
verify that these operations are well defined and that they give U−1M the structure
of an U−1R-module.
Proposition 17.1. Let R be a commutative ring, and let U be a multiplicative set.
Let M and N be R-modules and let φ : M → N be an homomorphism. The function
U−1(φ) : U−1M → U−1N defined as U−1(φ)([m/u]) = [φ(m)/u] is an homomorphism
of U−1R-modules.
MAS439/MAS6320 COMMUTATIVE ALGEBRA AND ALGEBRAIC GEOMETRY 42
Proof. We should first verify that U−1(φ) is well defined. Let m,m′ ∈ M , u, u′ ∈ U
be such that [m/u] = [m′/u′]. Pick a t ∈ U such that tu′m = tum′ and write[
φ(m)
u
]
=
[
tu′φ(m)
tu′u
]
=
[
φ(tu′m)
tu′u
]
=
[
φ(tum′)
tu′u
]
=
[
tuφ(m′)
tu′u
]
=
[
φ(m′)
u′
]
The verification of the U−1R-linearity of U−1(φ) is equally simple.
18. Exact sequences
Definition 18.1. Let R be a commutative ring, let L,M,N be R-modules and let
f : L → M and g : M → N be homomorphisms of R-modules. We say that the
sequence L
f−→M g−→ N is exact if ker g = Im f .
A sequence of R-modules and homomorphisms
. . .Mi−1
fi−1−−→Mi fi−→Mi+1 fi−→ . . .
is an exact sequence if, for all i for which fi and fi−1 are in the sequence, ker fi =
Im fi−1.
A short exact sequence is an exact sequence of the form 0 → L f−→ M g−→ N → 0.
Notice that this implies that f is injective and that g is surjective.
Example. Let R be a commutative ring, let M be an R-module and let N be an
R-submodule of M . We have an exact sequennce
0→ N i−→M g−→M/N → 0
where i is the inclusion map and g is the quotient map.
Example. The sequence 0 → L f−→ M is exact if and only if f is injective. The
sequence 0→M g−→ N is exact if and only if f is surjective.
Example. Assume that R is a field. If 0 → L f−→ M g−→ N → 0 is a short exact
sequence of finite dimensional R-vector spaces, dimM = dimL+ dimN .
Let 0→Mn →Mn−1 → · · · →M1 → 0 be an exact sequence of finite dimensional
R-vector spaces. Then dimM1 − dimM2 + dimM3 + · · ·+ (−1)n+1 dimMn = 0.
MAS439/MAS6320 COMMUTATIVE ALGEBRA AND ALGEBRAIC GEOMETRY 43
19. The exactness of localization
Theorem 19.1. Let R be a commutative ring, U a multiplicative set and let L
f−→
M
g−→ N be an exact sequence of R-modules. Then U−1L U−1f−−−→ U−1M U−1g−−−→ U−1N
is an exact sequence of U−1R-modules
Proof. Let m ∈M and u ∈ U .
[m/u] ∈ kerU−1g ⇔ [g(m)/u] = 0
⇔ tg(m) = 0 for some t ∈ U
⇔ g(tm) = 0 for some t ∈ U
⇔ tm = f(`) for some t ∈ U and some ` ∈ L
⇔ [m/u] = U−1f [`/ut] for some ` ∈ L
⇔ [m/u] ∈ ImU−1f

Theorem 19.2. Let R be a commutative ring, U a multiplicative set.
(a) Let M be an R-module and let N ⊆ M be a submodule. Then U−1N is
an U−1R-submodule of U−1M and the natural map U−1 (M) /U−1 (N) →
U−1 (M/N) sending [m/u] + U−1N to [m+N/u] is an isomorphism.
(b) Let M be an R-module and let M1, . . . ,Mn ⊆ M be submodules. We have
U−1 (M1 ∩ · · · ∩Mn) = U−1M1 ∩ · · · ∩ U−1Mn.
(c) Let {Mλ}λ∈Λ be a set of R-submodules of M . The U−1R-modules U−1
(∑
λ∈Λ
Mλ
)
and
∑
λ∈Λ
U−1Mλ are (naturally) isomorphic.
Proof. (a) The short exact sequence of R-modules 0→ N ↪→M q−→M/N → 0 where
q is the quotient homomorphism gives a short exact sequence of U−1-modules
0→ U−1N ↪→ U−1M U−1q−−−→ U−1(M/N)→ 0
so the first isomorphism theorem implies
U−1(M/N) = ImU−1q ∼= U−1M/U−1N.
MAS439/MAS6320 COMMUTATIVE ALGEBRA AND ALGEBRAIC GEOMETRY 44
(b) Using an easy induction we can reduce to the case n = 2. It is easy to see that
U−1 (M1 ∩M2) ⊆ U−1 (M1) ∩ U−1 (M2). Now any [m/u] ∈ U−1 (M1) ∩ U−1 (M2) can
be written as [m/u] = [m1/u1] = [m2/u2] with m1 ∈ M1,m2 ∈ M2 and u1, u2 ∈ U ,
and so there exist a t ∈ U such that tu2m1 = tu1m2 ∈M1 ∩M2; we deduce that
m1
u1
=
tu2m1
tu2u1
∈ U−1 (M1 ∩M2) .
(c) is an easy exercise.

Definition 19.3. Let R be a commutative ring, let U be a multiplicative set and let
φ : R→ U−1R be the localization map. For any ideal I ⊆ R we shall write U−1I for
the ideal of R generated by φ(I).
20. Local properties
Definition 20.1. Let R be a commutative ring and let M be an R-module. The
annihilator of M in R is the ideal
AnnRM := {r ∈ R | rm = 0 for all m ∈M}.
Proposition 20.2. Let R be a commutative ring, let U ⊆ R be a multiplicative
set and let M be a finitely generated R-module. Then U−1M = 0 if and only if
U ∩ AnnRM 6= ∅.
Proof. If we can find a u ∈ U ∩ AnnRM then for all m ∈M and all t ∈ U we have[m
t
]
=
[um
ut
]
=
[
0
ut
]
=
[
0
1
]
.
Assume that U−1M = 0 and pick generators m1, . . . ,ms for M . Since [mi/1] = 0
for all 1 ≤ i ≤ s, there exist u1, . . . , us ∈ U such that uimi = 0 for all 1 ≤ i ≤ s. Now
u = u1 · · · · · us ∈ U kills M .
Theorem 20.3. Let R be a commutative ring and let M be an R-module. The
following statements are equivalent:
(a) M = 0;
(b) MP = 0 for all prime ideals P ⊂ R;
(c) Mm = 0 for all maximal ideals m ⊂ R.
MAS439/MAS6320 COMMUTATIVE ALGEBRA AND ALGEBRAIC GEOMETRY 45
Proof. Clearly (a)⇒(b)⇒(c) so we only show (c)⇒(a).
Assume that M 6= 0 but Mm = 0 for all maximal ideals m ⊂ R. We can pick an
m ∈ M \ {0}, and we let I = AnnRRm = {r ∈ R | rm = 0}. Notice that 1 /∈ I, so
I 6= R and we may pick a maximal ideal m ⊇ I.
Proposition 20.2 now implies that Mm 6= 0, a contradiction.
Theorem 20.4. Let R be a commutative ring, let M and N be R-modules and let
φ : M → N be an homomorphism of R-modules.
The following statements are equivalent:
(a) φ is injective;
(b) φP : MP → NP is injective for all prime ideals P ⊂ R;
(c) φm : Mm → Nm is injective for all maximal ideals m ⊂ R.
The following statements are equivalent:
(a) φ is surjective;
(b) φP : MP → NP is surjective for all prime ideals P ⊂ R;
(c) φm : Mm → Nm is surjective for all maximal ideals m ⊂ R.
Proof. We only prove the first set of equivalences, the proof of the second is very
similar.
(a)⇔ the sequence 0→M φ−→ N is exact⇒ the sequence 0→MP φP−→ NP is exact
for all prime ideals P ⊂ R ⇔ (b).
Clearly (b) ⇒ (c), and we conclude the proof by showing (c) ⇒ (a).
Assume (c) and write K = kerφ. We have a exact sequence 0 → K ↪→ M φ−→ N
which for all maximal ideals m ⊂ R gives an exact sequence 0 → Km → Mm φm−→ Nm
from which we deduce that Km = kerφm = 0. Since this is true for all maximal ideals
m ⊂ R we deduce that K = 0, i.e., that (a) holds.
Roughly speaking, being zero, injective and surjective are local properties!
21. Further properties of localization
Theorem 21.1. Let R be a commutative ring, let U be a multiplicative set and let
φ : R→ U−1R be the localization map.
(a) Let I ⊆ R be an ideal; U−1I = U−1R if and only if U ∩ I 6= ∅.
(b) For every ideal J ⊆ U−1R there exists an ideal I ⊆ R such that J = U−1I.
(c) If Q ⊂ R is a prime ideal disjoint from U , U−1Q is a prime ideal of U−1R.
MAS439/MAS6320 COMMUTATIVE ALGEBRA AND ALGEBRAIC GEOMETRY 46
(d) There exists a bijection between the prime ideals of U−1R and the prime ideals
of R disjoint from U given by P 7→ φ−1(P ) and Q 7→ U−1Q.
