The Brauer–Nesbitt theorem that many literatures cite is stated as follows:
(BS.1): Let $KG$ be a group algebra of a finite group $G$ over a field $K$. Let $M,N$ be $KG$-modules. Suppose [A condition that will be quoted later], then $M$ and $N$ have the same composition factors iff for each $g \in G$, the matrices $M(g), N(g)$ have the same characteristic roots (counted according to their multiplicities).
This is quoted from (30.16) of:
[1] Charles W. Curtis, Irving Reiner - Representation Theory of Finite Groups and Associative Algebras -John Wiley & Sons Inc (1962)
However, there is another Brauer–Nesbitt theorem that I met:
(BS.2) Two Galois representations $\rho, \rho^{\prime}: G_K \rightarrow GL_n(L)$ have isomorphic semisimplifications if and only $\rho(g), \rho^{\prime}(g)$ have the same characteristic polynomials for each $g \in G_K$. [Other claims omitted.]
This is quoted from (2.5) of this note, whose author said that this is (BS.1).
[2] http://wwwf.imperial.ac.uk/~tsg/Index_files/ArizonaWinterSchool2013.pdf
Question 1: Since the absolute Galois group is profinite (not necessarily finite), how can we directly apply (BS.1) to show (BS.2)?
My attempts: I have read the proof quite thoroughly and feel that the finite-dimensionality of the group algebra $KG$ is essential. The finiteness is used to show that the characters of simple $FG$-modules are linearly independent, when $FG$ is split over $F$. (I used the Wedderburn-Artin Theorem to prove the linear independence)
I have also found a "purely algebraic" result on this, which seems to be quite broad:
(BS.3): Let $k$ be any field and $A$ a $k$-algebra. Let $M, N$ be two $A$-modules which are finite-dimensional as $k$-vector spaces. If for all $a \in A$, the characteristic polynomials on $M$ and $N$ are equal, then $M$ and $N$ have the same composition factors.
We see that (BS.2) is a direct corollary of (BS.3). This is quoted from this note:
Question 2: In [3], the proof is essentially the same as the one in [1]. But in [3], the author said that "For this question we may and do assume that $A$ is a semisimple finite dimensional k-algebra". It seems that this claim is a gaint leap for me! If $A$ were a finite dimensional $k$-algebra, I know that by quotient out the Jacobson radical, $A$ could indeed be assumed as semisimple. BUT again, why can we assume that $A$ is a finite dimensional $k$-algebra?
Sorry for such a lengthy post, and sincerely hope that someone may help me out! Thank you for commenting or answering!
P.S. The condition omitted in (BS.1) in the braket "[]":
there exists an extension field $L$ of $K$ such that $L$ is a splitting field for $G$, with the property that if $V$ is a completely reducible $KG$-module, then $V^L$ is a completely reducible $LG$-module. (We may take $L=K$ if $K$ is already a splitting field.)
With the OK of the OP, I'm just going to turn my comment into an answer.
The question that he is after is answered in Theorem 5.40 of this (Waybackup).