Commutative Algebra Notes, following Introduction to Commutative Algebra Atiyah & Macdonald (version 24 Feb 2007)

19 Pages·2007·0.193 MB·English

by  Adam Boocher

Commutative Algebra Notes
Introduction to Commutative Algebra
Atiyah & Macdonald
Adam Boocher
1 Rings and Ideals
1.1 Rings and Ring Homomorphisms
AcommutativeringAwithidentity isasetwithtwobinaryoperations(addition
and multiplication so that for all x,y,z ∈A:
1. A is an abelian group with respect to addition (so it contains a zero ele-
ment, 0, and every x∈A has an additive inverse −x.
2. Multiplicationisassociative((xy)z =x(yz))anddistributiveoveraddition
(x(y+z)=xy+xz)
3. Multiplication is commutative (xy =yx)
4. There is an identity element 1 with x1=1x=x
The identity element is unique.
Comment. If 1=0 then for any x∈A,
x=x1=x0=0
and we call A the zero ring denoted by 0.
A ring homomorphism is a mapping f of a ring A into a ring B such that
for all x,y ∈ A, f(x+y) = f(x)+f(y), f(xy) = f(x)f(y) and f(1) = 1. The
usual properties of ring homomorphisms can be proven from these facts.
A subset S of A is a subring of A if S is closed under addition and multi-
plication and contains the identity element of A. The identity inclusion map
f :S →A is then a ring homomorphism.
The composition of two homomorphisms is a homomorphism.
1
1.2 Ideals, Quotient Rings
A subset a of A is an ideal of A if a is closed under addition and Aa=aA=a.
(Meaning that ra ∈ a for all r ∈ A,a ∈ a). The quotient group A/a is then a
ring by the obvious multiplication (a+a)(b+a) = (ab+a). We call this ring
the quotient ring A/a. The map π : A → A/a defined by π(x) = x+a is a
surjective ring homomorphism.
Proposition1.1. Thereisa1-1correspondencebetweentheidealsbcontaining
a and the ideals b of A/a.
Proof. We prove this with careful detail. Let b be an ideal containing a. Then
define φ(b) = {x+a : x ∈ b}, an ideal in A/a. Conversely let b be an ideal of
A/a. Then take ψ(b)={x∈A:x+a∈b}. This is an ideal of A and contains
a. Ifwecancheckthatφ◦ψ =idandψ◦φ=idthenwearedone. Thisiseasy,
though. Forthefirst,LetbbeanidealofA/a. Thenψ(b)={x∈A:x+a∈b}.
Then
φ(ψ(b))={x+a:x∈ψ(b)}={x+a:x∈{x∈A:x+a∈b}}
whichtranslatestosayingthatφ(ψ(b))={x+a:x+a∈b}. Sothatφ(ψ(b))=
b. The other composition is similar.
If f : A → B is any ring homomorphism, the kernel of f, that is is the set
of all a∈A so that f(a)=0 is an ideal of A and the image of f is a subring of
B, and f induces a ring isomorphism
A/kerf ∼=Imf.
Which you might like to call the first isomorphism theorem for rings. The
notation x≡y (mod a) means that x−y ∈a.
1.3 Zero-Divisors, Nilpotent Elements, Units
Azero-divisor inaringAisanelementxsothatthereexistsanonzeroelement
y ∈ A with xy = 0. A ring with no nonzero zero-divisors is called an integral
domain.
An element x ∈ A is nilpotent if there exists and n > 0 so that xn = 0. A
nilpotent element is a zero-divisor (unless A=0) since xn−1x=0.
A unit in A is an element x so that there exists an element y ∈ A with
xy =1. Thisy isdetermineduniquelyandisdenotedx−1. TheunitsofAform
a multiplicative group.
The multiples ax of an element x∈A is called the principal ideal generated
by x and is denoted (x). Note that x is a unit iff (x)=A=(1).
A field is a ring A in which 1 6= 0 and every non-zero element is a unit.
Every field is an integral domain. (If xy =0, then y =x−1xy =0).
Proposition 1.2. Let A be a nonzero ring. Then the following are equivalent:
2
1. A is a field
2. the only ideals in A are 0 and (1);
3. every homomorphism of A into a non-zero ring B is injective.
Proof. 1 =⇒ 2: Let a be a nonzero ideal with x ∈ a nonzero. Then x is a unit
and thus a⊇(x)=(1).
2 =⇒ 3. Let f : A → B be a homomorphism and B nonzero. Then kerf is
an ideal of A and is either (1) or 0. If it is (1) then f = 0 which is impossible
since f(1)=1. Thus kerf =0 and f is injective.