Proof. (a) If a ∈ I ∩U then [1/1] = [a/a] = [1/a][a/1] ∈ U−1I. On the other hand, if
[1/1] = [a/u] for some a ∈ I and u ∈ U , t(a−u) = 0 for some t ∈ U and so ta ∈ I∩U .
(b) Let I = {r ∈ R | [r/1] ∈ J}; I is an ideal. Clearly, φ(I) ⊆ J and since for any
[r/u] ∈ J , [r/u] = [1/u][r/1], J ⊆ 〈φ(I)〉.
(c) Let Q ⊆ R be a prime ideal, i.e., R/Q is a domain. Consider the U−1R-module
S = U−1 (R/Q) ≡ U−1R/U−1Q. This is a ring in its own right, and is a localization of
a domain. It is straightforward to verify that localizations of domains, either vanish
or are again domains. Since U ∩Q = ∅, S 6= 0 so S is a domain and U−1Q ⊆ U−1R
is a prime ideal.
(d) Let Q ⊂ R be a prime ideal; then φ−1(U−1Q) is prime and if a ∈ φ−1(U−1Q),
i.e., if [a/1] ∈ U−1Q then [a/1] = [q/u] for some q ∈ Q and u ∈ U , and so tua = tq ∈ Q
for some t ∈ U . Since tu /∈ Q, we deduce that a ∈ Q so φ−1(U−1Q) = Q.
Let P ⊂ U−1R be a prime ideal; then U−1φ−1(P ) is a prime ideal and clearly all
elements in P of the form [b/1] are in U−1φ−1(P ). These elements generate P so
U−1φ−1(P ) = P .
22. Noetherian rings
Definition 22.1. Let R be a commutative ring. An R-module M is called Noetherian
if every submodule of M is finitely generated.
Theorem 22.2. Let R be a commutative ring and let M be an R-module. The
following are equivalent.
(a) M is Noetherian.
(b) If M1 ⊆ M2 ⊆ M3 ⊆ . . . is an infinite ascending chain of submodules of M ,
there exists a N ≥ 1 such that Mn = MN for all n ≥ N .
(c) Any non-empty set of submodules of M contains a submodule maximal with
respect to inclusion.
Proof. (a) ⇒ (b) Let M1 ⊆ M2 ⊆ M3 ⊆ . . . be an infinite ascending chain of
submodules of M , and let M ′ be its union; M ′ is an R-submodule of M and hence
generated by a finite set a1, . . . , an ∈M ′. Now find N ≥ 1 such that a1, . . . , an ∈MN
and deduce that Mn = MN for all n ≥ N .
MAS439/MAS6320 COMMUTATIVE ALGEBRA AND ALGEBRAIC GEOMETRY 47
(b) ⇒ (c) Let S be a non-empty set and assume it has no maximal element with
respect to inclusion. Define a sequence (Mn)n≥1 of submodules of M in S inductively
as follows: pick M1 ∈ S arbitrarily; if Mn ∈ S was chosen for n ≥ 1, use the fact
that Mn is not maximal in S with respect to inclusion to find an N ∈ S such that
N ) Mn and set Mn+1 = N . This defines a ascending chain of ideals which is not
eventually constant, contradicting (b).
(c)⇒ (a) Let N ⊆M be a submodule. Let S be the set of submodules of N which
are finitely generated R-modules. Notice that each element in S is also a submodule
of M so we may pick an element K ∈ S maximal with respect to inclusion. We must
have K = N , otherwise there exists a a ∈ N \K and K + aR is a submodule of N
which is finitely generated and is strictly larger than K, contradicting the maximality
of K.
Definition 22.3. We call a commutative ring R is Noetherian if it is a Noetherian
module over itself.
Notice that Theorem 22.2 applied to a commutative ring R implies that the fol-
lowing are equivalent.
(a) R is Noetherian.
(b) If I1 ⊆ I2 ⊆ I3 ⊆ . . . is an infinite ascending chain of ideals of R, there exists
a N ≥ 1 such that In = IN for all n ≥ N .
(c) Any non-empty set of ideals contains a ideal maximal with respect to inclusion.
Proposition 22.4. Let R and S be commutative and rings and let φ : R → S is a
surjective homomorphism of rings. If R is Noetherian, so is S. In particular, R/I is
Noetherian for all ideals I ⊂ R.
Proof. Let J ⊆ S be any ideal. Let G be the finite set of generators of φ−1(J). Now
J is generated by φ(G).
Recall the following (§18)
Definition 22.5. Let R be a commutative ring, let L,M,N be R-modules and let
f : L → M and g : M → N be homomorphisms of R-modules. We say that the
sequence L
f−→M g−→ N is exact if ker g = Im f .
A sequence of R-modules and homomorphisms
. . .Mi−1
fi−1−−→Mi fi−→Mi+1 fi−→ . . .
is an exact sequence if, for all i for which fi and fi−1 are in the sequence, ker fi =
Im fi−1.
MAS439/MAS6320 COMMUTATIVE ALGEBRA AND ALGEBRAIC GEOMETRY 48
A short exact sequence is an exact sequence of the form 0 → L f−→ M g−→ N → 0.
Notice that this implies that f is injective and that g is surjective.
Proposition 22.6. Let R be a commutative ring and let 0→ K ↪→ L φ−→ M → 0 be
a short exact sequence of R-modules. Then L is Noetherian if and only if K and L
are Noetherian.
Proof. Assume first that L is Noetherian. Every submodule of K is a submodule of
L, hence K is Noetherian. If M ′ is a submodule of M , φ−1(M ′) is a submodule of L
and hence generated by some m1, . . . ,mα. Now φ(m1), . . . , φ(mα) generate M
′.
Assume now that K and M are Noetherian and let L1 ⊆ L2 ⊆ . . . be an ascending
chain of submodules of L. We can find N 1 such that Ln ∩ K = LN ∩ K and
φ(Ln) = φ(LN) for all n ≥ N .
Pick any n ≥ N ; we will show that Ln = LN . Let ` ∈ Ln. Since φ(`) ∈ φ(Ln) =
φ(LN) there exists `
′ ∈ LN such that φ(`) = φ(`′). Now φ(` − `′) = 0, i.e., ` − `′ ∈
K ∩ Ln = K ∩ LN so ` ∈ LN .
Proposition 22.7. Let R be a Noetherian ring.
(a) For all n ≥ 1, Rn is a Noetherian R-module.
(b) Any finitely generated R-module is Noetherian.
Proof. (a) Proceed by induction on n. The case n = 1 is obviously true and, if n > 1,
apply Proposition 22.6 to the short exact sequence 0→ R→ Rn → Rn−1 → 0.
(b) Let N be a finitely generated R-module. Let g1, . . . , gn be generators for a
finitely generated R-module N . Let φ : Rn → N be the map which sends the ith free
generator ei of R
n to gi and extend it to a surjective homomorphism. Now the result
follows from (a).
23. Hilbert’s Basis Theorem
Theorem 23.1 (Hilbert’s Basis Theorem). If R is a Noetherian ring, so is R[x].
MAS439/MAS6320 COMMUTATIVE ALGEBRA AND ALGEBRAIC GEOMETRY 49
Proof. We will show that every ideal I ⊆ R[x] is finitely generated; fix such an ideal.
Every element f ∈ R[x] has the form f = anxn + an−1xn−1 + · · ·+ a0 with an 6= 0;
we define the leading coefficient of f to be an and denote it with lc(f).
Consider J = {lc(f) | f ∈ I} ⊆ R; it is not hard to see that J is an ideal:
(a) For a ∈ J and r ∈ R, pick f ∈ I such that a = lc(f). Now ra = lc(rf) and
rf ∈ I.
(b) For a, b ∈ J pick f, g ∈ I such that a = lc(f) and b = lc(g). Assume, with no
loss of generality, that deg(f) ≥ deg(g). Now a+ b = lc(f + xdeg f−deg gg) and
f + xdeg f−deg gg ∈ I.
Since R is Noetherian, J is finitely generated, say J = 〈a1, . . . , an〉. For each
generator ai we may pick a fi ∈ I such that ai = lc(fi); write D = max{deg(fi) | 1 ≤
i ≤ n} and let K = 〈f1, . . . , fn〉. Now view I as an R-module and let M be the
R-submodule of R[x] generated by 1, x, . . . , xD−1.
I claim that every f ∈ I can be written as f = g + h where g ∈ I ∩M and h ∈ K.
Suppose that this is not true and pick a counter example f ∈ I with minimal degree.
If deg f < D then f ∈ I∩M , so assume that deg f ≥ D. Write lc(f) = r1a1+· · ·+rnan
with r1, . . . , rn ∈ R and consider
f ′ = f −
n∑
i=1
rix
deg f−deg fifi.
Notice that f ′ ∈ I and that deg f ′ < deg f , so the minimality of deg f implies that
f ′ = g′ + h′ with g′ ∈ I ∩M and h′ ∈ K and so
f = g′ +
(
h′ +
n∑
i=1
rix
deg f−deg fifi
)
contradicting the choice of f .
