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?)?


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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    $\begingroup$ 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? $\endgroup$ – tomasz Oct 22 '12 at 21:45
  • $\begingroup$ 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. $\endgroup$ – Georges Elencwajg Jan 10 '13 at 21:42

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