The ring Z/pZ[Z/pZ] (p is a prime) is not a semi-simple ring.

Question

A ring A is semi-simple if and only if it is Artinian and radical(intersection of all of its maximal ideal) of A is zero. As the ring Z/pZ[Z/pZ] is finite,so it is Artinian.So now it is enough to show that radical of Z/pZ[Z/pZ] is non-zero. Any help would be appreciated.

Answer 1

I think(?) you're talking about $(\Bbb Z/p\Bbb Z)[G]$ where $G$ is the cyclic group of order $p$.

As you commented, it's surely an Artinian ring, but it's also commutative. Commutative semisimple rings have to be finite products of fields, and those don't have any nilpotent elements.

It's easy to check that $x=\sum_{g\in G} g$ is a nonzero nilpotent element. (It squares to zero.)

You might be interested in looking up Maschke's theorem which tells us when group rings over finite rings are semisimple.