Now Proposition 22.7 implies that M is a Noetherian R-module, and so I∩M ⊆M
is a finitely generated R-module, say I ∩M = 〈s1, . . . , s`〉. We showed that every
element in I is a sum of an R-linear combination of s1, . . . , s` and an R[x]-linear
combination of f1, . . . , fn, so every element in I is an R[x]-linear combination of
s1, . . . , s`, f1, . . . , fn, i.e., the ideal I ⊆ R[x] is generated by s1, . . . , s`, f1, . . . , fn.
Corollary 23.2. If R is Noetherian, so is R[x1, . . . , xn].
24. Primary decomposition
Throughout this section R will denote a Noetherian ring.
Definition 24.1. We call an ideal I ⊂ R irreducible if it cannot be written as I1∩ I2
where I1 and I2 are proper ideals of R which strictly contain I.
MAS439/MAS6320 COMMUTATIVE ALGEBRA AND ALGEBRAIC GEOMETRY 50
Proposition 24.2. Every proper ideal I ⊂ R is the intersection of finitely many
irreducible ideals.
Proof. Let S be the set of all proper ideals of R which are not the intersection of
finitely many irreducible ideals. Assume the proposition is false, i.e, that S 6= ∅, and
use the fact that R is Noetherian to pick an ideal J ∈ S maximal with respect to
inclusion.
Since J is not irreducible, we can find ideals J ( K ( R and J ( L ( R such that
J = K ∩ L. Now the maximality of J implies that we can write K = K1 ∩ · · · ∩Ks
and L = L1 ∩ · · · ∩ Lt where K1, . . . , Ks, L1, . . . , Lt are irreducible and so we obtain
J = K ∩ L = K1 ∩ · · · ∩Ks ∩ L1 ∩ · · · ∩ Lt, a contradiction.
Definition 24.3. We call a proper ideal I ⊂ R primary if the zero-divisors of R/I
are nilpotent, i.e., if, whenever ab ∈ I for some a, b ∈ R, a ∈ I or bn ∈ I for some
n ≥ 1.
Example. Any prime ideal is primary.
The primary ideals of Z are those generated by a power of a prime integer.
Proposition 24.4. If I ⊂ R is a primary ideal, √I is a prime ideal
Proof. Let a, b ∈ R be such that ab ∈ √I, i.e., anbn ∈ I for some n ≥ 1. Since I is
primary, an ∈ I or (bn)m = bmn ∈ I for some m ≥ 1 and so a ∈ √I or b ∈ √I.
Proposition 24.5. Irreducible ideals are primary.
Proof. Let I ⊂ R be an irreducible ideal and suppose that ab ∈ I for some a ∈ R \ I
and b ∈ R. We need to show that bN ∈ I for some N ≥ 1.
Consider the ascending chain of ideals {(I : bnR)}∞n=1 where (I : bnR) is defined as
the ideal {r ∈ R : rnn ∈ I}. Since R is Noetherian, there exists a N ≥ 1 such that
(I : bn) = (I : bN) for all n ≥ N . Now write K = I +Ra and L = I +RbN and notice
that K is a an ideal which strictly contains I.
We now show that I = K ∩ L. Since I ⊆ K and I ⊆ L, I ⊆ K ∩ L. To prove the
reverse inclusion, pick any f ∈ K ∩L and write it at f = i+ ra = j + sbN for i, j ∈ I
and r, s ∈ R. Multiply the equation by b to obtain fb = bi+rab from which we deduce
MAS439/MAS6320 COMMUTATIVE ALGEBRA AND ALGEBRAIC GEOMETRY 51
that fb ∈ I and fb = bj+sbN+1 from which we deduce that s ∈ (I : bN+1) = (I : bN).
Now f = j + sbN implies f ∈ I.
Now I = K ∩ L, K ) I, so L = I and i.e., bN ∈ I.
We can now combine Propositions 24.2 and 24.5 to obtain the following.
Proposition 24.6. Every proper ideal of R can be written as an intersection of
finitely many primary ideals.
Definition 24.7. Let P ⊂ R be a prime ideal. We call an ideal I ⊂ R P -primary if
it is primary and
√
I = P .
Example. Let (R,m) be a local ring. If I ⊂ R is an ideal such that √I = m, then I
is m-primary. To see this consider the local quotient ring S = R/I: the set of zero-
divisors of S consist of the image of m is S, and, since
√
I = m, any such element is
nilpotent.
Example. Let φ : R → S be a homomorphism of rings, let P ⊂ S be prime and let
Q ⊂ S be P -primary. Then q := φ−1(Q) is p := φ−1(P )-primary: If ab ∈ q then
φ(a)φ(b) = φ(ab) ∈ Q and either φ(a) ∈ Q (in which case a ∈ q or φ(bn) = φ(b)n ∈ Q
for some n (in which case bn ∈ q.
The verification that
√
q = p is straightforward.
Proposition 24.8. Let q1, . . . , qs be P -primary ideals of a commutative ring R. Then
q1 ∩ · · · ∩ qs is P primary.
Proof. Notice that
√
q1 ∩ · · · ∩ qs = √q1 ∩ · · · ∩ √qs = P so we need only verify that
q1 ∩ · · · ∩ qs is primary. Assume that a, b ∈ R are such that ab ∈ q1 ∩ · · · ∩ qs. If
a /∈ q1 ∩ · · · ∩ qs, pick 1 ≤ j ≤ s for which a /∈ qj; now ab ∈ qj so b ∈ √qj = P =√
q1 ∩ · · · ∩ qs.
Definition 24.9. Let R be any commutative ring and let I ⊆ R be an ideal. A
primary decomposition I = q1 ∩ . . . qs is minimal if √q1, . . .√qs are distinct and qi
does not contain ∩1≤j≤s,j 6=iqj. for all 1 ≤ i ≤ s.
Proposition 24.8 can be used to turn a primary decomposition into a minimal one.
Lemma 24.10. Let q be a primary ideal in a commutative ring R. For any a ∈ R,√
(q : aR) =
{
R if a ∈ q√
q if a /∈ q
MAS439/MAS6320 COMMUTATIVE ALGEBRA AND ALGEBRAIC GEOMETRY 52
Proof. Immediate from definition.
Theorem 24.11. Let R be any commutative ring and let I ⊆ R be an ideal. If
I = q1 ∩ . . . qs and I = q′1 ∩ . . . q′s are minimal primary decompositions of I then s = t
and {√q1, . . . ,√qs} = {
√
q′1, . . . ,
√
q′t}.
Proof. For every 1 ≤ i ≤ s we can pick a
ai ∈ (∩1≤j≤s,j 6=iqj) \ qi
Now √
(I : ai) =
√
(∩sj=1qj : ai)
=
√
∩sj=1(qj : ai)
= ∩sj=1
√
(qj : ai)
=
√
qi
and so ∩sj=1
√
(q′j : ai) is also
√
qi hence
√
q′j =
√
qi for some 1 ≤ j ≤ t. We deduce
that {√q1, . . . ,√qs} = {
√
q′1, . . . ,
√
q′t}.

Definition 24.12. Let I = q1 ∩ · · · ∩ qs be a minimal primary decomposition. The
set of associated primes of I, denoted Ass I is the set {√q1, . . . ,√qs}.
Proposition 24.13. For any ideal I ⊂ R, the zero divisors of R/I are given by the
image of ∪ ass I in R/I.
Proof. Let q1∩· · ·∩qn be a minimal primary decomposition of I The zero divisors are
the images of ∪a/∈I(I : a) = ∪a/∈I
√
(I : a). Now
√
(I : a) = ∩{√qj | 1 ≤ j ≤ n, a /∈ qj}
so the zero divisors are contained in the images of ∪ ass I.
But we also saw that any associated prime P is the radical of the annihilator of
some element in R/I, say P =
√
I : a. Pick any r ∈ P and choose n minimal such
that rna ∈ I. Now n ≥ 1 (otherwise P = R) and r(rn−1a) ∈ I so r is zero-divisor on
R/I.
Definition 24.14. Let R be a commutative ring and let I ⊂ R be an ideal. A prime
ideal P ⊂ R is a minimal ideal of I if P ⊇ I and there is no prime ideal Q such that
I ⊂ Q ( P .
MAS439/MAS6320 COMMUTATIVE ALGEBRA AND ALGEBRAIC GEOMETRY 53
Proposition 24.15. Let R be a Noetherian ring and let I ⊂ R be an ideal. The
minimal primes of I consist of the minimal elements of Ass I with respect to inclusion.
In particular, I has finitely many minimal primes.
Proof. If P is a prime containing I, it also contains
√
I = ∩{Q |Q ∈ Ass I} and hence
by Proposition 6.6 it must contain an associated prime. All associated primes of I
contain I, so, if P is a minimal prime of I, it must then be one of the associated
primes.
Proposition 24.16. Let R be a Noetherian ring, let I ⊂ R be an ideal and let P
be a minimal prime of I. In any minimal primary decomposition of I the P -primary
component is given by IRP ∩R.
Proof. Let I = q ∩ q2 ∩ · · · ∩ qn be a minimal primary decomposition of I where q is
P primary. Now for all 2 ≤ i ≤ n, qi " P because otherwise √qi $ P contradicting
the minimality of P .
