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The Brauer-Nesbitt theorem (well, one of them) says that if $k$ is a field and I have two semisimple representations (on finite-dimensional $k$-vector spaces) $r_1$ and $r_2$ of a $k$-algebra $A$ with the property that the char polys of $r_1(a)$ and $r_2(a)$ coincide for all $a\in A$, then the representations are isomorphic.

Is it the case that if the char poly of $r_1(a)$ divides the char poly of $r_2(a)$ for all $a\in A$, then the smaller representation is a direct summand of the bigger?

[I came up against this recently, but fortunately in the case I was considering $A$ was commutative and $k$ had characteristic zero, and I convinced myself it was surely fine in this case (base change up to an alg closure of $k$ and convince yourself that the semisimple representations kill all the nilpotent elements, so WLOG $A$ is etale and now do it by hand). If $k$ is finite then I'm still not sure which way to bet. If $A$ were a group ring and we knew only that one char poly divided the other for all elements of the group, then my gut feeling is that this isn't enough in characteristic $p$ but maybe I'm wrong. If $k$ has characteristic zero then I am betting on yes but then again I'm no algebraist.]

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3 Answers 3

up vote 4 down vote accepted

The result is true, regardless of characteristic.

Lemma: Let A be a k algebra and M a semi-simple A-module which is finite dimensional as a k-algebra. Then the image of A in $\mathrm{End}_k(M)$ is a semi-simple ring.

So, by the Artin-Wedderburn theorem, this image is a direct sum of matrix algebras over division rings.

Proof: Call this image S. Since S is finite dimensional, it is artinian. Let $M= \bigoplus V_i$ and let $t=(t_i)$ lie in the Jacobson radical of S. For each $V_i$, the condition that $V_i$ is a simple A-module implies that it is a simple S-module. So t must act trivially on $V_i$, and thus $t_i=0$. But we have proved this for all i, so $t=0$ and the Jacobson radical of S is trivial. QED.

We can now reduce to the case that A=S, and is a direct sum of division rings. Say $A = \bigoplus \mathrm{Mat}_{n_i}(\Delta_i)$. So every A-module is of the form $$\bigoplus (\Delta_i^{n_i})^{k_i}$$ for some $k_i$ and the corresponding characteristic polynomial is $$\chi(\lambda, g) = \prod \chi_i(\lambda, g_i)^{k_i}$$ where, for $h \in \mathrm{Mat}_{n_i}(\Delta_i)$, the polynomial $\chi_i(\lambda, h)$ is the characteristic polynomial of $h$ acting on $\Delta_i^{n_i}$.

A much better argument, suggested by buzzard's comment below. (I am not sure whether or not the original can be fixed.) Let $M = \bigoplus (\Delta_i^{n_i})^{k_i}$ and $N = (\Delta_i^{n_i})^{\ell_i}$. Suppose M is not a summand of N, so $k_i > \ell_i$ for some $i$. Let g be 1 on the i component and 0 everywhere else. Then the characteristic polynomials of M and N are of the form $(\lambda-1)^{n_i k_i} \lambda^{\bullet}$ and $(\lambda-1)^{n_i \ell_i} \lambda^{\bullet}$. So the former does not divide the latter.

We just need to show that the polynomials $\chi_i$, as polynomial functions on $\overline{k} \times A$, are relatively prime to each other. This is easy enough. Let $t_a$ be a basis for the $k$-linear functions on $\mathrm{Mat}_{n_i}(\Delta_i)$ and $u_b$ a basis for the $k$-linear functions on $\mathrm{Mat}_{n_j}(\Delta_j)$. Then $\chi_i(\lambda, g_i)$ is a homogenous polynomial in $\lambda$ and $t_a$; while $\chi_j$ in homogenous in $\lambda$ and $u_b$. Their GCD must be homogenous in both ways, hence, it is of the form $\lambda^m$.

But, if $\lambda | \chi_i(\lambda)$, this means that there is no $g$ which acts invertibly on $\Delta_i^{n_i}$; contradicting that the identity does so act. So the GCD is 1, and the result is true.

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"We just need to show that the polynomials chi_i are relatively prime to each other.". That's a bit vague. The chi_i(lambda.1) aren't relatively prime to each other, for example. I think that to finish you should just be taking elements of a that are 1 in one matrix group and 0 in the others, right? –  Kevin Buzzard Nov 23 '09 at 18:24
You are right, I have edited. –  David Speyer Nov 23 '09 at 18:51
The proof you give is the same proof given in the Duke paper I hint about in my own answer. Well spotted; good natural proof. –  Kevin Buzzard Nov 23 '09 at 20:26

By a weird coincidence I've found the answer to my question. I was trying to generalise some ideas of Chenevier; I was reading his Jacquet-Langlands paper---a version I'd got from his website. For other reasons I actually went to Duke's website to get the official version---and the official version has got the lemmas proved in more generality! See Proposition 3.2 of the published version for some pertinent comments on this issue...

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Not an answer to the whole question, but I think when $A$ is a group ring it's not enough to check on elements of the group, even in the $G$ commutative/$\text{char}(k)=0$ case where buzzard can prove his Brauer-Nesbitt variant. Consider the example where $G$ is an abelian group, $r_1$ is the trivial character, and $r_2$ is the sum of all the non-trivial characters (since every $g \in G$ is in the kernel of some nontrivial character).

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"every g in G is in the kernel of some non-trivial character"---not in general, I shouldn't think. But I agree that checking on G should definitely be weaker than checking on A. –  Kevin Buzzard Nov 23 '09 at 15:25
Whoops, meant to type "non-cyclic abelian group". –  D. Savitt Nov 23 '09 at 15:33
(and that's in char(k)=0, of course; if char(k)=p then take the prime-to-p part to be non-cyclic) –  D. Savitt Nov 23 '09 at 15:41
Gotcha. Yes, agreed, so even in char 0 knowing divisibility for elements of the group isn't enough. –  Kevin Buzzard Nov 23 '09 at 15:43
Thanks FC. Given that it's only number theorists, not algebraists, answering, why I don't I add this: In fact I was on the automorphic, not the Galois, side, when thinking about the above: A was a Hecke algebra and I was thinking about Chenevier-like ideas for Jacquet-Langlands. –  Kevin Buzzard Nov 23 '09 at 16:57

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