# Is $R^n$ projective as a $M_n(R)$-module?

It is clear that $R^n$ is not free over $M_n(R)$. But is it projective? I suspect that it should be projective because we can probably come up with a projective basis, but I'm not sure how to find the basis.

Moreover, ideally if $R^n$ were projective over $M_n(R)$, we would get a big class of projective modules that are not free which would be interesting, I guess.

Whenever $e$ is an idempotent element in a ring $S$, the left ideal $Se$ of $S$ is, when viewed as a module, projective. Indeed, one can easily check that $f=1-e$ is also idempotent and that $S=Se\oplus Sf$.

Now if $S=M_n(R)$ is a matrix ring over some other ring $R$, the elementary matrix $e=E_{1,1}$ (the one whose components are all zero except for the one in the position $(1,1)$, which is equal to $1$) is idempotent. You should check that the left ideal $Se$ is isomorphic to $R^n$ as an $S$-module.

Note that $M_n(R) \cong (R^{n})^{\oplus n}$ by considering a matrix's columns as tuples in $R^n$. (This is an isomorphism of $M_n(R)$-modules by the definition of matrix multiplication: to multiply two matrices $A$ and $B$, we apply $A$ to each of the columns of $B$.) Then $R^n \oplus (R^n)^{\oplus n-1} \cong M_n(R)$, so $R^n$ is a direct summand of a free module, hence is projective.

Yes: $R^n$ is a finitely generated projective generator of $\textrm{Mod-}R$, so it is a finitely generated projective generator as a module over its endomorphism ring, which is the ring of matrices $M_n(R)$.

This is quite easy in general. Let $P_R$ be a finitely generated projective generator of $\mathrm{Mod\text{-}}R$ and let $S=\operatorname{End}(P_R)$. Then $P$ is a left $S$-module. Let's prove it is a finitely generated projective generator.

Consider a (split) epimorphism $R^n\to P$. By applying $\operatorname{Hom}_R(-,P)$, we get the split monomorphism $$\operatorname{Hom}_R(P,P)\to\operatorname{Hom}_R(R^n,P)$$ The domain is isomorphic to $S$ as a left module, the codomain is isomorphic to $P^n$ as $S$-modules. Thus $S$ is a direct summand of ${}_SP^n$ and so ${}_SP^n$ is a generator of $S\textrm{-Mod}$, which implies ${}_SP$ is a generator as well.

Since $P_R$ is a generator, there is a split epimorphism $P^n\to R$. Then, applying $\operatorname{Hom}_R(-,P)$, we get a split monomorphism $\operatorname{Hom}_R(R,P)\to\operatorname{Hom}_R(P^n,P)$. The domain is isomorphic to ${}_SP$ and the codomain is isomorphic to ${}_SS^n$. Thus ${}_SP$ is a direct summand of ${}_SS^n$ and therefore it is finitely generated projective.

• I don't understand the first line. What is $\mathrm{Mod}-R$? And what do you mean by a finitely generated projective generator? Dec 10, 2016 at 17:18
• @math.n00b $\textrm{Mod-}R$ is the category of right $R$-modules. A generator $P$ is a module such that for every module $M\ne0$ there exists a nonzero homomorphism $P\to M$, which is equivalent to the existence of an epimorphism $P^n\to R$. Dec 10, 2016 at 17:22
• I don't understand why you brought in generatingness :-/ Showing that something is a projective generator in order to show that it is projective seems like a weird detour! Dec 11, 2016 at 2:48
• @MarianoSuárez-Álvarez As you see from the proof, being a generator is essential for the module being projective over its endomorphism ring. Dec 11, 2016 at 9:21
• Yes, I understand that. But that route to projectiveness seems to be a local maximum with respect to length and complexity! Dec 11, 2016 at 9:34