Showing that $\mathbb{Z}/2\mathbb{Z}[\mathbb{Z}/2\mathbb{Z}]$ is semi-simple

Question

Showing that $\mathbb{Z}/2\mathbb{Z}[\mathbb{Z}/2\mathbb{Z}]$ is semi-simple.

Then I denote that first $\mathbb{Z}/2\mathbb{Z}=\{0,1\}$ and the second one as $\{e,g\}$.

Then I have that $\mathbb{Z}/2\mathbb{Z}[\mathbb{Z}/2\mathbb{Z}]=\{0,e,g,e+g\}=\{0,e\}\oplus \{0,g\}\oplus \{0,e+g\}$

Which is a direct sum of semi-simple rings hence it is semi-simple.

Is the above correct?

Answer 1

$e+g$ is in the span if the first two summands, so that sum isn't direct ( and Alexander already pointed out that would make too many elements anyway.)

Yeah if I just leave the last summand off I think I am ok right?

No, because $\{0,e\}$ is not an ideal of the ring.

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Actually, Maschke's theorem tells us we should not expect this ring to be semisimple.

Since $(\sum g)^2=n(\sum g)$ where n is the order of the group, and $n=0$ In this case because of the characteristic, you can see that here $\sum g$ is a nonzero nilpotent element, and hence lies in the radical.

Answer 2

If you think about it, your direct sum decomposition can't really make sense because $\{0, e\} \oplus \{0, g \} \oplus \{0, e+g\}$ formally has $8$ elements, but you've written your ring as only having $4$ elements. Something similar to this should work though.