3 =⇒ 2. Let x be a nonzero element of A. Suppose that x is not a unit so
that(x)isnotequalto(1). ThenA/(x)isanonzeroidealandhencethenatural
homomorphismπ :A→A/(x)isinjective. Butthismeansthat(x)=ker(f)=0
which is a contradiction.
1.4 Prime Ideals and Maximal Ideals
An ideal p in A is prime if p6=(1) and if xy ∈p then either x∈p or y ∈p. An
ideal m in A is maximal if m6=(1) and if there is no ideal a such that
m⊂a⊂(1) (strict inclusions).
Equivalently we have the following mini-proposition/definition:
p is prime ⇐⇒ A/p is an integral domain;
m is maximal ⇐⇒ A/m is a field
Should we prove this? This first is quite easy, so we’ll work out the second. Let
m be maximal. Then since there is a correspondence between ideals of A/m
and ideals containing m, the maximality of m says that A/m has no non-trivial
ideals and Proposition 1.2 thus guarantees that A/m is a field. Conversely, if
A/m is a field, then by the correspondence again, there are no ideals between
m and A.
Comment. Note that this implies that maximal ideals are prime, but not nec-
essarily vice versa. Also note that the zero ideal is prime ⇐⇒ A is an integral
domain.
Proposition 1.3. If f : A → B is a ring homomorphism (henceforth abbrev.
HM) and b is a prime ideal in B then f−1(b) is a prime ideal in A.
Proof. This is very direct. The fact that f−1(b) is an ideal is immediate. Let
xy ∈ f−1(b). Then f(xy) ∈ b and therefore f(x)f(y) ∈ b it follows since b is
prime that either f(x) or f(y)∈b and thus either x or y ∈f−1(b).
Comment. The corresponding statement about maximal ideals is not true in
general, since if A is any ring that is not a field and F is any field, then 0 is
maximal in F but its inverse image in A may not be maximal at all. (Just let
f be injective!)
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Theorem 1.1. Every ring A6=0 has at least one maximal ideal.
The proof will use Zorn’s Lemma so remind you first what it says. Let S be
a partially ordered set (sometimes called a poset), that is, one with a relation
≤ that is reflexive, transitive and anti-symmetric. A subset T of S is called a
chain if any two elements of T are comparable. That is to say that if x,y ∈ T
then either x ≤ y or y ≤ x. An upper bound for a T in S is an element x ∈ S
such that t≤x for every t∈T. Finally, a maximal element in S is an element
x∈S so that for all y such that x≤y, we have x=y.
Theorem 1.2 (Zorn’s Lemma). If every chain T of S has an upper bound in
S then S has at least one maximal element.
Zorn’s Lemma is equivalent to the axiom of choice.
Proof. (of Theorem) Let Σ be the set of all ideals not equal to (1) in A. Order
Σ by inclusion. Σ is not empty, since 0 ∈ Σ. We must show that every chain
in Σ has an upper bound in Σ. Thus we are inspired to let (a ) be a chain of
α
ideals in Σ, so that for each pair of indices α,β we have either
a ⊆a or a ⊆a .
α β β α
S
Let a = a . We claim that a is an ideal. Indeed, a is clearly closed under
α α
multiplication by A, so we show closure under addition. Let x,y ∈ a. Then
x∈a , y ∈a for some α,β. Then one of these ideals contains the other, since
α β
they are elements of a chain and we therefore have x,y contained in the same
ideal and thus x+y ∈ a. Note that 1 ∈/ a since 1 ∈/ a for all α. Hence a ∈ Σ
α
and a is an upper bound of the chain. Thus by Zorn’s Lemma, Σ contains a
maximal element. What is a maximal element in Σ? It is an ideal that does
not contain 1 so that it there is larger ideal in Σ containing it; a maximal ideal
in A.
Corollary 1.1. If a 6= (1) is an ideal of A, there exists a maximal ideal of A
containing a.
Proof. Note that A/a has a maximal ideal m by the above theorem. Denote
by n the corresponding ideal of A containing a. We claim that n is maximal
in A. The claim is justified as follows: suppose that there is an ideal p strictly
between (1) and n. Then the ideal p={x+a:x∈p}=6 (1) is an ideal of A/a
that strictly contains m which is a contradiction.
Corollary 1.2. Every non-unit of A is contained in a maximal ideal. (Just let
a=(x)).
Comment. Thereexistringswithexactlyonemaximalideal,forexamplefields.
A ring A with exactly one maximal ideal m is called a local ring and the field
k =A/m is called the residue field of A.
Proposition 1.4. Let A be a ring and m6=(1) and ideal of A such that every
x∈A−m is a unit in A. Then A is a local ring and m its maximal ideal.
