There are many ways to define the propery of semisimple for a ring $R$. My favourite is the "left global dimension zero approach": a ring $R$ is left semisimple if every left $R$-module is projective, which is just the same thing as saying that every left $R$-module is injective. In particular, ideals are direct summands, and an easy application of Zorn's lemma shows that $R$ can be written as a direct product of minimal left ideals, which is actually a finite sum because $R$ contains $1$.
An attack of Schur's lemma yields the famous Wedderburn-Artin theorem: a ring $R$ is semisimple if and only if it is the finite direct product of matrix rings over division rings.
Since $R$ can be written as a finite direct product of minimal left ideals, we see that $R$ must be Noetherian and Artinian. Is the converse true?
Of course not! Here is a minimal counterexample: $\Z/4$. This ring cannot be semisimple. Indeed if it were, by the Wedderburn-Artin theorem, it would be a direct product of fields since it is commutative. It is not a field so it is not $\F_4$, and the only other possibility is $\Z/2\times\Z/2$, which it is also not isomorphic to since $\Z/4$ is cyclic.
We don't have to appeal to the Wedderburn-Artin theorem however: the reduction map $\Z/4\to\Z/2$ makes $\Z/2$ into a $\Z/4$-module. If $\Z/4$ were semisimple, then $\Z/2$ would be a projective $\Z/4$-module, and hence at the very least as abelian groups, $\Z/2$ would be a direct summand of $\Z/4$, which is also nonsensical.
Can you think of a noncommutative example?