$\DeclareMathOperator\Tr{Tr}$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?



Is there a converse to the Brauer–Nesbitt theorem?
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**](http://dx.doi.org/10.1080/00927877408548718), Commun. Algebra 1, 491-501 (1974). [ZBL0288.16019](https://zbmath.org/?q=an:0288.16019).




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**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 $n$.


