4. The nilradical and nilpotent ideals. Recall that a ring R is left Artinian if every descending chain of left ideals is eventually stationary and it is Artinian if it is left and right Artinian. The aim of the next two chapters is to study the structure of Artinian rings, and to a lesser extent Noetherian rings. Here is a typical example to illustrate the main results in the Artinian case: Let C C C U = U3(C) = 0 C C 0 0 C be the ring of complex upper triangular matrices. Then the ideal 0 C C I = 0 0 C 0 0 0 satisfies I3 = 0 and C 0 0 ∼ ∼ (4.1) U/I = 0 C 0 = C ⊕ C ⊕ C 0 0 C is just a direct sum of simple rings (even fields in this case). We aim to prove something similar in this chapter—given any left Artinian ring R then R has an ideal N with N r = 0 for some r ≥ 0 and such that ∼ R/N = Mn1 (D1) ⊕ Mn2 (D2) ⊕ · · · ⊕ Mnℓ (Dℓ) is a direct sum of matrix rings over division rings. Exercise. Prove Equation 4.1. Probably the best way of doing this is to find the (rather obvious!) homo- morphism from U to the second ring in (4.1) and check that it really is a homomorphism and that it has kernel I. Now apply the first isomorphism theorem for rings. We begin with some relevant definitions. First, given abelian subgroups A, B of a ring R then, as in Definition 2.29, we write AB = { i aibi : ai ∈ A, b i ∈ B}. The same definition applies if (say) A is a subgroup of R and B is a subgroupP of a left R-module M. The following simple facts about this concept are left as an exercise. Exercise 4.1. (1) If A is a left ideal and B is a right ideal of a ring R then AB is a two-sided ideal of R. In particular, if A and B are both ideals then so is AB . (2) if A is an ideal of R and B is a submodule of a left R-module M then AB is also a submodule of B. (3) Given abelian subgroups A1, A 2, A 3 of R then t (A1A2)A3 = A1(A2A3) = aibici : t ∈ N, a i ∈ A1, b i ∈ A2, c i ∈ A3 . (i=1 ) X By part (3) and induction we can also define, without ambiguity, A1A2 · · · Ar for subgroups Aj of R. 48 Definition 4.2. Let R be a ring. An element a ∈ R is nilpotent if an = 0 for some n ≥ 1. An abelian subgroup A of R is nilpotent if An = 0 for some n. Note that, by distributivity one has (4.2) The subgroup A is nilpotent ⇐⇒ a1a2 · · · an = 0 for all aj ∈ A. The subgroup A is nil if each element a ∈ A is nilpotent. Examples : (1) 0 is a nilpotent element in any ring. (2) In an integral domain, 0 is the only nilpotent element. 0 1 0 (3) 0 0 1 is a nilpotent element in M3(Z). 0 0 0 2 3 (4) Consider R = C[x1, x 2, · · · ]/(x1, x 2, x 3,... ). Then I claim that the maximal ideal M = ( x1, x 2, x 3,... ) is nil but not nilpotent. In order to prove this claim (which I leave as an exercise) the important thing to prove is that M is nil. To prove this you should first prove part (1) of the next lemma. Lemma 4.3. (1) If an = bm = 0 in a commutative ring R then (a + b)n+m−1 = 0 . (2) If I is a nilpotent left ideal in a ring S then I is contained in a nilpotent two-sided ideal. Proof. Part (1) is an exercise. (If you need a hint, see Lemma 4.5, below.) For part (2) note that, if In = 0 then IR is an ideal by Exercise 4.1(1) and so (IR )n = ( IR )( IR ) . (IR ) = I(RI )( RI ) . RI ⊆ In = 0 by Exercise 4.1(3) and induction. The first fundamental fact about Artinian rings is that, for such rings, the distinction between nil and nilpotent left ideals is illusory. Theorem 4.4. If I is a nil left ideal in a left Artinian ring R then I is nilpotent. Proof. We have a descending chain of left ideals I ⊇ I2 ⊇ I3 · · · . As R is left Artinian this chain must be eventually stationary; say In = In+1 = · · · . If In = 0 we are done, so assume not and set J = In. Then J is still nil but now J 2 = I2n = In = J. Now let S = {left ideals A : A ⊆ J and JA 6= 0 }. Note that S 6= ∅ as J ∈ S . Thus by Theorem 3.2 there exists a minimal element M ∈ S . Notice that this means there exists x ∈ M, so Rx ⊆ M, such that Jx 6= 0 hence such that J(Rx ) 6= 0. Thus M = Rx by minimality of M. Also, J(Jx ) = J 2x = Jx 6= 0 and so Jx = Rx = M. But this means that x = 1 x ∈ Rx = Jx ; say x = ax for some a ∈ J. This in turn implies that x = ax = a(ax ) = a2x = a3x = · · · = amx 49 for any m ≥ 1. As J is nil, am = 0 for some m and hence x = xa m = 0, giving the required contradiction. n Lemma 4.5. Given nilpotent left ideals Ij in a ring R then j=1 Ij is also nilpotent. P Proof. It suffices to prove the result for two left ideals, say I and J. If In = J m = 0 consider ( I + J)n+m−1. This consists of a sum of sets of the form a1 b1 a2 ar br X = I J I · · · I J where aj + bℓ = n + m − 1. X X b1 b2 br m In each such expression, either bℓ ≥ m in which case X ⊆ J J · · · J ⊆ J = 0 or aj ≥ n in which + −1 case X ⊆ Ia1 Ia2 · · · Iar J br ⊆ InPJ br = 0. In either case X = 0 and hence so is ( I + J)n mP . Note that this lemma fails for elements or indeed abelian subgroups of R; for example take 0 0 0 C C C I1 = and I2 = inside R = = M2(C). C 0 0 0 C C 0 1 Then of course I1 and I2 are nilpotent but ( 1 0 ) ∈ I1 + I2. Definition 4.6. Let R be a ring, and let N(R) = {I : I is nilpotent ideal of R}, X i.e. let N(R) be the sum of all nilpotent ideals of R. The ideal N(R) is called the nilradical of R. Clearly N(R) ⊆ {I : I is nilpotent left ideal of R}. X But the reverse inclusion also holds since every nilpotent left ideal I is contained in a nilpotent ideal IR (see Lemma 4.3). Hence N(R) = {I : I is nilpotent left ideal of R} = {I : I is nilpotent right ideal of R}. X X Proposition 4.7. Let R be a ring. Then N(R) is a nil ideal of R. Proof. Let x ∈ N(R). Then x ∈ I1 + I2 + · · · + It some finite set of nilpotent ideals Ij. By Lemma 4.5 I1 + I2 + · · · + It is a nilpotent ideal. So x is nilpotent and N(R) is a nil ideal. The example after Definition 4.2 shows that N(R) need not be nilpotent in an arbitrary ring R. However, once again, things are nicer for Artinian rings: Theorem 4.8. Let R be a left Artinian ring. Then (1) N(R) is a nilpotent ideal of R, (2) R/N (R) is a left Artinian ring with no nonzero nil or nilpotent left ideals. 50 Proof. (1) By Proposition 4.7 and Theorem 4.4, N(R) is a nilpotent ideal. (2) R/N (R) is a left Artinian ring by Theorem 3.6. Suppose that R/N (R) has a nil left ideal I. By the Correspondence Theorem for Rings, I has the form I/N (R), where I is a left ideal of R containing N(R). Let a ∈ I. Then there is a positive integer m such that ( a + N(R)) m = 0; equivalently am ∈ N(R). Since N(R) is nil, there is a positive integer n such that amn = ( am)n = 0. Therefore I is a nil left ideal and hence is nilpotent by Theorem 4.4. Thus I ⊆ N(R) and I/N (R) = 0. This shows that R/N (R) has no nonzero nil left ideals and hence no nonzero nilpotent left ideals. Remark: The theorem shows that, for a left Artinian ring R, N(R) is the unique largest nilpotent ideal of R. By Lemma 4.3 it is therefore also the largest nilpotent right ideal and the largest nilpotent left ideal. Example 4.9. (formerly Example 4.10) (1) Consider the ring of upper triangular matrices U = U3(C) from the beginning of the chapter, with ideal I of strictly upper triangular matrices. Then certainly I ⊆ N(U). If I 6= N(U), then N(U)/I would be a non-zero nil ideal of U/I . On the other hand (4.1) shows that U/I has no nilpotent elements, contradicting Theorem 4.8(2). Hence I = N(U). ni (2) Given the ring Z = Z/a Z for some a ∈ Z, write a = i∈I pi for distinct primes pi and let b = i∈I pi.