# Generalizing Artin's theorem on independence of characters

Artin's theorem says that for any field $K$ and any (semi) group $G$, the set of homomorphisms from $G$ into the multiplicative group $K^*$ is linearly independent over $K$.

Can this theorem be generalized into higher dimensions? That is, are there any simple restrictions to put on a set of (finite-dimensional) representations of a given (semi?) group $G$ over a fixed (algebraically closed?) field $K$ so as to assure that their characters are linearly independent? It is natural to assume that the representations are irreducible (otherwise, obviously the character of $\pi$ and $\pi\oplus \pi$ are linearly dependent, and the latter would even turn out to be $0$ if $\operatorname{char} K=2$), and in case of $K=\bf C$ and finite group $G$ I suppose irreducibility is enough by Schur's orthogonality (so I guess this is also true for algebraically closed $K$ of characteristic $0$ or large enough for a given $G$ by some model-theoretical argument).

This question arose out of curiosity about the theorem as stated in a commutative algebra course, and I have little to no idea about modular representation theory, or even any non-$\bf C$ representation theory.

Summing it all up, is there a known theorem that generalizes Artin's theorem, and if not, is there any reason that there isn't (perhaps the reason being that it is trivial from some viewpoint?)?

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Bourbaki's generalization of Artin's theorem is as follows:

Let $L/K$ be a field extension and $A$ be a $K$-algebra.
Then the set $Alg_K(A,L)$ of $K$-algebra morphisms $A\to L$ is linearly independant in the $L$-vector space $\mathcal L_{K-lin}(A,L)$ of $K$-linear maps $A\to L$
(And, yes, these $K$-linear maps $\mathcal L_{K-lin}(A,L)$ form an $L$-vector space, even though $A$ is not an $L$-vector space: this is a bit confusing!)

Artin's theorem is obtained by choosing for $A$ the group algebra $K[G]$, taking $L=K$ and remembering the isomorphism of $K$-vector spaces $$\mathcal L_{K-lin}(K[G],K)\xrightarrow \cong K^G:u\mapsto (u(g))_{g\in G}$$ sending $Alg_K(K[G],K)$ to $Hom_{groups}(G,K^*)=Char (G)$

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Isn't it simpler to say that $Alg_K(A,L)$ is linearly independent in the space $L^A$ of functions from $A$ into $L$? Or is the linear structure something different? –  tomasz Oct 22 '12 at 21:45
Dear @tomasz, what you say is indeed equivalent to what I wrote since $\mathcal L_{K−lin}(A,L)$ is an $L$- subspace of your $L$-vector space $L^A$. However I used $\mathcal L_{K−lin}(A,L)$ because it is more geometric: for example the finite-dimensional $K$-algebra $A$ is diagonalized by $L$ if and only if the set $Alg_K(A,L)$ generates the $L$-vector space $\mathcal L_{K-lin}(A,L)$. This means that the map $Spec(A) \to Spec(K)$ is étale and becomes trivial over $Spec(L)$. I didn't want to drag this into my answer but as an algebraic geometer I always have this kind of picture in mind. –  Georges Elencwajg Jan 10 '13 at 21:42