Download Artinian Commutative Rings

Artinian Commutative Rings
Mathematics 582, Spring 2013
In this note we prove that a commutative ring A with identity is Artinian if and only if
A is Noetherian and dim(A) = 0.
Lemma 1. Let A be an Artinian integral domain. Then A is a field.
Proof. Let a ∈ A be nonzero. Then there is a decreasing chain of ideals (a) ⊇ (a2 ) ⊇ · · · .
Since A is Artinian, this chain stabilizes, say at n. Then (an ) = (an+1 ). Then an ∈ (an+1 ).
Thus, there is b ∈ A with an = an+1 b. Cancelling an yields ab = 1, so a is invertible. Thus,
A is a field.
Recall that dim(A) is the supremum of the length of chains of prime ideals of A. Also
recall that dim(A) = 0 if and only if each prime ideal of A is maximal.
Corollary 2. Let A be Artinian. Then dim(A) = 0.
Proof. Let P be a prime ideal of A. Then A/P is a domain. Since A is Artinian, A/P is
Artinian. Thus, A/P is a field by the previous lemma. Thus, P is maximal. Consequently,
dim(A) = 0.
Lemma 3. Let A be Artinian. Then A has finitely many maximal ideals.
Proof. Let S be the set of all finite intersections of maximal ideals of A. Then S has a minimal
element, say M1 ∩ · · · ∩ Mn . We claim that the Mi are all the maximal ideals of A. To see
this, let M be a maximal ideal of A. Then M1 ∩ · · · ∩ Mn ∩ M ⊆ M1 ∩ · · · ∩ Mn . Minimality
implies M1 ∩ · · · ∩ Mn ∩ M = M1 ∩ · · · ∩ Mn . Thus, M1 · · · Mn ⊆ M1 ∩ · · · ∩ Mn ⊆ M . Since
M is prime, this forces Mi ⊆ M for some i. But, since Mi is maximal, we get M = Mi .
Let N (A) be the nil radical of A. That is,
N (A) = {a ∈ A : an = 0 for some n ∈ N}.
By Problem 4 of Homework 6, N (A) is equal to the intersection of all prime ideals of A.
More generally, if I is an ideal of A, we set
√
I = {a ∈ A : an ∈ I for some n ∈ N}.
1
p
√
Then N (A) = (0) and I is the pullback
of N (A/I) to an ideal of A. An argument similar
√
to that of Problem 4 will show that I it the intersection of all prime ideals of A containing
I.
Lemma 4. Let A be Artinian. Then N (A)n = 0 for some n.
Proof. We write N (A) = N for simplicity. We have a decreasing chain N ⊇ N 2 ⊇ · · · . Since
A is Artinian, there is n with N n = N n+1 = · · · . We prove that I := N n is 0. To see this,
suppose that I 6= 0 and let S be the set of all ideals J with IJ 6= 0. Then S is nonempty
since (1) ∈ S. Let K be a minimal element of S. Since IK 6= 0 there is a ∈ K with Ia 6= 0.
Then (a) = K by minimality of K. However, since N n = N 2n , we see that I = I 2 . Thus,
Ia = (I(I)a = I(Ia). Therefore, Ia ∈ S and Ia ⊆ (a). Minimality of K = (a) then yields
Ia = (a). Thus, there is x ∈ I with xa = a. Multiplying by x gives x2 a = x(xa) = xa = a.
Repeating this, we see that xn a = a for each n. However, x ∈ I implies that x ∈ N = N (A),
so x is nilpotent. Thus, xn = 0 for some n. Consequently, a = xn a = 0. This contradiction
proves that I = N (A)n = 0.
Let I be an ideal of a commutative ring R. Then I is irreducible if whenever I = J ∩ K
for two ideals J, K, then either J = I or J = K. We note that prime ideals are irreducible,
since if P is prime and P = J ∩ K, then JK ⊆ P . Since P is prime, either J ⊆ P or K ⊆ P .
However, P ⊆ J and P ⊆ K since P = J ∩ K. Therefore, either P = J or P = K.
The notion of irreducibility is in some sense a generalization of prime power. For, an
ideal (a) of Z is irreducible if and only if (a) = (pn ) for some prime p.
Lemma 5. Let A be a Noetherian ring.
√ n
1. Let I be an ideal of A. Then
I ⊆ I for some integer n.
2. Each ideal of A is a finite intersection of irreducible ideals.
√
3. If I is an irreducible ideal, then I is a prime ideal.
√
√
Proof. (1). Let I be an ideal.√Since A is Noetherian, I is finitely generated; say I =
ni
(a1 , . . . , at ). By definition
√ofn I for each i there is an integer ni with ai ∈ I. Let n =
n1 + · · · + nt . The ideal
I is generated by
ai11 · · · aitt : i1 + · · · + it = n .
