# 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 M-primary; (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 A-module. Then B is a Noetherian ring. 829

25 Proof: By the immediately preceding observation, B is a Neotherian A-module. But all ideals of B are also A-submodules of B (though not necessarily conversely). Since A-submodules 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 Z-module, 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 A-module, M is finitely generated, by 1, x,..., x d. But A is Noetherian, so M is a Noetherian A-module (by the Theorem on page 766). But then J M, being an A-submodule of M 838

34 must be finitely generated as an A-module 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 A-algebra, 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 A-algebra 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 well-behaved rings A and C is itself well-behaved. 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 A-algebra; C is finitely generated as a B-module. Then B is finitely generated as an A-algebra. 845

41 Proof: Let x 1,..., x m generate C as an A-algebra, and y 1,..., y n generate C as a B-module. 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 A-subalgebra 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 A-algebra, so, B is also finitely generated as an A-algebra, and the Theorem is proved. 850

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