Now localize at P :
IRP = (q∩ q2∩ · · · ∩ qn)RP = qRP ∩ q2RP ∩ · · · ∩ qnRP = qRP ∩RP ∩ · · · ∩RP = qRP
and so IRP ∩R = qRP ∩R.
We now show that qRP ∩R = q: r ∈ qRP ∩R if and only if r/1 ∈ qRP if and only
if there exists an s ∈ R \P such that sr ∈ q. But, as q is P -primary the zero devisors
of R/q are
√
q/q = P/q, and since s /∈ P , we conclude that r ∈ q.

25. Artinian rings
Definition 25.1. A commutative ring R is Artinian if its ideals satisfy the descending
chain condition: every descending chain of ideals I1 ⊇ I2 ⊇ . . . stabilizes.
Proposition 25.2. An Artinian ring R has finitely many prime ideals and these are
all maximal ideals.
MAS439/MAS6320 COMMUTATIVE ALGEBRA AND ALGEBRAIC GEOMETRY 54
Proof. Let P ⊂ R be a prime ideal and consider the domain D = R/P ; notice that D
is also Artinian. For any a ∈ D\{0} the descending chain aD ⊇ a2D ⊇ . . . stabilizes,
and so an = an+1d for some n ≥ 1 and d ∈ D. Now an(da − 1) = 0 which implies
that a is a unit. Thus D is a field and P is maximal.
If P1, P2, . . . is an infinite sequence of distinct primes then P1 ∩ . . . Pn ∩ Pn+1 =
P1∩. . . Pn for some n ≥ 1, i.e., Pn+1 ⊇ P1∩. . . Pn and we must have Pn+1 ⊇ Pj for some
1 ≤ j ≤ n (cf. Proposition 6.6). We conclude that Pn+1 = Pj a contradiction.
Definition 25.3. Let R be a commutative ring. We say that an R-module M has
finite length if there exists a chain M0 ( M1 ( · · · ( Mn = M of submodules of
M where for all 1 ≤ i ≤ n− 1 the quotients Mi+1/Mi are simple, i.e., isomorphic to
R/mi for some maximal ideal mi ⊂ R.
Theorem 25.4. Let (R,m) be a Noetherian local commutative ring. The following
are equivalent:
(a) R is Artinian,
(b) m is the only prime ideal of R,
(c) mα = 0 for some α ≥ 1,
(d) R has finite length as a module over itself.
Proof. The previous proposition shows that (a)⇒ (b).
Notice that (b) implies that all elements in m are nilpotent, and since m is a finitely
generated ideal, (c) follows.
Assume now (c) and write K = R/m. Notice that Nakayama’s Lemma implies
that m/m2 is a finitely dimensional vector space and one can show that if a1, . . . , ad
is a K-vector space of m/m2 then the image in mi/mi+1 of the monomials of degree i
in a1, . . . , ad span the K-vector space m
i/mi+1. So the factors in the chain R ⊃ m ⊃
m2 ⊃ · · · ⊃ mα = 0 are finite dimensional K-vector spaces and can be refined to a
finite chain with simple factors.
Assume now (d). Any descending chain can be refined to a descending chain with
simple factors, so any finite-length module is Artinian
26. The height on an ideal
Definition 26.1. Let R be a commutative ring.
(a) The height of a prime ideal P ⊂ R, denoted htP , is the length of the longest
chain of prime ideals
P0 ( P1 ( · · · ( Ph = P.
MAS439/MAS6320 COMMUTATIVE ALGEBRA AND ALGEBRAIC GEOMETRY 55
If there are arbitrarily long such chains we set ht p =∞.
(b) The height of an ideal I ⊂ R, denoted ht I, is defined to be the supremum
over all prime ideals P ⊇ I of htP .
Remark. Heights of ideals in Noetherian rings may be infinite!
Let R be a commutative ring and let I ⊂ R be an ideal. Recall that prime ideal
P ⊂ R is a minimal ideal of I if P ⊇ I and there is no prime ideal Q such that
I ⊂ Q ( P (cf. Definition 24.14). The height of I is the maximal height of a the
minimal primes of I.
The dimension of a local ring (R,m) is defined to be the height of m.
Theorem 26.2 (Krull’s principal ideal Theorem). The height of a minimal prime of
a principal ideal in a Noetherian ring R is at most one.
Proof. Let xR be the principal ideal, let Q be a minimal prime of xR and assume
that we have a chain of prime ideals P0 ( P ( Q. We may localize at Q without
affecting this setup and so we can assume that Q is a the unique maximal ideal of
R. We may further replace R with R/P0 without affecting any issues an hence we
assume that (R,Q) is a local domain. Next define P (t) = P tRP ∩ R for all t ≥ 1.
Since P tRP has radical PRP , it is primary to PRP (see Example 24) and so each P
(t)
is primary to P (see Example 24).
Now Q/xR is the only prime ideal in R/xR and so R/xR is Artinian. The images of
the descending chain
{
P (t)
}
t≥1 in R/xR must stabilize, and so there exists an N ≥ 1
such that P (α) +xR = P (α+1) +xR for all α ≥ N and in particular P (α) ⊆ P (α+1) +xR
for all α ≥ N . Pick any α ≥ N and pick any vα ∈ P (α); write vα = vα+1 + xr with
vα+1 ∈ P (α+1) and r ∈ R. Now xr = vα+1 − vα ∈ P (α+1), but x /∈ P , otherwise Q
would not be a minimal prime of Q and so x /∈ P =
√
P (α+1) so r ∈ P (α+1) because
P (α+1) is a primary ideal. We deduce that vα ∈ P (α+1) and so P (α) = P (N) for all
α ≥ N .
Write J = ∩i≥1P (i); recall that we are assuming that R is a domain and so the
localization map R→ RP is injective. Now J ⊆ JRP ⊆ ∩i≥1P (i)RP = ∩i≥1P iRP = 0
and hence PN ⊆ P (N) = 0. But R is a domain and so P = 0.
Lemma 26.3. Let Ps ⊃ Ps−1 ⊃ · · · ⊃ P0 be a chain of prime ideals and let x ∈ Ps.
There exists a chain of primes Ps = Qs ⊃ Qs−1 ⊃ · · · ⊃ Q1 ⊃ Q0 such that x ∈ Q1.
Proof. If x ∈ P1 we are done; otherwise choose minimal i ≥ 1 such that x ∈ Pi+1 \Pi.
Using induction on i it is sufficient to show that there exists a prime Q with Pi+1 ⊃
Q ⊃ Pi such that x ∈ Q. Replace R with R/Pi and assume that R is a domain. Now
MAS439/MAS6320 COMMUTATIVE ALGEBRA AND ALGEBRAIC GEOMETRY 56
replace R with RPi+1 and assume that (R,m) is local, x ∈ m \ 0 and we need to find
a prime ideal 0 6= Q ( m such that x ∈ Q. Choose Q to be any minimal prime of x:
m ⊃ Q because htm ≥ 2.
Theorem 26.4 (Krull’s height Theorem). Let R be a Noetherian ring R and let I be
an ideal generated by n elements. Then ht I ≤ n.
Proof. If n ≤ 1 we are done, so assume n ≥ 2. Choose a maximal chain of primes
P0 ⊂ · · · ⊂ Ps ending at a minimal prime Ps of I and choose it so that xn ∈ P1. Now
consider the images in R/xnR; the image of I is generated by n− 1 elements and the
Ps/xnR is a minimal prime of I/xnR so s− 1 ≤ n− 1
27. Matrix algebra over commutative rings
Matrix algebra can be generalized to matrices with entries in a commutative ring R
(rather than a field). The notions of addition, multiplication and scalar multiplication
extend naturaly to this more general setting. Also, the zero and identity matrices
can be viewed as matrices with entries and R, and play the role of additive and
multiplicative identities, respectively.
The determinant of a square matrix also extends to matrices with entries in a
commutative ring as well as the following notion of classical adjoints.
Definition 27.1. Let M be an n× n matrix with entries in a commutative ring R.
For each 1 ≤ i, j ≤ n let Mij be the (n− 1)× (n− 1) matrix obtained by deleting the
ith row and jth column of M .
Define the adjoint matrix ofM to be the n×nmatrix whose ij entry is (−1)i+j detMji.
Proposition 27.2. Let M = (mij) be an n×n matrix with entries in a commutative
ring R.
M(adjM) = (adjM)M = (detM)I
where I is the n× n identity matrix.
Proof. Compute the ij entry aij in M(adjM) as follows. If i = j,
aij =
n∑
k=1
(−1)j+kmik detMjk = detM
because the middle expression is merely the expansion of detM along the ith row.
MAS439/MAS6320 COMMUTATIVE ALGEBRA AND ALGEBRAIC GEOMETRY 57
If i 6= j, let N be the matrix obtained from M by replacing its jth row with a
copy of its ith row. Note that detN = 0. For each 1 ≤ i, j ≤ n let Nij be the
(n− 1)× (n− 1) matrix obtained by deleting the ith row and jth column of N . Now
aij =
n∑
k=1
(−1)j+kmik detMjk =
n∑
k=1
(−1)j+kmik detNjk = ±
n∑
k=1
(−1)j+kmik detNik = detN = 0.
