Let $R$ be a $k-$algebra and $M,N$ two irreducible $R-$modules, isomorphic as vector spaces. If we know that for every $r\in R$ we have the same eigenvalues on $M$ and $N$ (with multiplicities) is it true that $M\simeq N$?


I have two thoughts:


1. Denoting $St(m)=\{ rm=m \mid r\in R\}$, if we have $St(m)=St(n)$ for some $m\in M, n\in N$, we could find an isomorphism that sends $m$ to $n$ and is extended by $f(rm)=rf(m)$ (every element of $M$ is $rm$ for some $r$ by irreducibility). I cannot prove the existence of such a pair though.
2. If $R$ was commutative, we could simultaneously diagonalize all the actions according to some basis and take the isomorphism of vector spaces that sends the one basis to the other. This will then be an isomorphism of modules. I cannot extend that to the general case.



Isomorphism of irreducible R-modules
If $k$ is a field and $M$ and $N$ are finite-dimensional, then the answer is yes. That is part of (one version of) the Brauer-Nesbitt theorem. Having the same eigenvalues with the same algebraic multiplicities is equivalent to having the same characteristic polynomials.


For convenience define the characteristic polynomial of a $k$-linear endomorphism $\alpha:V\to V$ as $\chi\_\alpha:=\det(1-T\alpha) \in k[T]$. The convenience is that $\chi\_\alpha$ always has constant term 1 and therefore can be considered as an element of the multiplicative group $1+T\cdot k[[T]] \subseteq k[[T]]^\times$.


The Brauer-Nesbitt theorem then states more generally that the group homomorphism $\chi: G\_0(R) \to (k[[T]]^\times)^R$ (where $G\_0(R)$ is the Grothendieck group of finite-dimensional $R$-modules) induced by $[M]\mapsto(\chi\_{r:M\to M})\_{r\in R}$ is injective.


One can prove more: For example you don't need to have the characteristic polynomials for *all* elements of $r$, a $k$-basis is sufficient. For example the classic statement of Brauer-Nesbitt is that $G\_0(kG) \to (k[[T]]^\times)^G$ is injective for finite groups $G$. This is a modular analogue of the characteristic-zero theorem that the character map $G\_0(\mathbb{C}G) \to\mathbb{C}^G$ that comes from taking traces instead of characteristic polynomials is injective.


Using that $1+Tk[[T]]$ is the additive group of the Witt-ring of $k$ and playing around with the multiplicative structure on $W(k)$ one can even show that much smaller subsets than bases are also sufficient for injectivity.