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Proof. Note that m is clearly maximal in A since if an ideal contained m and
anyotherelement, itwouldcontainaunit. Also, allideals6=(1)consistofnon-
units. Note that m contains all non-units so that m contains all ideals 6= (1),
and is therefore the only maximal ideal.
Proposition 1.5. Let A be a ring with m a maximal ideal of A, such that every
element of 1+m is a unit in A. Then A is a local ring.
Proof. Let x ∈ A−m. Then the ideal generated by m and x is (1) since m is
maximal. It follows then that there exist m∈m and t∈A so that m+xt=1.
This implies that
xt=1−m∈1+m
andthusxtisaunit. Thisalsoimpliesthatxisaunit. (Doyouseewhy?) Thus
we can apply the previous proposition and conclude that A is a local ring.
A ring with only finitely many maximal ideals is called semi-local.
Example 1.1. See Atiyah & MacDonald for examples.
Proposition 1.6. A principal ideal domain (PID) is an integral domain A
where every ideal is principle. In such a domain, every nonzero prime ideal is
maximal.
Proof. Supposethat(x)isaprimeideal,butnotmaximal. Let(y)⊃(x). Then
since (x) sits inside of (y), we see that x = yt for some t ∈ A. Thus yt ∈ (x)
and y ∈/ (x). Thus it follows that t ∈ (x) since (x) is prime and hence t = xw
for some w ∈A. Substituting this, we see that
x=yt=yxw =⇒ yw =1
and y is a unit so that (y)=(1) and (x) is maximal.
1.5 Nilradical and Jacobson Radical
Proposition 1.7. The set R of all nilpotent elements in a ring A is an ideal
and A/R has no nilpotent elements 6=0.
Proof. If x is nilpotent then so is ax for all a ∈ A. Now let x,y be nilpotent
elements, say xn,ym =0. Then
(x+y)n+m−1 =0
since each term of the expansion must contain either a power of x greater than
n−1 or a power of y greater than m−1. To see that A/R has no nilpotent
elements, note that x+R∈A/R is nilpotent if and only if xn+R=0 in A/R
which is equivalent to saying that xn ∈R which would imply that x∈R.
TheidealRdefinedaboveiscalledthenilradical ofA. Thefollowingpropo-
sition gives another definition of R.
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Proposition 1.8. The nilradical of A is the intersection of all the prime ideals.
Proof. Let R0 denote the intersection all prime ideals of A. Then if f is nilpo-
tent, then fn = 0 for some n > 0. Since 0 ∈ p for all ideals and p is prime, we
have that f ∈p for all prime ideals p and hence f ∈R0.
Conversely, we will show that if f is not nilpotent, then it is not in the
intersection of all prime ideals. Suppose that f is not nilpotent. Then let Σ be
the set of all ideals a such that no power of f is in a. Ordering Σ by inclusion
we can apply Zorn’s Lemma to conclude that it has a maximal element, p. We
shall show that p is prime by showing that x,y ∈/ p implies xy ∈/ p. Indeed, if
x,y ∈/ p then p+(x) and p+(y) properly contain p and thus are not elements
of Σ by the maximality of Σ. Thus it follows that there exist some n,m so that
fn ∈p+(x)
fm ∈p+(y)
which clearly imply that fn+m ∈p+(xy) which implies that xy ∈/ p. Thus p is
prime and does not contain f as required.
The Jacobson radical R of A is the intersection of all the maximal ideals of
A. It can be characterized as follows:
Proposition 1.9. x∈R ⇐⇒ 1−xy is a unit in A for all y ∈A.
Proof. ⇒: Suppose 1−xy is not a unit for some y ∈A. Then by Corollary 1.2
1−xy is contained in some maximal ideal m of A. But since x ∈ R ⊂ m we
have 1−xy ∈m =⇒ 1∈m which is absurd.
⇐: Suppose x ∈/ m for some maximal ideal m. Since x and m generate A
we have m+xy = 1 for some elements m ∈ m and y ∈ A. Thus 1−xy ∈ m
contradicting the fact that 1−xy is a unit.
2 Operations of Ideals
Wedefinethesum oftwoidealsaandbtobethethesetofallx+y wherex∈a
and y ∈ b. We denote the sum by a+b. It is the smallest ideal containing a
P
and b. In general we define the sum a of any family (possibly infinite) of
i∈I i
P
ideals a of A; its elements are all sums x , where x ∈a for all i∈I and all
i i i i
but finitely many of the x are zero. It is the smallest ideal of A which contains
i
all of the ideals a .
i
The intersection of any family of ideals is an ideal. (Proof is easy)
The product of two ideals a,b in A is the ideal generated by all products xy
P
where x∈a and y ∈b. It is the set of all finite sums x y where x ∈a and
i i i
y ∈ b. We can similarly define the product of any finite family of ideals. In
i
particular, the powers an of an ideal are defined. We also make the convention
that a0 =(1).
