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

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.
