Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

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.]

share|improve this question
add comment

3 Answers

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.

share|improve this answer
1  
"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
add comment

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...

share|improve this answer
add comment

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).

share|improve this answer
    
"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
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.