I know free modules aren't always semisimple($\mathbb{Z}$ module over $\mathbb{Z}$ for example), but the converse seems to be true? My proof:
Let $U=S_1\bigoplus…\bigoplus S_n$ be a semisimple module over a ring, and $S_1,…,S_n$ simple submodules of $U$, and let $s_i$ be a non zero element of $S_i$ for each $i$. Each $u\in U$ can be written as a sum of elements of each $S_i$, and these are simple modules so each element is a scalar multiple of $s_i$. The set $\{s_1,…,s_n\}$ is linearly independent (sum is direct) and generates $U$, so it's a basis.
If this proof is correct, then how come $\mathbb{Z_p}$ (p prime) as a $\mathbb{Z}$ module is simple (implies semisimple) but not free?
Best Answer
This is incorrect in $\mathbb{Z}_p$, as $p \cdot 1 = 0$.