Let $F$ be a field, and let $A$ be a finite-dimensional associative $F$-algebra with multiplicative identity 1. If $M$ is a finite-dimensional module of $A$, define the character $\chi_M:A\rightarrow F$ given by $a\mapsto\mathrm{Trace}(a|_M)$.

If $\chi_i$ for $1\leq i\leq n$ are characters of pairwisely non-isomorphic finite-dimensional simple $A$-modules $M_i$ for $1\leq i\leq n$, are $\chi_i$ linearly independent?

It is true if $A$ is a finite group and the characteristic of $F$ does not divide the order of $A$, or if $A$ is a finite-dimensional semisimple algebra with 1.

Is it true for arbitrary finite-dimensional algebra?

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    $\begingroup$ The modules have o be non-isomorphic. In the group group case, you excluded some cases of positive characteristic, so if you consider the corresponding group algebras for the excluded cases you should have your answer. $\endgroup$ – Mariano Suárez-Álvarez Dec 17 '15 at 16:04
  • $\begingroup$ @MarianoSuárez-Alvarez Thank you for your comment! I get your ideal, If $G$ is a group with its order divided by the characteristic of $F$, the statement is false with an obvious example. But what if the characteristic of $F$ is 0 and $A$ is an arbitrary algebra (not necessarily group algebra)? $\endgroup$ – Hebe Dec 17 '15 at 16:44
  • $\begingroup$ Then you can look at the book by Curtis and Reiner, where this is discussed. (There must be more modern references, but still0 $\endgroup$ – Mariano Suárez-Álvarez Dec 17 '15 at 16:51
  • $\begingroup$ @MarianoSuárez-Alvarez Yes, I looked up this book one hour ago, but I could not find the content of this part. I shall be grateful if you tell me the chapter and the section. Thanks! $\endgroup$ – Hebe Dec 17 '15 at 16:59
  • $\begingroup$ I don't remember. Browse it: it's a win-win situation! $\endgroup$ – Mariano Suárez-Álvarez Dec 17 '15 at 17:15

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