Posts Tagged ‘nilpotent left ideals’

Every nil, multiplicatively closed set of matrices is nilpotent

Posted: January 26, 2024 in Basic Algebra, Matrices, Rings
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Throughout this post, k is a field and M_n(k) is the ring of n \times n matrices with entries in k.

We have already seen the concepts of nil and nilpotency in a ring in this blog several times, but let’s see it here again. Let R be a ring, and S a subset of R. We say that S is nil if every element of S is nilpotent, i.e. for every s \in S there exists a positive integer n such that s^n=0. We say that S is nilpotent if there exists a positive integer n such that S^n=(0), i.e. s_1s_2 \cdots s_n=0 for any s_1, s_2, \cdots , s_n \in S. It is clear that S is nilpotent if and only if S^n=(0) for all positive integers n that are large enough.

If S is nilpotent, then S is clearly nil too but, as the following example shows, the converse is not true.

Example. Let \{e_{11},e_{12},e_{21},e_{22}\} be the standard k-basis for M_2(k), and let S:=\{e_{12},e_{21}\} \subset M_2(k). Then e_{12}^2=e_{21}^2=0 and so S is nil. But S is not nilpotent because (e_{12}e_{21})^n=e_{11}^n=e_{11} \ne 0, for any positive integer n.

Notice that the set S in the above Example is not multiplicatively closed because e_{12}e_{21}=e_{11} \notin S. We are now going to show that if S \subset M_n(k) is both nil and multiplicatively closed, then S is nilpotent. In fact, we show that S^n=(0). But first a couple of useful lemmas.

Lemma 1. Any finite sum of nilpotent left ideals of a ring R is nilpotent.

Proof. i) By induction, we only need to prove that for two ideals. So let I,J be two nilpotent left ideals of R. So I^m=J^n=(0) for some positive integers m,n. We claim that (I+J)^{m+n-1}=(0). So we need to show that if a_i \in I, b_i \in J, \ 1 \le i \le m+n-1, then

(a_1+b_1) \cdots (a_{m+n-1}+b_{m+n-1})=0.

The product on the left-hand side is a sum of terms of the form c_1c_2 \cdots c_{m+n-1}, where for each i, either c_i \in I or c_i \in J. So we just need to show that c_1c_2 \cdots c_{m+n-1}=0. Since there are m+n-1 of c_i, either at least m of c_i are in I or at least n of c_i are in J. We will assume that at least m of c_i are in I since the argument for the other case is similar. So we can write

c_1c_2 \cdots c_{m+n-1}=u_1v_1u_2v_2\cdots u_mv_mu_{m+1},

where each v_i is in I and each u_i is in I \cup J. Notice that there might be nothing before v_1 or after v_m or between v_i and v_{i+1} for some i, but that won’t cause any problem for our argument. Now, u_iv_i \in I for all i because v_i \in I and I is a left ideal. We also have that I^m=(0). Thus

c_1c_2 \cdots c_{m+n-1}=(u_1v_1)(u_2v_2)\cdots (u_mv_m)u_{m+1} \in I^mu_{m+1}=(0),

and so c_1c_2 \cdots c_{m+n-1}=0. \ \Box

Next is to generalize this basic fact that if A \in M_n(k) is nilpotent, then A^n=0.

Lemma 2. If S is a nilpotent subset of M_n(k), then S^n=(0).

Proof. We will assume S \ne (0), since there is nothing to prove in this case. Let k^n be the k-vector space of all n \times 1 vectors with entries in k and let

V_j:=\{v \in k^n: \ S^jv=(0)\}, \ \ \ \ j \ge 1.

So V_j is the annihilator of S^j in k^n. Clearly each V_j is a k-vector subspace of k^n and V_1 \subseteq V_2 \subseteq \cdots. Since S is nilpotent, there exists the smallest positive integer m \ge 2 such that S^m=(0). So V_m=k^n, and V_{m-1} \ne k^n.

Claim. V_j \ne V_{j+1} for all 1 \le j \le m-1.

Proof. By contradiction. Let j \le m-1 be the largest positive integer such that V_j=V_{j+1}. That means if S^{j+1}v=(0), for some v \in k^n, then S^jv=(0). Notice that in fact we have j \le m-2 because V_{m-1} \ne V_m=k^n. But then S^{j+2}v=S^{j+1}(Sv)=0, \ v \in k^n, implies that S^{j+1}v=S^j(Sv)=(0) and so V_{j+1}=V_{j+2}, contradicting the maximality of j. This completes the proof of the Claim.

So, by the Claim, V_1 \subset V_2 \subset \cdots \subset V_{m-1} \subset V_m=k^n. Therefor

\dim_k V_1 < \dim_k V_2 < \cdots < \dim_k V_{m-1} < \dim_k V_m=n,

and so m \le n, which gives S^n=(0) because S^m=(0). \ \Box

We are now ready to prove the main result.

Theorem. Let S be a nil, multiplicatively closed subset of M_n(k). Then S^n=(0).

Proof. Since there is nothing to prove if S=(0), we will assume that S \ne (0). Let V be the k-vector subspace of M_n(k) generated by S. We are done if we show that V^n=(0) because S \subseteq V. Notice that V is also a ring because S is multiplicatively closed.
The proof of V^n=(0) is by induction over \dim_k V. If \dim_k V=1, then V=ks for some s \in S and so V^n=ks^n=(0).
Suppose now that \dim_k V > 1. Let s \in S and define the map f: V \to Vs by f(v)=vs, for all v \in V. Clearly f is an onto k-linear map. Also, \ker f \ne (0) because if s=0, then \ker V=V and if s \ne 0, then 0 \ne s^{m-1} \in \ker f, where m is the smallest positive integer such that s^m=0. So, as k-vector spaces, V/\ker f \cong Vs and hence \dim_k Vs=\dim_k V-\dim_k \ker f < \dim_k V. Therefore since Vs is the k-vector space generated by Ss \subseteq S and Ss is a nil multiplicatively closed subset of M_n(k), because S is so, we have (Vs)^n=(0), by our induction hypothesis. So we have shown that (Vs)^n=0 for all s \in S. Now, V=\sum_{j=1}^m Vs_j for some positive integer m and so s_j \in S, because \dim_k V < \infty. Each Vs_j is clearly a left ideal of V and we just showed that they are all nilpotent. Thus, by Lemma 1, V is nilpotent and so, by Lemma 2, V^n=(0). \ \Box

The Theorem in fact holds in any finite-dimensional k-algebra not just M_n(k). To prove that, we need the following simple yet important lemma.

Lemma 3 (Cayley). Every n-dimensional k-algebra with identity R can be embedded into M_n(k).

Proof. Since \dim_k R=n, the ring of matrices M_n(k) is just \text{End}_k(R), the ring of k-linear maps R \to R. Define the map f: R \to \text{End}_k(R) by f(r)(x)=rx for all r,x \in R, and see that f is an injective ring homomorphism. \ \Box

Corollary. Let R be an n-dimensional k-algebra with identity, and let S be a nil, multiplicatively closed subset of R. Then S^n=(0).

Proof. By Lemma 3, S is a subset of M_n(k) and so, by the Theorem, S^n=(0). \ \Box

Note. The reference for this post is Chapter 1, Section 3, of the book Polynomial Identities in Ring Theory by Louis Rowen.

Exercise. Try to fully understand the proof of Lemma 3.