Simple K[G] module has dimension 1 over K Let G be a finite abelian group, K an algebraically closed field, K[G] the group algebra of G over K and M a K[G] module. I would like to show that if M is simple, it has dimension 1 over K.
I think that I can interpret this as showing that M is isomorphic to K as a K[G] module. You could interpret K itself as a module over K[G]. Take as addition the usual field addition and as scalar multiplication by $x \in K[G], x = \sum_{g \in G} a_g$ the multiplication by $a_e$, where e is the group identity.
Now, if I can show that there is a module homomorphism between K and M which is not zero than $K \simeq M$ by Schur's Lemma.
Is this a helpful interpretation? How would a homomorphism look like?
 A: If you could prove that $M$ was isomorphic to $K$ as a $K[G]$-module, you would have the result you're looking for, but I think even this is too strong. Instead, try this:
For $h\in G$, let $\phi_h:M\to M$ be the linear transformation given by $\phi_h(m)=hm$ for all $m\in M$. Since $K$ is algebraically closed, $\phi_h$ must have an eigenvalue $\lambda\in K$. Consider the linear transformation $\phi_h-\lambda I$, where $I$ is the identity on $M$. Since $G$ is Abelian, 
\begin{align*}
(\phi_h-\lambda I)gm&=\phi_h (gm)-\lambda gm=hgm-\lambda g m = ghm-\lambda g m \\&= g\phi_h(m)-\lambda g m =g(\phi_h(m)-\lambda m)=g(\phi_h-\lambda I)m,
\end{align*}
so $\phi_h-\lambda I$ is a $K[G]$-module homomorphism, but, since $\lambda$ is an eigenvalue of $\phi_h$, it is not invertible, so by Schur's lemma it must be zero. It follows that $\phi_h=\lambda I$, or, in other words, $hm=\lambda m$ for every $m\in M$. Since this is true for every $h\in G$, it follows that every one-dimensional subspace of $M$ is a $K[G]$-submodule. This would contradict the assumption that $M$ is simple module unless $M$ were itself one-dimensional.
A: By Maschke's theorem, (in the special case of a commutative semisimple ring over an algebraically closed field,) $K[G]=R$ is isomorphic to a finite direct product of copies of $K$. Each simple $R$ module is isomorphic to $R/M$ for some maximal ideal $M$ of $R$, and of course the maximal ideals of $R$ are just those sets of all elements which are zero on a particular fixed position. 
This clearly makes all simple modules isomorphic to $K$ as $K$ modules, but they are pairwise nonisomorphic as $K[G]$ modules. If all simple modules were isomorphic to $K$ as $K[G]$ modules, then $G$ would only have one element. 
