4.4 Noetherian Rings


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1 4.4 Noetherian Rings Recall that a ring A is Noetherian if it satisfies the following three equivalent conditions: (1) Every nonempty set of ideals of A has a maximal element (the maximal condition); (2) Every ascending chain of ideals is stationary (the ascending chain condition (a.c.c.)); (3) Every ideal of A is finitely generated. 806
2 Later in this section we will prove Hilbert s Basis Theorem which says that a polynomial ring in one indeterminate over a Noetherian ring is itself Noetherian. In particular, by iteration, the polynomial ring F[x 1,...,x n ] over a field F is Noetherian. It will follow quickly that all finitely generated rings are Noetherian. 807
3 But first we will prove that all proper ideals in Noetherian rings have primary decompositions, and simplify the First Uniqueness Theorem concerning the uniqueness of associated prime ideals. Call an ideal I of a ring A irreducible if, for all ideals J, K of A, I = J K = ( I = J or I = K ). Lemma: Every ideal of a Noetherian ring is a finite intersection of irreducible ideals. 808
4 Proof: Suppose the set Σ = { J A J is not a finite intersection of irreducible ideals } is nonempty. By the maximal condition, Σ has a maximal element M. Certainly M is not irreducible, so M = J K for some ideals J, K such that J M K. 809
5 But M J and M K, so, by maximality of M in Σ, J Σ and K Σ. Hence J and K are both finite intersections of irreducible ideals, so M = J K also is such an intersection, contradicting that M Σ. Hence Σ =, and the Lemma is proved. 810
6 Lemma: Every proper irreducible ideal of a Noetherian ring is primary. Proof: Let A be a Noetherian ring. If I is a proper ideal then I is irreducible iff the zero ideal of A/I is irreducible. It suffices then to suppose A is nonzero and {0} is irreducible, and prove that {0} is primary. Let x,y A such that xy =
7 We show x = 0 or y n = 0 for some n 1. Consider the chain of ideals Ann (y) Ann (y 2 )... Ann (y n )... which is stationary, since A is Noetherian. Hence ( n 1) Ann (y n ) = Ann (y n+1 ) =
8 We show x y n = {0}. Suppose z x y n. Then z = vx = wy n ( v, w A), so wy n+1 = (wy n )y = vxy = v0 = 0, so 813
9 w Ann (y n+1 ) = Ann (y n ), yielding z = wy n = 0. This proves x y n = {0}. By irreducibility of {0}, we get x = {0} or y n = {0}, whence x = 0 or y n = 0, and we are done. 814
10 The previous two lemmas prove: Theorem: Every proper ideal of a Noetherian ring has a primary decomposition. We can refine the First Uniqueness Theorem for primary decompositions, in this context, but first prove: Proposition: Every ideal of a Noetherian ring contains a power of its radical. 815
11 Proof: Let I be an ideal of a Noetherian ring A, so r(i) = x 1,..., x k for some x 1,...,x k A. Then ( i = 1,...,k)( n i 1) x n i i I. Put m = k (n i 1) + 1. i=1 816
12 Observe that ( r(i) ) m = x j x j k k k i=1 j i = m. k But if j i = m then j l n l for some i=1 l {1,...,k}). Hence ( ) each generator of m r(i) I. ( r(i) ) m lies in I, so 817
13 Corollary: The nilradical is nilpotent in a Noetherian ring. Proof: If A is Noetherian, then N = r({0}) is the nilradical of Proposition, A, so, by the previous N m {0} for some m 1, whence equality holds, which proves N is nilpotent. 818
14 Exercises: (1) Find an example of a Noetherian ring whose Jacobson radical does not equal the nilradical. (2) Show that if a ring satisfies the d.c.c. on ideals then the nilradical and Jacobson radical are equal. (3) Find an example of an ideal I of a ring A which does not contain a power of its radical r(i) (so necessarily A is not Noetherian). 819
