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.]



Version of Brauer-Nesbitt for summands
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.