Example 2.1.
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1. If A=Z,a=(m),b=(n) then a+b is the ideal generated by gcd(m,n).
This follows since the gcd of any two numbers can always be represented
as an integer combination of the two numbers; a∩b is the ideal generated
bytheirlcm(proofeasy);andab=(mn)(proofeasy)Thusitfollowsthat
ab=a∩b ⇐⇒ m,n are coprime.
2. A = k[x ,...,x ], a = (x ,...,x ) = ideal generated by (x ,...,x ).
1 n 1 n 1 n
Then am is the set of all polynomials with no terms of degree <m.
The above operations are all commutative and associative. We also have a
distributive law for products.
a(b+c)=ab+ac
In general, ∩ and + are not distributive over each other. The best we can do is
the modular law
a∩(b+c)=a∩b+a∩c if a⊇b or a⊇c.
Indeed, if a⊇b and x∈a∩(b+c) then x=b+c for some b∈b⊆a and c∈c.
Since x∈a we have b+c∈a and thus that c∈a. The rest is easy to do.
Finally, (a+b)(a∩b)⊆ab. (since(a+b)(a∩b)=(a(a∩b)+b(a∩b)⊆ab).
Also we clearly have that ab⊆a∩b, hence
a∩b=ab provided a+b=(1).
We say that two ideals a,b are coprime if a+b=(1). Thus for coprime ideals
we have a∩b=ab. Two ideals are coprime iff there exist x∈a and y ∈b such
that x+y =1.
Let A ,...,A be rings. Their direct product
1 n
n
Y
A= A
i
i=1
is the set of all sequences x = (x ,...,x ) with x ∈ A with componentwise
1 n i i
addition and multiplication. A is a commutative ring with identity (1,...,1).
Wehaveprojectionsp :A→A definedbyp (x)=x ;theyareringhomomor-
i i i i
phisms.
Let A be a ring and a ,...,a ideals of A. Define a homomorphism
1 n
n
Y
φ:A→ (A/a )
i
i=1
by the rule φ(x)=(x+a ,...,x+a ).
1 n
T
Proposition 2.1. i. If a ,a are coprime whenever i6=j, then Πa = a .
i j i i
ii. φ is surjective ⇐⇒ a ,a are coprime whenever i6=j.
i j
7
T
iii. φ is injective ⇐⇒ a =(0).
i
Proof. i). Wewillinductonn,thenumberofideals. Whenn=2theargument
is handled above. Suppose we have a ,...,a and the result is true for any set
1 n
of n−1 ideals. Let b=Qn−1a =Tn−1a (by assumption). Since a +a =1
i=1 i i=1 i i n
for each i we know that we have equations
x +y =1 with x ∈a , y ∈a
i i i i i n
Note that
n−1 n−1
Y Y
x = (1−y )≡1(mod a ).
i i n
i=1 i=1
Hence a +b=(1) so they are coprime. Thus
n
n n
Y \
a =ba =b∩a = a .
i n n n
i=1 i=1
ii). ⇒: We will show, for example, that a and a are coprime. We ac-
1 2
complish this by asserting noting that since φ is surjective, there is an x so
that
φ(x)=(1+a ,0,...,0).
1
This means that x≡1 mod a and x≡0 mod a . This means that
1 2
1=(1−x)+x∈a +a
1 2
so that a and a are coprime.
1 2
⇐: It is enough to show that we can obtain (1+a ,0,...,0) for example.
1
Since a ,a are coprime for i>1 we have a set of equations
1 i
c +y =1 where c ∈a , y ∈a .
i i i 1 i i
Let x=Qn y . Then x is 0 mod a when i>2. On the other hand,
i=2 i i
n
Y
x= (1−c )
i
i=2
which is 1 mod a . Thus φ(x)=(1+a ,0,...,0).
1 1
T
iii). Note that a =ker(φ) which obviously implies the result.
i
Ingeneral,theunionoftwoidealsisnotanideal. Thereasonforthisisthat
the sum of two elements need not be contained in the ideal. For an example,
consider, Z with the union of the ideals (3),(5). This is the set of all integers
whicharemultiplesofeither3or5. Itisnotclosedunderadditionsince3+5=8
is not in the union. We can do better than that, however, with the following
“Prime Avoidance Lemma”.