15 Theorem: Let A be Noetherian, Q and M ideals of A with M maximal. TFAE (i) Q is Mprimary; (ii) r(q) = M ; (iii) M n Q M ( n > 0). Proof: (i) = (ii): follows by definition. (ii) = (i): follows (regardless of whether A is Noetherian) by an earlier result (page 684). 820
16 (ii) = (iii): follows from the previous Proposition. (iii) = (ii): if M n Q M for some n > 0 then M = r(m n ) r(q) r(m) = M, so r(q) = M. Our final observation about primary decompositions of ideals in Noetherian rings is a refinement of the First Uniqueness Theorem (page 701): 821
17 Theorem: Let A be a Noetherian ring and I a proper ideal. Then the prime ideals belonging to I are the prime ideals in { (I : x) x A }. n Proof: Let I = i=1 Q i decomposition of I, and put be a minimal primary P i = r(q i ) ( i). 822
18 Then, by the First Uniqueness Theorem, { P 1,..., P n } = { prime ideals P ( x A) P = r(i : x) } If x A and (I : x) is prime, then r(i : x) = (I : x) is prime, so (I : x) {P 1,...,P n }. Conversely, let i {1,...,n}. By the earlier 823
19 Proposition (page 815), ( m 1) Q i P m i. Put R = j i Q j. Then R P m i R P m i R Q i = I. Let m 0 be the least integer such that R P m 0 i I. 824
20 If m 0 = 0 then R I R, so I = R, which contradicts minimality of the primary decomposition of I. Hence m 0 1, and so we may choose some x R P m 0 1 \ I. In particular x R\I, so x Q i. As in the 825
21 proof of the First Uniqueness Theorem (see pages ), we have (I : x) = (Q i : x) and r(i : x) = r(q i : x) = P i. Certainly then (I : x) r(i : x) P i. 826
22 Also, P i x R P m 0 1 i P i = R P m 0 I, so P i (I : x), whence equality holds. This shows {P 1...,P n } = { prime ideals (I : x) x A }, and completes the proof of the Theorem. 827
23 We now investigate the preservation of the property of being Noetherian under certain natural constructions. We have already observed (on page 766) that homomorphic images of Noetherian rings are Noetherian, and (on page 766) that 828
24 a finitely generated module over a Noetherian ring is Noetherian. Theorem: Let A be a subring of a ring B. Suppose that A is Noetherian and B is finitely generated as an Amodule. Then B is a Noetherian ring. 829
25 Proof: By the immediately preceding observation, B is a Neotherian Amodule. But all ideals of B are also Asubmodules of B (though not necessarily conversely). Since Asubmodules satisfy the a.c.c., so do ideals of B, so B is a Noetherian ring. Example: The ring Z[i] of Gaussian integers is a finitely generated Zmodule, and Z is Noetherian. By the previous Theorem, Z[i] is a Noetherian ring. 830
26 Theorem: Rings of fractions of Noetherian rings are Noetherian. Proof: Let A be a Noetherian ring and S a multiplicatively closed subset. Let J S 1 A, so J = S 1 I ( I A). since all ideals of S 1 A are extended. But A is Noetherian, so I = x 1,..., x n 831
27 for some x 1,... x n A, whence J = x 1 /1,..., x n /1. Thus all ideals of S 1 A are finitely generated, which shows S 1 A is Noetherian. Hilbert s Basis Theorem: Let A be a Noetherian ring. Then the polynomial ring A[x] is Noetherian. 832
28 Proof: We prove that all ideals of A[x] are finitely generated. Consider {0} J A[x], and put I = { leading coefficients of polynomials in J }. It is easy to show that I A, so I = a 1,..., a n for some a 1,...,a n A, since A is Noetherian. 833
29 Then, for each i = 1,...,n, there is a polynomial p i (x) = a i x d i + ( terms of lower degree ) in J, for some d i 0. Put J = p 1 (x),..., p n (x) and d = max { d 1,..., d n }. 834
