Is the $k$-algebra $k[G]$ semisimple?

Question

Let $k$ be an algebraically closed field of characteristic $p$, and $G$ a cyclic group of order $p$. I was asked to show that $k[G]$ is not semisimple.

But I get confused since for any $k_1g_1 + k_2 g_2 \in k[G]$, $ (k_1g_1 + k_2 g_2)^p = k_1^p g_1^p + k_2 g_2 ^p = k_1 ^p + k_2 ^p $ since $k$ is of characteristic $p$. But isn't that then implies that every element of $k[G]$ is invertible so it's actually a field. Thus it's a simple module of itself since it does not have any non-trivial ideals. Then how could it be not semisimple?

Can anyone help me with this? Thanks a lot.

Answer 1

I don't know where to find a reference for the claim I made in the comments, so here is a proof. It feels annoyingly complicated to me; I don't know if a substantially simpler proof is possible.

Theorem (Maschke): Let $k$ be a field and let $G$ be a finite group. Then $k[G]$ is semisimple iff the characteristic of $k$ does not divide $|G|$.

Proof. First suppose that the characteristic of $k$ divides $|G|$; call it $p$. We will use the fact that a finite-dimensional $k$-algebra is semisimple iff its Jacobson radical vanishes (see e.g. this blog post). Our goal will be to exhibit an element of $k[G]$ in its Jacobson radical. Let

$$s = \sum_{g \in G} g \in k[G].$$

Note that $gs = s$ for all $g \in G$ and that $s^2 = |G| s = 0$; in particular, $s$ is nilpotent, but moreover the left ideal generated by $s$, which consists of scalar multiples of $s$, consists of nilpotent elements. It follows that $s$ is a nonzero element of the Jacobson radical of $k[G]$, and hence that $k[G]$ is not semisimple.

Now suppose that the characteristic of $k$ does not divide $|G|$; call it $p$. Recall that if $A$ is any finite-dimensional $k$-algebra, then there is a canonical linear functional on $A$ called the trace

$$\text{tr}(a) = \text{tr}(L_a), L_a : A \ni x \mapsto ax \in A.$$

That is, $\text{tr}(a)$ is the trace of $a$ acting on $A$ by left multiplication. The trace induces a symmetric bilinear form on $A$ called the trace form $\langle a, b \rangle = \text{tr}(ab)$. Recall that a bilinear form on a vector space is nondegenerate if $a \neq 0$ implies that there exists some $b$ such that $\langle a, b \rangle \neq 0$.

Lemma: If the trace form on $A$ is nondegenerate, then $A$ is semisimple.

Proof. Let $j \in J(A)$ be an element of the Jacobson radical of $A$. On a finite-dimensional $k$-algebra, $J(A)$ can be characterized as the largest nilpotent right ideal of $A$. Hence if $b \in A$, then $jb \in J(A)$ is nilpotent. By passing to an algebraic closure of $k$ and upper-triangularizing the action of $jb$ on $A$ by left multiplication we see that the eigenvalues of $jb$ are all zero, so $\text{tr}(jb) = \langle j, b \rangle = 0$. It follows that if $j \neq 0$ then the trace form is degenerate. $\Box$

The trace on $k[G]$ takes a very simple form: $\text{tr}(1) = |G|$, and $\text{tr}(g) = 0$ for all non-identity elements $g$. By assumption, $|G| \neq 0$ in $k$, so it follows that if $a = \sum a_g g, a_g \neq 0$, then

$$\text{tr}(g^{-1} a) = |G| a_g \neq 0$$

and hence the trace form is nondegenerate. The conclusion follows. $\Box$

Answer 2

Your mistake is assuming that $k_1^p + k_2^p \neq 0$.

Hint: As $G$ is cyclic, say with generator $g$, we have $$k[G] = k[g]/(g^p - 1).$$ Show that the map $$k[x]/x^p \to k[G]$$ defined by $x \mapsto g - 1$ is an isomorphism. Then it shouldn't be hard to see that $k[x]/x^p$ is not semisimple.

Answer 3

For the very reason you indicate, you can see that $k[G]$ contains a non-zero nilpotent element: let $g$ be a generator of $G$: the $(1_{G}-g)^{p} = 1_{G} - g^{p} = 0.$ In particular, if $M$ is an irreducible $k[G]$-module, then $0$ is the only eigenvalue of $1_{G}-g,$ and $1$ is the only eigenvalue of $g$ on $M$. Now as $G$ is Abelian, $M(1_{G}-g)$ is a $k[G]$-submodule of $M$, so is $\{0\}$, as $1_{G}-g$ is not invertible.Since $g$ generates $G,$ we see that $M$ must be the trivial module. Hence $k[G]$ is not semisimple, since the regular module is certainly not a direct sum of trivial modules.

Answer 4

Let us see that $R=k[G]$ is semisimple when $G$ is a finite group and $k$ is a field whose characteristic does not divide $|G|$.

Let $M$ an $R$-module and $N$ an $R$-submodule. We observe that if we can find an onto map $\psi\in Hom_R(M,N)$, then we'll get $M\simeq N\oplus K$ as $R$-modules where $K=ker\, \psi$ and so we will be done.

As $N$ is an $R$-submodule, we have an action of $G$ on the set $Hom_k(M,N)$ given by $g.\varphi=g^{-1}\varphi \,g$. This is the function which sends $x\in M$ to $g^{-1}.\varphi\,(g.x)\in N$. It is easy to see that $\varphi$ is fixed point if and only if $\varphi\in Hom_R(M,N)$.

We get a map $a:Hom_k(M,N)\to Hom_R(M,N)$ given by $a(\varphi)=\sum_{g\in G}\,g.\varphi$.

As char $k$ does not divide $|G|$, then $\alpha=|G|.1\neq 0$ where $1$ denotes the unity of $k$. So we get a map $b:Hom_R(M,N)\to Hom_R(M,N)$ given by multiplication by $\alpha^{-1}$.

Let $c:Hom_k(M,N)\to Hom_R(M,N)$ be the map $b\circ a$.

We can find a $k$-vector subspace $P$ of $M$ such that $M=N\oplus P$ as $k$-vector spaces. Let $\varphi:M\to N$ be the projection to $N$ along $P$. Then $\varphi(x)=x$ for each $x\in N$.

So for any $x\in N$ we get that $a(\varphi)(x)=\alpha x$ and so $c(\varphi)(x)=x$. Thus, we have obtained an onto map $\psi\in Hom_R(M,N)$.