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Let $R$ be a regular local ring and $n>0$ be an integer. Consider the matrix ring $A:=M_n(R)$. I am wondering if $\text{Hom}_R(A,R)$ is a projective $A$-module? By Lemma 2.15 it is projective as a left $A$-module if and only if it is projective as a right $A$-module so we do not have to worry about sides.

Please help.

uno
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    Just curious (and my non-commutative algebra is not great), do you know how to do it when $R$ is a field? Is it simply that in this case $A$ is a simple ring and therefore every $A$-module is projective? – Aphelli Jun 25 '25 at 23:30
  • @Aphelli: you're basically on point with the field case. Any matrix ring over a division ring (e.g. a field) is a semisimple ring, and every module over semisimple ring is both projective and injective. P.S. My non-commutative algebra isn't great either :( – uno Jun 26 '25 at 00:10

2 Answers2

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This is true for every commutative ring $R$.

Let $A= M_n(R)$. For an $R$-module $U$, let $U^\ast$ denote $\operatorname{Hom}_R(U,R)$.

Let $V=R^n$, considered as an $R$-module, so we can identify $\operatorname{End}_R(V)$ with $A$, and $V$ naturally becomes an $A$-$R$-bimodule. Then $$A\cong V\otimes_RV^\ast\cong\operatorname{Hom}_R(V,V)$$ as $A$-bimodules, where the left and right $A$-module structures are induced by the left action of $A$ on $V$.

So we have isomorphisms of $A$-bimodules $$\operatorname{Hom}_R(A,R)\cong\operatorname{Hom}_R(V\otimes_RV^\ast,R) \cong\operatorname{Hom}_R(V,V)\cong A,$$ since $V^{\ast\ast}\cong V$.

Jeremy Rickard
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I can't post a comment, so I will delete this if it is not useful or it is wrong.

Being projective is preserved under Morita equivalence (as being projective is defined via the morphisms of the category), so as $R$-$\operatorname{Mod}$ is equivalent to $M_n(R)$-$\operatorname{Mod}$, it is equivalent to ask if $\operatorname{Hom}_R(A,R)$ is a projective $R$-module. Then if $A$ is a projective $R$-module, $\operatorname{Hom}_R(\operatorname{Hom}_R(A,R),R) \cong A$ by this post, and so $\operatorname{Hom}_R(\operatorname{Hom}_R(A,R),-)$ would be an exact functor.

Now, $M_n(R)$ is not a simple module (as $M_n(\mathfrak{m})$ is a nontrivial $R$-submodule) so we can't conclude the same way as we could if $R = k$ is a field, but we can consider $M_n(R)\otimes_R R/\mathfrak{m} \cong M_n(R/\mathfrak{m})$. As $\oplus$ and $\otimes$ commute it is easy to see that if $M_n(R)$ is $R$-projective then $M_n(R/\mathfrak{m})$ is $R/\mathfrak{m}$-projective, but the converse is not always true (the usual $R = \mathbb{Z}$, $S = \mathbb{Q}$ and $M = \mathbb{Z}/2\mathbb{Z}$ where $M\otimes_{\mathbb{Z}}\mathbb{Q}$ shows why). The converse is true if $M$ were faithfully flat, so if $A$ is a faithfully-flat $R$-module the result holds. But my understanding of the non-commutative case is too limited to conclude. I hope though that if this chain of reasoning is correct that it may be helpful in answering the question.

Chiarandini
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