30 Let M = { q(x) A[x] degree of q(x) d }. Claim: J = (J M) + J. Clearly (J M) + J J. Conversely let 0 p(x) J. Then p(x) = ax m + ( terms of lower degree ) for some m
31 If m d then p(x) J M. Suppose m > d. Since a I Put a = n i=1 u i a i ( u 1,...,u n A). q(x) = p(x) n i=1 u i x m d i p i (x). Then q(x) is an element of J of degree < m. 836
32 By an inductive hypothesis, so q(x) (J M) + J, p(x) = q(x) + n i=1 u i x n d i p i (x) (J M) + J. 837
33 This proves J (J M) + J, whence equality holds, and the Claim is proved. Regarded as an Amodule, M is finitely generated, by 1, x,..., x d. But A is Noetherian, so M is a Noetherian Amodule (by the Theorem on page 766). But then J M, being an Asubmodule of M 838
34 must be finitely generated as an Amodule by, say, q 1 (x),..., q k (x). The ideal of A[x] generated by q 1 (x),..., q k (x), p 1 (x),..., p n (x) therefore contains J M and J, and is contained in J, so equals (J M) + J = J. 839
35 Thus J is finitely generated, completing the proof that A[x] is Noetherian. Corollary: If A is Noetherian then so is A[x 1,...,x n ] for every n 1. Proof: Theorem. immediate by iterating Hilbert s Basis 840
36 Corollary: If A is a Noetherian ring and B is a finitely generated Aalgebra, then B is a Noetherian ring. In particular, every finitely generated ring, and every finitely generated algebra over a field, is Noetherian. Proof: If A is Noetherian and B is generated as an Aalgebra by b 1,..., b n, 841
37 then B is a homomorphic image of the polynomial ring A[x 1,...,x n ] under the map p(x 1,...,x n ) p(b 1,...,b n ), so must be Noetherian, since A[x 1,...,x n ] is Noetherian (by the previous Corollary), and homomorphic images of Noetherian rings are Noetherian. 842
38 The last statement of the Corollary follows because every field is Noetherian, and every ring is an algebra over its subring generated by 1, which is isomorphic to either Z n, for some n, or Z, both of which are Noetherian. 843
39 The next (tricky) result says that, under certain conditions, an intermediate ring B, sandwiched between wellbehaved rings A and C is itself wellbehaved. C B A 844
40 Theorem: Suppose A B C is a chain of subrings and that (i) (ii) (iii) A is Noetherian; C is finitely generated as an Aalgebra; C is finitely generated as a Bmodule. Then B is finitely generated as an Aalgebra. 845
41 Proof: Let x 1,..., x m generate C as an Aalgebra, and y 1,..., y n generate C as a Bmodule. Then ( i = 1,...,m) ( b i1,...,b in B ) n x i = b ik y k k=1 846
42 and ( i, j {1,...,m})( c ij1,..., c ijn B ) n y i y j = c ijk y k. k=1 Let B 0 be the Asubalgebra of B generated by { } b ik, c ijk i,j {1,...,m}, k {1,...,n}. By the previous Corollary, B 0 is a Noetherian ring. 847
43 Further is a chain of subrings. A B 0 B C Claim: C is generated by y 1,..., y n as a B 0 module. Each element c of C can be expressed as a polynomial in x 1,...,x m with coefficients from A. 848
44 But each x i and each product y i y j is a linear combination of y 1,...,y n with coefficients from B 0, so that, after multiplying out, we can express c as a linear combination of y 1,...,y n with coefficients from B 0, which proves the Claim. By the Claim and the observation that B 0 is Noetherian, we conclude that C is a Noetherian B 0 module. 849
45 But B is a B 0 submodule of C, so B is a Noetherian B 0 module, so is finitely generated. But B 0 finally, is finitely generated as an Aalgebra, so, B is also finitely generated as an Aalgebra, and the Theorem is proved. 850
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