We conclude that M(adjM) = (detM)I. The equality (adjM)M = (detM)I is
proved in a similar way.
28. Algebras
Definition 28.1. Let R be a commutative ring.
(a) We call S an R-algebra if it is a commutative ring and there exists a ring
homomorphism φ : R→ S.
(b) We call a subring of S which is also an R-algebra a R-subalgebra of S.
Proposition 28.2. Let R be a commutative ring, let S be an R-algebra and let
{aλ}λ∈Λ be a set of elements of S. There exists a minimal R-algebra of S which
contains {aλ}λ∈Λ. (We call this R-algebra the R-subalgebra of S generated by {aλ}λ∈Λ
and denote it with R [{aλ}λ∈Λ].)
Proof. Let C be the set of all R-subalgebras of S which contain {aλ}λ∈Λ. Notice that
S ∈ C so C 6= ∅. It is easy to verify that ∩T∈CT ∈ C and it must be minimal.
Example. Take R = K a field, S = K[x, y, z] and consider the K subalgebra K[xy, xz]
of S. It consists of all P (xy, xz) ∈ S for all polynomials P ∈ K[U, V ].
Example. Let A = Q[x, y] and I = y2A ⊂ A. x2 + I is not a zero divisor in A/I.
Let B = Q[x2, y2, x2y, y3] be the Q-subalgebra of A generated by x2, y2, x2y and
y3. Let J = y2B, x2 + J is a zero divisor in B/J .
Definition 28.3. Let R be a commutative ring and let S be an R-algebra. We call S
a finitely generated R-algebra if there exist s1, . . . , sn ∈ S such that S = R[s1, . . . , sn].
Example. An R-algebra S has a natural structure of R-module given by r ·s = φ(r)s.
Finitely generated R-modules which are also R-algebras are finitely generated R-
algebras. Finitely generated R-algebras are usually not finitely generated R-modules.
MAS439/MAS6320 COMMUTATIVE ALGEBRA AND ALGEBRAIC GEOMETRY 58
Definition 28.4. Let R be a commutative ring and let S, T be R-algebras. An
homomorphism of R algebras φ : S → T is an homomorphism of rings for which
φ(rs) = rφ(s) for all r ∈ R and s ∈ S. (Notice that if R is a subring of S and T , we
have φ(r) = r for all r ∈ R.)
Proposition 28.5. Let R be a commutative ring and let S be an R-algebra. For any
s1, . . . , sn ∈ S there exists an unique homomorphism of R-algebras φ : R[x1, . . . , xn]→
S for which φ(xi) = si for all 1 ≤ i ≤ n.
Proof. For any f(x1, . . . , xn) ∈ R[x1, . . . , xn], define φ(f(x1, . . . , xn)) = f(s1, . . . , sn).

Proposition 28.6. Let R be a commutative ring, and let S be an R-algebra. S is a
finitely generated R-algebra if and only if there exists a n ≥ 0 and a homomorphism
of R algebras φ : R[x1, . . . , xn]→ S which is surjective.
29. Integral extensions
Definition 29.1. Let R be a commutative ring and let S be an R-algebra.
(a) We say that S is module finite over R if it is a finitely generated R-module.
(b) If R ⊆ S, we say that an element s ∈ S is integral over R if there exists a
monic polynomial f(X) = Xd + ad−1Xd−1 + · · ·+ a1X + a0 ∈ R[X] such that
f(s) = 0.
Proposition 29.2. Let R be a commutative ring. Let S be an R algebra and T be
an S algebra. If S is module finite over R and T is module finite over S then T is an
R-algebra which is module finite over R.
Proof. It is easy to see that T is an R-algebra.
Assume that s1, . . . , sm ∈ S generate S as an R-module and that t1, . . . , tn ∈ T
generate T as an S-module. Any a ∈ T can be written as
a = b1t1 + · · ·+ bntn
MAS439/MAS6320 COMMUTATIVE ALGEBRA AND ALGEBRAIC GEOMETRY 59
with b1, . . . , bn ∈ S. Also, for all 1 ≤ i ≤ n, each bi can be written as
bi = ci1s1 + · · ·+ cimsm
with ci1, . . . , cim ∈ R so a =
∑n
i=1
∑m
j=1 cijtisj and we deduce that {tisj | 1 ≤ i ≤
n, 1 ≤ j ≤ m} generate T as an R-module.
Theorem 29.3. Let R be a commutative ring and let S be an R-algebra containing
R. The following are equivalent:
(a) S is module finite over R,
(b) there exist θ1, . . . , θn ∈ S which are integral over R such that S = R[θ1, . . . , θn],
(c) S is a finitely generated R-algebra and every element in S is integral over R.
Proof. Clearly (c) ⇒ (b).
We now assume (b) and prove by induction on n that S is module finite over R.
We start with n = 1 and write θ = θ1; now S = R[θ] and
(6) θd = −ad−1θd−1 − · · · − a1θ − a0
for some a0, . . . , ad−1 ∈ R . The fact that S = R[θ] implies that {θi}∞i=0 generate
S as an R-module; I claim that {1, θ, . . . , θd−1} generate S as an R-module, and to
show that it is enough to show that for all k ≥ d, θk can be written as an R-linear
combination of {1, θ, . . . , θd−1}. We proceed by induction on k: the case k = d follows
from equation (6) and if
θk = bd−1θd−1 + bd−2θd−2 + · · ·+ b1θ + b0
for some b0, . . . , bd−1 ∈ R then
θk+1 = bd−1θd + bd−2θd−1 + · · ·+ b1θ2 + b0θ
= bd−1
(−ad−1θd−1 − · · · − a1θ − a0)+ bd−2θd−1 + · · ·+ b1θ2 + b0θ.
Now for n > 1, we deduce that R[θ1, . . . , θn] = R[θ1, . . . , θn−1][θn] is module finite
over R[θ1, . . . , θn−1] and the induction hypothesis implies that R[θ1, . . . , θn−1] is mod-
ule finite over R. Proposition 29.2 now implies that R[θ1, . . . , θn] is module finite over
R.
We now prove (a)⇒ (c). Pick a set of generators {s1, . . . , sn} for S as an R-module.
Now every s ∈ S is an R-linear combination of s1, . . . , sn, so s can certainly be written
as a polynomial with coefficients in R of s1, . . . , sn, i.e., s1, . . . , sn generate S as an
R-algebra. Add a generator, if necessary, and assume that s1 = 1.
MAS439/MAS6320 COMMUTATIVE ALGEBRA AND ALGEBRAIC GEOMETRY 60
Fix any s ∈ S; we shall show that f(s) = 0 for some monic polynomial f(X) ∈
R[X]. For each 1 ≤ i ≤ n we can write ssi = ai1s1 + · · ·+ ainsn. Write
σ =
 s1...
sn

and let A be the n × n matrix whose (i, j)-entry is aij. We have Aσ = sσ, i.e.,
(A − sI)σ = 0 where I is the n × n identity matrix. Multiplying by the adjoint
matrix we obtain
adj(A− sI)(A− sI)σ = det(A− sI)Iσ = 0
and since s1 = 1, det(A− sI) = 0. Now (−1)n det(A− sI) is a monic polynomial in
s of degree n.
We end with the following proposition which we shall need later.
Proposition 29.4. Let R be a subring of the commutative ring S and assume that
all elements in S are integral over R. If a ∈ R is a unit in S then it is a unit in R.
Consequently, if S is a field, so is R.
Proof. Write as = 1 for some s ∈ S and use the fact that s is integral over R to
find a d ≥ 1 and r0, . . . , rd−1 ∈ R such that sd + rd−1sd−1 + · · · + r0 = 0 and assume
that these were chosen so that d is minimal. If d = 1, s ∈ R and we are done.
Assume by way of contradiction that d > 1. Multiply throughout by a to obtain
sd−1 + rd−1sd−2 + · · ·+ (r1 + ar0) = 0 contradicting the minimality of d.
30. Noether’s Normalization Theorem
Definition 30.1. Let R be a commutative ring. A polynomial f(x) ∈ R[x] is es-
sentially monic if its leading coefficient is invertible in R. When R = K[x1, . . . , xn]
for some field K, we think of R as K[x1, . . . , xn−1][xn] and we call f ∈ R essentially
monic if
f = adxn + ad−1xd−1n + · · ·+ a1xn + a0
where a0, . . . , ad−1 ∈ K[x1, . . . , xn−1] and ad ∈ K \ {0}.
Lemma 30.2. Let R = K[x1, . . . , xn] for some field K, and let f ∈ R. There exists
an homomorphism of K-algebras φ : R→ R which is a isomorphism of rings and for
which φ(f) is essentially monic.
MAS439/MAS6320 COMMUTATIVE ALGEBRA AND ALGEBRAIC GEOMETRY 61
Proof. Pick B ∈ N bigger that any exponent of any variable occurring in any non-zero
term of f . Let φ : R → R be the unique homomorphism of K-algebras for which
φ(xn) = xn and φ(xi) = xi + x
Bi
n for all 1 ≤ i ≤ n− 1. Also define ψ : R → R to be
the unique homomorphism of K-algebras for which ψ(xn) = xn and ψ(xi) = xi− xBin
for all 1 ≤ i ≤ n − 1. Since ψ (φ(xj)) = φ (ψ(xj)) = xj for all 1 ≤ j ≤ n, ψ ◦ φ and
ψ ◦ φ are the identity maps on R hence φ is an automorphism.
