TITLE: Is there a converse to the Brauer–Nesbitt theorem?
QUESTION [7 upvotes]: $\DeclareMathOperator\Tr{Tr}$Say that we have an algebra $R$ over $\mathbb{C}$. If, for two finitely generated (edit: and semisimple) $R$-modules $M, N$ we know that $\Tr_M(r)=\Tr_N(r)$ for all $r\in R$, where we consider $r$ as the linear endomorphism of the corresponding module, then we know that $M\cong N$ by the Brauer–Nesbitt theorem. 
My question is, assuming that we have specific values for traces (some homomorphism $g: R \rightarrow \mathbb{C}$ with $g(rr')=g(r'r)$ for every $r, r'\in R$), do we know that there exists some module $M$ with $Tr_M(r)=g(r)$ for every $r$? If not, can we somehow efficiently describe the functions that can appear as traces?

REPLY [8 votes]: Not always — e.g. $g(1)$ should be an integer. The desired description is given in 
Helling, H., Eine Kennzeichnung von Charakteren auf Gruppen und assoziativen Algebren, Commun. Algebra 1, 491-501 (1974). ZBL0288.16019.

Edit as requested: To a linear $g:R\to\mathbf C$ with $g(rs)=g(sr)$ Helling attaches $g_n:R^n\to\mathbf C$ by
$$
g_n(r_1,\dots,r_n)= \sum_{\sigma\in\mathfrak S_n}(-1)^\sigma\prod_{(i_1,\dots,i_l):\text{ cycle of }\sigma}g(r_{i_1}\!\cdots r_{i_l})
$$
and proves (Satz p. 496): $g$ is a character iff $g_n=0$ for some integer $n$.