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Proposition 2.2. (Prime Avoidance Lemma)
Let p ,...,p be prime ideals and let a be an ideal contained in Sn p .
1 n i=1 i
Then a⊆p for some i.
i
Proof. This is a somewhat complicated argument so we first walk through the
steps. We would like to show that
n
[
a⊆ p =⇒ a⊆p for some i.
i i
i=1
We will accomplish this by proving the contrapositive:
n
[
a*p for all i =⇒ a* p .
i i
i=1
We will do this by induction in fact! The statement is certainly true when n is
1. Wenowproveitforn>0. Assumethattheresultistrueforn−1. Suppose
a*p for all i=1...n. Then for each i the remaining n−1 ideals satisfy the
i
induction hypothesis so we can say
n
[
a* p \p .
j i
j=1
This is the same as saying that there is an element x ∈ a so that x is not in
i i
any of the p when j 6= i. If for any i we have that x ∈/ p then we will have
j i i
succeeded in showing the statement for n and are through. Thus suppose that
x ∈p for all i. Consider the element
i i
n
X
y = x x ···x x x ···x .
1 2 i−1 i+1 1+2 n
i=1
Then y ∈ a (clearly) and y ∈/ p (1 ≤ i ≤ n) (since y is the sum of n−1 terms
i
in p and one term not contained in p . In fact we need the primeness of p to
i i i
assert this latter part). Thus a*Sn p .
i=1 i
Proposition 2.3. Let a ,...,a be ideals and let p be a prime ideal containing
1 n
T
their intersection. Then p⊇a for some i. If p= a , then p=a for some i.
i i i
Proof. This proof is straightforward. Suppose the statement is false. Then for
Q Q T
each i there exists and x ∈a such that x ∈/ p. Then x ∈ a ⊆ a ⊆p.
i i i i i i
But since p is prime, the product cannot be an element of p. Thus we have a
T
contradiction and p⊇a for some a . If p= a then p⊆a for all i and hence
i i i i
p=a for some i.
i
If a,b are two ideals in a ring A, their ideal quotient is
(a:b)={x∈A:xb⊆a}
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This is an ideal: (This is clear) In particular, (0:b) is called the annihilator of
bandisalsodenotedAnn(b): itisthesetofallx∈Asuchthatxb=0. Inthis
notation, the set of all zero-divisors in A is
[
D = Ann(x)
x6=0
where (x) is the principal ideal generated by x. In fact, if b=(x) is a principal
ideal, we will just use (a:x) instead of the fancier (a:(x)).
Example 2.2. Let A=Z, a=(m), b=(n), where
m=2µ23µ35µ5··· , n=2ν23ν35ν5···
Then (a : b) is the set of all numbers which when multiplied with n give a
multiple of m. It is clear that (a:b)=(q) where
q =2γ23γ35γ5···
and γ =max(µ −ν ,0). In other words, q =m/gcd(m,n).
p p p
Exercise 1.12 i) a⊆(a:b). (Clear)
ii) (a:b)b⊆a. If x∈(a:b) then xb⊆a which is what this sentence is saying.
iii) ((a:b):c)=(a:bc)=((a:c):b). This argument is done by noticing that
x∈((a:b):c) =⇒ xc⊆(a:b)
=⇒ xcb⊆a =⇒ x(bc)⊆a =⇒ x∈(a:bc)
All the other directions are the same.
T T
iv)( a :b)= (a :b). These are both easy to see.
i i i i
P T P
v)(a : b ) = (a : b ). Suppose that x( b ) ⊆ a. Then in particular,
i i i i i i
T T
xb ⊆ a for each i. Thus x ∈ (a : b ). Conversely, if x ∈ (a : b ), then
i i i i i
P
x( b )⊆a as required.
i i
If a is any ideal of A, the radical of a is
r(a)={x∈A:xn ∈a for some n>0}
If φ : A → A/a is the standard homomorphism, then r(a) = φ−1(R ) (the
A/a
nilradical of the quotient). Thus since it is the preimage of an ideal, it is an
ideal.
Exercise 1.13 i) r(a)⊇a. (Clear)
ii) r(r(a))=r(a).
Suppose x ∈ r(r(a)). Then for some n, xn ∈ r(a) and thus for some m,
(xn)m =xnm ∈a. The other inclusion follows from i).
iii) r(ab)=r(a∩b)=r(a)∩r(b).
We prove the first inequality. Suppose that x ∈ r(ab). Then for some n,
xn ∈ ab ⊆ a∩b so x ∈ r(a∩b). Conversely, suppose x ∈ r(a∩b). Then for
some n, xn ∈a,xn ∈b which implies that x2n ∈ab so x∈r(ab).
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