Pick any term t = λxα11 x
α2
2 . . . x
αn−1
n−1 x
αn
n in f ; we have
φ(t) = λ(x1 + x
B
n )
α1(x2 + x
B2
n )
α2 . . . (xn−1 + xB
n−1
n )
αn−1xαnn
and its expansion has a unique term of maximal degree in xn, namely
(7) λxαn+α1B+α2B
2+···+αn−1Bn−1
n .
The crucial point is to realize that different terms in f will produce different terms of
maximal degree in xn in φ(f), because B > αj for all 1 ≤ j ≤ n so αn+α1B+α2B2 +
· · ·+αn−1Bn−1 is the base-B expansion of the integer αn+α1B+α2B2+· · ·+αn−1Bn−1.
Now the leading term of φ(f) will be one of the terms (7), hence φ(f) is essentially
monic.
Definition 30.3. Let R be a finitely generated algebra over a field K. We call
u1, . . . , un ∈ R algebraically independent if, for all non-zero f ∈ K[x1, . . . , xn], f(u1, . . . , un) 6=
0.
Alteratively, u1, . . . , un ∈ R are algebraically independent if the unique homomor-
phism of K-algebras φ : K[x1, . . . , xn]→ R for which φ(xj) = uj for all 1 ≤ j ≤ n is
injective.
Theorem 30.4 (Noether’s Normalization Theorem). Let R be a finitely generated
algebra over a field K. There exist elements u1, . . . , ud ∈ R which are algebraically
independent and such that R is module-finite over K[u1, . . . , ud].
Proof. Pick a minimal n ≥ 0 for which there exists θ1, . . . , θn ∈ R for which R =
K[θ1, . . . , θn]. We proceed by induction on n. If n = 0, R = K and the theorem
holds. Assume now that n ≥ 1. If θ1, . . . , θn are algebraically independent take
d = n and u1 = θ1, . . . , ud = θd and we are done, so assume that θ1, . . . , θn are
not algebraically independent. There exists a non-zero f(x1, . . . , xn) ∈ K[x1, . . . , xn]
for which f(θ1, . . . , θn) = 0, and we may assume that xn occurs in f . Let φ be as
in the previous lemma; Let θ′1 = θ1 − θn, . . . , θ′n−1 = θn−1 − θBn−1n , θ′n = θn; now
θ1 = θ
′
1 + θ
′
n, . . . , θn−1 = θ
′
n−1 + θ
′Bn−1
n , θn = θ
′
n and so K[θ1, . . . , θn] = K[θ
′
1, . . . , θ
′
n].
MAS439/MAS6320 COMMUTATIVE ALGEBRA AND ALGEBRAIC GEOMETRY 62
Let g = φ(f) and recall that g is essentially monic in xn. Now
g(θ′1, . . . , θ
′
n) = φ(φ
−1(f))(θ1, . . . , θn) = f(θ1, . . . , θn) = 0.
Replace θ1, . . . , θn with θ
′
1, . . . , θ
′
n and f with g = φ(f) and assume that f is essen-
tially monic, hence θn is integral over K[θ1, . . . , θn−1]. Now the induction hypothesis
implies that there exist algebraically independent u1, . . . , ud ∈ K[θ1, . . . , θn−1] such
that K[θ1, . . . , θn−1] is module finite over K[u1, . . . , ud], and since K[θ1, . . . , θn−1, θn]
is module finite over K[θ1, . . . , θn−1] Proposition 29.2 implies that K[θ1, . . . , θn−1, θn]
is module finite over K[u1, . . . , ud].
Proposition 30.5. Let R be a subring of the commutative ring S and assume that
all elements in S are integral over R. If a ∈ R is a unit in S then it is a unit in R,
hence if S is a field, so is R.
Proof. Write as = 1 for some s ∈ S and use the fact that s is integral over R to
find a d ≥ 1 and r0, . . . , rd−1 ∈ R such that sd + rd−1sd−1 + · · · + r0 = 0 and assume
that these were chosen so that d is minimal. If d = 1, s ∈ R and we are done.
Assume by way of contradiction that d > 1. Multiply throughout by a to obtain
sd−1 + rd−1sd−2 + · · ·+ (r1 + ar0) = 0 contradicting the minimality of d.
31. Hilbert’s Nullstellensatz
In this section we describe all the maximal ideals of K[x1, . . . , xn] when K is an
algebraically closed field. Clearly, for any λ1, . . . , λn ∈ K, (x1−λ1)R+· · ·+(xn−λn)R
is a maximal ideal; we shall see that when K is algebraically closed, all maximal ideals
of K[x1, . . . , xn] have this form.
Theorem 31.1 (Zariski’s Lemma). Let K be a field and let R be a finitely generated
K-algebra. If R is also a field then R is module finite over K, (i.e., a finite field
extension of K.)
Proof. Apply Noether’s Normalization Theorem to find algebraically independent
θ1, . . . , θd ∈ R such that R is module finite over A = K[θ1, . . . , θd] ⊆ R. But R
is a field, so Proposition 29.4 implies that A must be a field too, hence d = 0.
Definition 31.2. A field K is algebraically closed if every polynomial in K[X] has
a root in K. (This implies that every polynomial in K[X] factors into a product of
linear factors.)
MAS439/MAS6320 COMMUTATIVE ALGEBRA AND ALGEBRAIC GEOMETRY 63
Example. None of Q,R and Z/pZ (prime p) are algebraically closed. The funda-
mental theorem of algebra states that C is algebraically closed.
Proposition 31.3. Let K be an algebraically closed field and let L ⊇ K be a field
extension. If L is module finite over K then K = L.
Proof. Any a ∈ L is integral over K, say f(a) = 0 for f ∈ K[X]. But all roots of f
are in K, hence a ∈ K.
Theorem 31.4 (Hilbert’s Nullstellensatz– weak form). Let K be an algebraically
closed field,let R = K[x1, . . . , xn] and let f1, . . . , fs ∈ R. If f1R+ · · ·+fsR 6= R, there
exists a (a1, . . . , an) ∈ Kn such that fi(a1, . . . , an) = 0 for all 1 ≤ i ≤ s.
Proof. Assume f1R+ · · ·+fsR 6= R, and pick a maximal ideal m ⊂ R which contains
f1R + · · · + fsR. The composition of K ⊂ R → R/m is a ring homomorphism
whose kernel is either K or 0. Since 1 ∈ K is not in the kernel we deduce that
K ⊆ R/m. Furthermore, R/m is a finitely generated K algebra, because R is a
finitely generated K algebra, so Zariski’s Lemma implies that R/m is module finite
over the algebraically closed field K, hence K = R/m.
Now we see that for all 1 ≤ i ≤ n we can find ai ∈ K such that xi + m =
ai +m and so (x1 − a1)R+ · · ·+ (xn − an)R ⊆ m and since both ideals are maximal
(x1 − a1)R + · · ·+ (xn − an)R = m and fi(a1, . . . , an) = 0 for all 1 ≤ i ≤ s.
32. The Going Up Theorem
Theorem 32.1 (The going up Theorem). Let S ⊇ R be an integral extension.
(a) For any prime P ⊂ R there exists a prime Q ⊂ S with P = Q ∩R.
(b) If Q ⊆ Q′ ⊂ S are primes with Q ∩R = Q′ ∩R, then Q = Q′.
(c) For any chain of primes P0 ⊂ · · · ⊂ Pn of R and prime Q0 ⊂ S such that
Q0 ∩R = P0 there exist primes Q0 ⊂ Q1 ⊂ · · · ⊂ Qn in S with Pi = Qi ∩R.
Lemma 32.2. Let R be a subring of S, and let P ⊂ R be prime. The following are
equivalent:
(a) There exists a prime Q ⊂ S such that P = Q ∩R,
MAS439/MAS6320 COMMUTATIVE ALGEBRA AND ALGEBRAIC GEOMETRY 64
(b) R \ P is disjoint from PS, and
(c) PS ∩R = P .
Proof. (b) ⇔ (c) is easy. If (a) then PS ∩ R ⊆ Q ∩ R = P . Assume (b) and note
that R \ P is a multiplicative system in S. There exists a (prime) ideal PS ⊆ Q ⊂ S
disjoint from R\P maximal with respect to inclusion, and for this Q, P = Q∩R.
Proof of going up (a). We show that if I ⊆ R is an ideal and u ∈ IS∩R then u ∈ √I.
In particular, if I is prime IS ∩R = I.
Write u =
∑d
t=1 jtθt where jt ∈ I, θt ∈ S, and let S ′ = R[θ1, . . . , θd]. Each θt is
integral/R, so S ′ is integral/R. We still have u ∈ IS ′ ∩ R, and so we may assume
S = S ′ is module finite/R.
