In the post Examples: Projective Modules that are Not Free, we saw nine examples of projective modules that are not free. On in particular was ‘the’ submodule $M = \oplus_{i=1}^\infty \mathbb{Z}$ of $\prod_{i=1}^\infty\mathbb{Z}$. Now, that’s a cool example to be sure, but the way we showed that $M$ was not free was to cite that $\prod_{i=1}^\infty\mathbb{Z}$ is uncountable. Actually, I like the argument a lot, but it’s possible to use the idea of that example and choose $M$ instead to be finite and different from all the examples in the aforementioned post. In fact, we’ll see a large class of examples that can be constructed from the ideas here.

The idea is to take an abelian group $A$ an consider $A$ as a module over its endomorphism ring $E = \mathrm{Hom}(A,A)$, where the endomorphisms are just homomorphisms $A\to A$ of abelian groups. Sometimes, $A$ can be projective over $E$. Actually, for a while it was believed that the projective dimension of $A$ over $E$ could only be $0$ or $1$, but eventually I.V. Bobylev showed in [1] that $A$ could have any projective dimension over $E$, including infinity!
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Posted by Jason Polak on 19. December 2014 · Write a comment · Categories: homological-algebra, modules · Tags:

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?