√
Consequently, for each
ai11 · · · aitt of I, some exponent ir ≥ nr . Thus, ai11 · · · aitt ∈
√generator
n
I. This proves that
I ∈ I.
(2). Suppose the result is false. Then the set S of ideals which are not finite intersections
of irreducible ideals is nonempty. Since A is Noetherian, S has a maximal element I. Then
I is not irreducible (since I ∈ S), so I = J ∩ K for ideals J, K both properly larger than I.
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Maximality implies that both J and K are finite intersections of irreducible ideals. Therefore
I is also a finite intersection of irreducible ideals. This contradiction proves (2). q
√
√
(3). Let I be irreducible. Since I is a prime ideal if and only if I/I =
0 is a
p
prime idealpof A/I, it suffices
to
assume
that
I
=
(0).
We
then
show
(0) is prime. Suppose
p
n
that xy ∈ (0) and y ∈
/ (0). Then (xy) = 0 for some n and y is not nilpotent. We have
the chain of ideals ann(xn ) ⊆ ann(x2n ) ⊆ · · · , which must stabilize since R is Noetherian.
Suppose that ann(xnm ) = ann(x(n+1)m ). We claim that (xnm ) ∩ (y n ) = (0). For, if a ∈ (y n ),
then axn = 0. Also, if a ∈ (xnm ), then a = bxnm for some b ∈ R. Putting these together
gives 0 = axn = bxnm xn = bx(n+1)m . Therefore, b ∈ ann(x(n+1)m ) = ann(xnm ). This means
a = bxnm = 0. This proves
(xnm ) ∩ (y n ) = (0) as desired. Since (0) is irreducible and
p
(y n ) p
6= (0) becausepy ∈
/ (0), we conclude that (xnm ) = (0). Therefore, xnm is nilpotent, so
x ∈ (0). Thus, (0) is prime, as desired.
Recall that if M ⊆ N are A-modules, then N is Noetherian (resp. Artinian) if and only
if both M and N/M are Noetherian (resp. Artinian). We use this fact in the proof of the
following lemma.
Lemma 6. Suppose that there are (not necessarily distinct) maximal ideals m1 , . . . mt of R
with m1 · · · mt = (0). Then A is Artinian if and only if A is Noetherian.
Proof. Consider the chain
A ⊇ m1 ⊇ m1 m2 ⊇ · · · ⊇ m1 · · · mt−1 ⊇ m1 · · · mt = (0)
of A-modules. The quotient module m1 · · · mi−1 /m1 · · · mi is an A/mi -module since it is an
A-module and is annihilated by mi . Thus the quotient is an R/mi -vector space because
A/mi is a field. Therefore, m1 · · · mi−1 /m1 · · · mi is an Artinian A/mi -module if and only
if it is a Noetherian R/mi -module. Since its A/mi -submodules are exactly the same as its
A-submodules, m1 · · · mi−1 /m1 · · · mi is an Artinian A-module if and only if it is a Noetherian
R-module. By a repeated application of the fact stated immediately before the lemma, we
conclude that R is an Artinian A-module if and only if A is a Noetherian A-module, so A is
an Artinian ring if and only if A is a Noetherian ring.
Theorem 7. Let A be a commutative ring. Then A is Artinian if and only if A is Noetherian
and dim(A) = 0.
Proof. Suppose that A is Artinian. We have already proved that each prime ideal of A is
maximal, so dim(A) =). We show that A is Noetherian. By Lemma 3 the ring A has only
finitely many maximal ideals. By Lemma 4, we may write (0) = m1 · · · mt with each mi
maximal. Thus, by Lemma 6, A is Noetherian.
For the converse, suppose that A is a Noetherian ring and that each prime ideal of A is
maximal. Lemma 5 shows that (0) is an intersection of irreducible ideals, say (0) = I1 ∩· · ·∩It .
√ n
By the first part of that lemma, there is a common integer n with
Ii ⊆ Ii . Therefore,
√
√ n
√
(0) =
I1 · · · It . Because each Ii is a prime ideal, by assumption on R, the ideal (0)
is then a product of maximal ideals. Lemma 6 shows that A is Artinian.
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Remark 8. The assumption that A has an identity is crucial. The Abelian group Z (p∞ )
of all elements of Q/Z of order a power of p is known to be Artinian but not Noetherian as
a Z-module. If we define multiplication by xy = 0 for all x, y ∈ Z (p∞ ), then Z (p∞ ) is a
commutative ring without 1, and ideals are the same thing as subgroups. Thus, as a ring,
Z (p∞ ) is Artinian but not Noetherian. In all the proofs above the assumption that A has
an identity was only used to obtain the existence of maximal ideals. It is worth noting that
Z (p∞ ) has no maximal subgroups, and so as a ring it has no maximal ideals.
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