Pick R-module generators s1 = 1, s2, . . . , sn ∈ S and for all 1 ≤ µ ≤ n write usµ =∑n
ν=1 iµνsν where iµν ∈ I. Now det (uIn×n − (iµν)) = 0 and expanding this gives
un ∈ I.
Lemma 32.3. Let R be a subring of a domain S. Any non-zero s ∈ S integral over
R has a non-zero multiple in R.
Proof. Consider a monic relation sd+rd−1sd−1 + · · ·+r1s+r0 = 0, of smallest degree.
We have r0 6= 0, otherwise sd−1 + rd−1sd−2 + · · ·+ r1 = 0 because S is a domain.
Proof of going up (b). We may replace R and S with R/P and S/Q and assume that
R and S are domains, and that Q′∩R = 0. This contradicts the previous lemma.
Proof of going up (c). Using induction on n we can reduce to the case n = 1. S/Q0
is integral over R/P0 and part (a) shows that there exists a prime Q1/Q0 of S/Q0
lying over P1/P0.
33. Dimensions of rings
We now arrive at one of the central concepts in this course
Definition 33.1. The dimension of a ring R, denoted dimR, is the length of the
longest chain of primes in R.
The dimension of an algebraic set X, denoted dimX, is the dimension of its coor-
dinate ring.
Example. Fields are zero dimensional.
Finite sets of points in An are zero dimensional.
Z is one dimensional and so is any PID.
Corollary 33.2. If R ⊆ S is an integral extension, dimR = dimS.
MAS439/MAS6320 COMMUTATIVE ALGEBRA AND ALGEBRAIC GEOMETRY 65
Proof. Any chain of primes Q0 ⊂ · · · ⊂ Qd in S gives a chain of primes Q0 ∩ R ⊂
· · · ⊂ Qd ∩R hence dimR ≥ dimS.
Conversely, and chain of primes P0 ⊂ · · · ⊂ Pd in R can be lifted to a chain of
primes in S of the same length, hence dimR ≤ dimS

Recall the proof of Noether normalization actually showed that every finitely gen-
erated K-algebra R is module finite over a polynomial subring in R.
Now we can compute the dimension of any finitely generated K-algebra:
Corollary 33.3. Let K be a field and R = K[x1, . . . , xn]; dimR = d.
Proof. Since we have a chain of primes
0 ( x1R ( x1R + x2R ( · · · ( x1R + . . . xnR
we have dimR ≥ n.
To prove that dimR ≤ n we proceed by induction on n ≥ 0.
We first note that for any f ∈ R of positive degree, dimR/fR ≤ n − 1. We may
assume that xn occurs in f and the images of x1, . . . , xn in R/fR are not algebraically
independent, and so Noether’s Normalization Theorem implies that and R/fR is
module finite over a polynomial ring in a subset of these. Hence dimR/fR ≤ n− 1.
Pick a maximal chain of primes
0 = P0 ( P1 ( P2 ( · · · ( Ph−1 ( Ph ( R
and f ∈ P1 of positive degree. If f is reducible, one of its irreducible factors g is in
P1 and we can replace f with g assume that f is irreducible, and hence P1 = fR.
Now dimR = h = dimR/fR + 1 ≤ n− 1 + 1 = n.

34. Algebraic sets, again
Throughout this section K will denote a field.
Definition 34.1. For any S ⊆ K[x1, . . . , xn] we define
V (S) = {(a1, . . . , an) ∈ Kn | f(a1, . . . , an) for all f ∈ S}.
An algebraic subset of Kn is the set of the form V (S) for some S ⊆ K[x1, . . . , xn].
Example.
(a) V (x21 + x
2
2 − 1) ⊆ R2 is a circle with center at the origin and radius 1.
(b) V (x1x2x3) ⊆ R3 is the union of the planes x1 = 0,x2 = 0,x3 = 0.
(c) The algebraic subsets of K are its finite subsets and the whole of K = V (0).
MAS439/MAS6320 COMMUTATIVE ALGEBRA AND ALGEBRAIC GEOMETRY 66
Proposition 34.2. Let R = K[x1, . . . , xn] and S ⊆ R be any subset. Then V (S) =
V (〈S〉) and if 〈S〉 is generated by f1, . . . , fm ∈ R, V (S) = V (f1, . . . , fm).
Proof. Notice that, since S ⊆ 〈S〉, V (S) ⊇ V (〈S〉).
On the other hand pick any f ∈ 〈S〉, write it as f = r1s1 + · · · + rαsα where
r1, . . . , rα ∈ R and s1, . . . , sα ∈ S and pick any a ∈ V (S). Now f(a) = r1(a)s1(a) +
· · ·+ rα(a)sα(a) = 0 and we deduce that a ∈ V (〈S〉).
Since R is Noetherian, 〈S〉 is finitely generated; the second statement follows from
a similar argument: replace s1, . . . , sα with f1, . . . , fm.
We shall henceforth think of algebraic sets as V (I) for ideals I ⊆ K[x1, . . . , xn].
Proposition 34.3. Write R = K[x1, . . . , xn].
(a) For any ideals I, J ⊆ R, I ⊆ J ⇒ V (I) ⊇ V (J).
(b) V (R) = ∅ and V (0) = Kn.
(c) For any ideals I, J ⊆ R, V (IJ) = V (I ∩ J) = V (I) ∪ V (J).
(d) For any set of ideals {Iλ}λ∈Λ of R, V (
∑
λ∈Λ Iλ) = ∩λ∈ΛV (Iλ).
Proof. (a) and (b) are straightforward.
(c) Since IJ ⊆ I ∩ J and I ∩ J is contained in both I and J , (a) implies that
V (IJ) ⊇ V (I ∩J) ⊇ V (I)∪V (J). On the other hand, if a /∈ V (I)∪V (J), there exist
f ∈ I and g ∈ J such that f(a) 6= 0 and g(a) 6= 0. Now fg(a) 6= 0 so a /∈ V (IJ). We
deduce that V (IJ) ⊆ V (I) ∪ V (J).
(d) Since
∑
λ∈Λ Iλ ⊇ Iµ for all µ ∈ Λ, V (
∑
λ∈Λ Iλ) ⊆ ∩λ∈ΛV (Iλ). On the other
hand, if a ∈ ∩λ∈ΛV (Iλ) and f ∈
∑
λ∈Λ Iλ, write f = fλ1+· · ·+fλm with λ1, . . . , λm ∈ Λ
and verify that f(a) = fλ1(a) + · · ·+ fλm(a) = 0.
35. Hilbert’s Nullstellensatz– strong form
Definition 35.1. Let K be a field and let R = K[x1, . . . , xn]. For any subset X ⊆ Kn
we define
I(X) = {f ∈ R | f(x) = 0 for all x ∈ X} .
Proposition 35.2. Let K be a field.
(a) For any U ⊆ V ⊆ Kn, I(U) ⊇ I(V ).
(b) I(∅) = K[x1, . . . , xn] and if K is infinite, I(Kn) = 0.
(c) For any collection {Uλ}λ∈Λ of subsets of Kn, I(∪λ∈ΛUλ) = ∩λ∈Λ I(Uλ).
MAS439/MAS6320 COMMUTATIVE ALGEBRA AND ALGEBRAIC GEOMETRY 67
(d) If X ⊆ Kn is an algebraic set, then V (I(X)) = X.
Proof. (a) and the first statement of (b) are easy exercises.
Assume now that K is infinite. Since 0 ∈ I(Kn), to prove that I(Kn) = 0 we need
to show that for any non-zero polynomial f ∈ K[x1, . . . , xn] there exists a a ∈ Kn
such that f(a) 6= 0. We proceed by induction on n, the case n = 1 being well known
(a non-zero polynomial of degree d has at most d roots.) Assume now that n > 1 and
consider f as an element in K[x1, . . . , xn−1][xn], i.e., write f = gdxdn + · · ·+ g0 where
g0, . . . , gd ∈ K[x1, . . . , xn−1]. If there exists a b ∈ Kn−1 for which gi(b) 6= 0 for some
0 ≤ i ≤ d then the case n = 1 produces an a ∈ K such that f(a, b) 6= 0. If no such b
exists, the induction hypothesis implies that g0, . . . , gd = 0 hence f = 0.
(Notice that I(Kn) = 0 fails when K is finite. E.g., let p be prime: f(x) = xp − x
vanishes on Z/pZ.)
As for (c),
f ∈ I(∪λ∈ΛUλ) ⇔ f(a) = 0 for all a ∈ ∪λ∈ΛUλ
⇔ for all λ ∈ Λ, f(a) = 0 for all a ∈ Uλ
⇔ f ∈ ∩λ∈Λ I(Uλ)
To prove (d) write X = V (J) for some ideal J ⊆ K[x1, . . . , xn]. Now I(V (J)) ⊇ J
and so V (I(V (J))) ⊆ V (J) = X. The reverse inclusion is straightforward.
Proposition 35.3. For any subset X ⊆ Kn, I(X) is a radical ideal.
Lemma 35.4. Let K be an algebraically closed field, let R = K[x1, . . . , xn] and let
J ⊆ R be an ideal. We have I(V (J)) ⊆ √J .
