# Yet Another non-Free Finitely Generated Projective

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!

But let's keep things simple and try and find our $A$ so that $A$ is projective, but not free over $A$. The simplest example of $A = \mathbb{Z}$ doesn't work, and neither does $A = \mathbb{Z}/m$ for $m > 0$. The group $A = \mathbb{Z}/m\oplus\mathbb{Z}/m$ does work, but it's not that interesting because we already covered it, as its endomorphism ring is semisimple. However, the slight generalisation $\mathbb{Z}/m\oplus\mathbb{Z}/n$ where $n | m$ also works, and the basic idea is one I learned from a paper by A.J. Douglas and H.K. Farhat [2], while I was browsing around for information on endomorphisms of abelian groups. In that paper, the authors observe two facts that have nice, short proofs that the reader can verify:

1. If $B$ is a direct summand of $A$ as an abelian group, then $\mathrm{Hom}(B,A)$ is a direct summand of $E$ as an $E$-module. Indeed, choose a splitting $B\oplus C\cong A$, and it's easy to verify that the map $E = \mathrm{Hom}(A,A)\to \mathrm{Hom}(B,A)\oplus \mathrm{Hom}(C,A)$ induced by the factor inclusions is an $E$-module isomorphism, and not just an abelian group isomorphism.
2. If $\mathbb{Z}/m$ is a direct summand of $A$ and $mA = 0$ then $A$ itself is a direct summand of $E$. This includes $m = 0$! One again has to check that $\mathrm{Hom}(\mathbb{Z}/m,A)\to A$ given by $f\mapsto f(1)$ is an isomorphism of $E$-modules, and not just of abelian groups.

Given these two, we see that e.g. $A = \mathbb{Z}/4\oplus\mathbb{Z}/2$ is a projective $E$-module. In order to show that it's not free, we need to calculate what $E$ actually is! Since $A$ is a direct sum, the endomorphism ring is just the $2\times 2$ matrix ring of the form
$$\begin{pmatrix} \mathbb{Z}/4 & \mathbb{Z}/2\\ \mathbb{Z}/2 & \mathbb{Z}/2 \end{pmatrix}$$
Thus, any nonzero free $E$-module must have at least $32$ elements. However, $A = \mathbb{Z}/4\oplus\mathbb{Z}/2$ only has $8$ elements, so it cannot be free. In order to see that this is really a new example, we need to show that this ring is not semisimple. How do we do that?! Assume that it is semisimple. Then it must be a product of matrix rings over division rings. The division rings have to be finite, because $E$ is finite, and hence they have to be fields. So if $E$ were semisimple, it would have to be a product of matrix rings over fields, and in this direct product, we need at least one $2\times 2$ matrix ring. Now $|E| = 32$ so the only possibility for this $2\times 2$ matrix ring is $M_2(\mathbb{Z}/2)$ which has $16$ elements. On the other hand, we can't have a product of two such matrix rings isomorphic to $E$ because $E$, as an abelian group, has an element of order $4$. However, the only possibility of stuff we can tack onto $M_2(\mathbb{Z}/2)$ to make it have the same order as $E$ then are central elements, which cannot work because the center of $E$ has eight elements.

[1] Bobylev, I. V. Endoprojective dimension of modules. Siberian Mathematical Journal 16.4 (1975): 508-523.
[2] Douglas, A. J.; Farahat, H. K. The homological dimension of an Abelian group as a module over its ring of endomorphisms. Monatsh. Math. 69 1965 294-305