8
$\begingroup$

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.

$\endgroup$
0

3 Answers 3

9
$\begingroup$

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.

$\endgroup$
8
$\begingroup$

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.

$\endgroup$
1
$\begingroup$

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.

$\endgroup$
5
  • $\begingroup$ I don't understand the first line. What is $\mathrm{Mod}-R$? And what do you mean by a finitely generated projective generator? $\endgroup$
    – math.n00b
    Dec 10, 2016 at 17:18
  • $\begingroup$ @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 $P^n\to R$. $\endgroup$
    – egreg
    Dec 10, 2016 at 17:22
  • $\begingroup$ 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! $\endgroup$ Dec 11, 2016 at 2:48
  • $\begingroup$ @MarianoSuárez-Álvarez As you see from the proof, being a generator is essential for the module being projective over its endomorphism ring. $\endgroup$
    – egreg
    Dec 11, 2016 at 9:21
  • $\begingroup$ Yes, I understand that. But that route to projectiveness seems to be a local maximum with respect to length and complexity! $\endgroup$ Dec 11, 2016 at 9:34

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.