Proof. Pick any f ∈ I(V (J)); we will show that f ∈ √J . This is clear when f = 0 so
assume that f 6= 0.
Recall that R is Noetherian so we may pick generators f1, . . . , fm for J . Now f ∈
I(V (J)) if and only if f vanishes at all points where f1, . . . , fm vanish simultaneously.
Let z be a new indeterminate and S = R[z]. Regard f, f1, . . . , fm as elements in S
and define f0 = zf−1 ∈ S. Notice that f0, f1, . . . , fm cannot all vanish simultaneously
at a point. The weak form of Hilbert’s Nullstellensatz implies that f0, f1, . . . , fm
generate the unit ideal in S so we can find polynomials
G0(x1, . . . , xn, z), G1(x1, . . . , xn, z), . . . , Gn(x1, . . . , xn, z) ∈ S
MAS439/MAS6320 COMMUTATIVE ALGEBRA AND ALGEBRAIC GEOMETRY 68
such that
(8) 1 = G0(x1, . . . , xn, z)(zf − 1) +G1(x1, . . . , xn, z)f1 + · · ·+Gn(x1, . . . , xn, z)fn.
Let T = Rf , the localization of R at the multiplicative set {f i}∞i=0 and let φ : S → T
be the unique map of R-algebras for which φ(z) = 1/f . Since φ fixes R, when we
apply it to (8) we obtain
1 = G0(x1, . . . , xn, 1/f)× 0 +G1(x1, . . . , xn, 1/f)f1 + · · ·+Gn(x1, . . . , xn, 1/f)fn
= G1(x1, . . . , xn, 1/f)f1 + · · ·+Gn(x1, . . . , xn, 1/f)fn.
We can find aN 0 such that eachGi(x1, . . . , xn, 1/f) has the form gi(x1, . . . , xn)/fN ;
now clear denominators by multiplying both sides by fN to obtain an expression of
the form
fN = g1f1 + · · ·+ gnfn
where now g1, . . . , gn ∈ R.
Theorem 35.5 (Hilbert’s Nullstellensatz– strong form). Let K be an algebraically
closed field and let R = K[x1, . . . , xn]. There exists an order reversing bijection
between the algebraic subsets of Kn and the radical ideals of R under which X maps
to I(X).
Proof. The map X 7→ I(X) is clearly order-reversing.
Proposition 35.2 shows that V (I(X)) = X for every algebraic set X.
Let J ⊆ R be a radical ideal. Clearly, I(V (J)) ⊇ J , so we need to show that
if f ∈ I(V (J)) then f ∈ J , and this follows from the previous lemma because J is
radical.
36. Irreducible algebraic sets, again
Throughout this section K will denote any field.
Definition 36.1. A non-empty algebraic set X ⊆ Kn is reducible if there exists
algebraic sets Y, Z ⊂ Kn properly contained in X for which X = Y ∪ Z.
An algebraic set which is not reducible is called irreducible.
Example. V (x2 − x21) ⊂ K2 is irreducible.
V (x1x2x3) ⊂ K2 is reducible.
Recall Lemma 8.7
Lemma. The algebraic set X ⊆ Kn is irreducible if and only if I(X) is a prime ideal
of K[x1, . . . , xn].
MAS439/MAS6320 COMMUTATIVE ALGEBRA AND ALGEBRAIC GEOMETRY 69
Theorem 36.2. Every algebraic set X ⊆ Kn is the union finitely many irreducible
algebraic sets.
Proof. Write X = V (J) for an radical ideal J ⊆ K[x1, . . . , xn]. Consider the primary
decomposition of J = P1 ∩ · · · ∩ Ps and notice that P1, . . . , Ps are prime. Now
X = V (P1∩· · ·∩Ps) = V (P1)∪· · ·∪V (Ps), and V (P1), . . . , V (Ps) are irreducible.
37. The dimension of an algebraic set
Definition 37.1. Let K be a field and let L ⊇ K be a field extension. The transcen-
dence degree of L over K, denoted tr-degK L, is the maximal n ≥ 0 for which there
exists elements a1, . . . , an ∈ L which are algebraically independent over K (cf. Defi-
nition 30.3.)
Example. If L is module finite over K, tr-degK L = 0.
Fact 37.2. Any two maximal sets of algebraically independent elements have the
same cardinality.
Example. LetK be a field and letK(x1, . . . , xn) be the field of fractions ofK[x1, . . . , xn]
(i.e., the localization at the prime ideal 0 ofK[x1, . . . , xn].) Now tr-degK K(x1, . . . , xn) =
n because x1, . . . , xn are algebraically independent over K and is a maximal such set.
Definition 37.3. Let K be a field and R = K[x1, . . . , xn].
(a) Let X ⊆ Kn be an irreducible algebraic set, i.e., X = V (P ) for some prime
ideal P ⊆ R. We define the dimension of X, denoted dimX, to be tr-degK L
where L is the field of fractions of the domain R/P , i.e., L = (R/P )〈0〉. If
X is not irreducible, we define dimX to be the maximal dimension of an
irreducible component of X.
(b) Let X ⊆ Kn be any algebraic set and let x ∈ X. Write X as a union of its
irreducible components X = X1 ∪ · · · ∪Xs. The dimension of X at the point
x, denoted dimxX, is
max{dimXi | 1 ≤ i ≤ s and x ∈ Xi}.
(c) The codimension of a algebraic set X ⊆ Kn is n− dimX.
Notice that this definition of dimension of an irreducible algebraic set X = V (P )
gives the same value as the dimension of the ring R/P : apply Noether Normalization
MAS439/MAS6320 COMMUTATIVE ALGEBRA AND ALGEBRAIC GEOMETRY 70
(Theorem 30.4) to S = R/P we find y1, . . . , yd ∈ S which are algebraically inde-
pendent over K, and the images of these in the fraction field of S are algebraically
independent over K.
Proposition 37.4. Let X ⊆ Kn be an irreducible algebraic set and let Z ( X be an
algebraic subset. Then dimZ < dimX.
Proof. Write R = K[x1, . . . , xn] and X = I(Q) for some prime ideal Q ⊂ R. It is
enough to show this when Z is irreducible, so assume that Z = V (P ) for some prime
ideal P ⊂ R which properly contains Q.
Let S = R/P . Write d = dimZ and let y1, . . . , yd ∈ R whose images in S are
algebraically independent. Pick any f ∈ P \ Q. We show that y1, . . . , yd, f are
algebraically independent in R/Q. If this fails, we have a polynomial relation
a0(y1, . . . , yd) + a1(y1, . . . , yd)f + · · ·+ am(y1, . . . , yd)fm = 0
in R/Q. If a0(y1, . . . , yd) = 0 in R/Q, we use the fact that R/Q is a domain to cancel
a power of f and obtain a relation in which a0(y1, . . . , yd) 6= 0 in R/Q. Now view the
relation modulo R/P (where the image of f is zero) to obtain a0(y1, . . . , yd) = 0 in
S, contradicting the choice of y1, . . . , yd.
Theorem 37.5. Let X ⊆ Kn be an irreducible algebraic set. Write R = K[x1, . . . , xn]
and let f ∈ R. Then
(a) X ∩ V (f) = X, or
(b) X ∩ V (f) = ∅, or
(c) every irreducible component of dimX ∩ V (f) has dimension dimX − 1.
Proof. Write X = V (P ) where P ⊂ R is prime, and let S = R/P . Either f ∈ P and
we are in case (a), or P + Rf = R and we are in case (b) or 0 ( fS ( S. In the
latter case we can apply Krull’s Principal Ideal Theorem 26.2 to deduce that fS has
height one. So for every minimal prime Q of P +Rf we have
dimR/Q = dim
R/P
Q/P
= dimX − htQ/P = dimX − 1.

Corollary 37.6. Let X ⊆ Kn be an irreducible algebraic set and let Z ( X be a
maximal proper irreducible algebraic subset. Then dimZ = dimX − 1
MAS439/MAS6320 COMMUTATIVE ALGEBRA AND ALGEBRAIC GEOMETRY 71
Proof. Write R = K[x1, . . . , xn], X = V (Q) and Z = V (P ) for some prime ideals
Q ( P ⊆ R. For any f ∈ P \Q we have Z ⊆ V (f) ∩X ⊆ X, but if V (f) ∩X = X,
X ⊆ V (f) and f ∈ Q, so we have Z ⊆ V (f) ∩X ( X.
Let Y1 = V (P1), . . . , Ys = V (Ps) be the irreducible components of X ∩ V (f). Since
Z ⊆ Y1 ∪ · · · ∪ Ys, P ⊇ P1 ∩ · · · ∩ Ps and so P ⊇ Pi for some 1 ≤ i ≤ s. Hence
Z ⊆ Yi and by the maximality of Z, we must have Z = Yi and part (c) of 37.5 implies
dimZ = dimX − 1.
Corollary 37.7. Let X ⊆ Kn be an irreducible algebraic set and let
0 6= X0 ( X1 ( · · · ( Xd−1 ( X
be a strictly ascending chain of irreducible algebraic sets of maximal length. Then
dimX = d.
Proof.
dimX = dimXd−1 + 1 = dimXd−2 + 2 = · · · = dimX0 + d.