Let $S \subset R$ be two non-commutative rings and assume that $R$ is free of finite rank as an $S$-module and that $S$ is central subring of $R$. What are the (minimal) conditions such that: $${\rm Hom}_S(R,S) \cong R\text{ as $R$-modules} ? $$

Here $R$ acts on ${\rm Hom}_S(R,S)$ via $r. \phi(x)=\phi(xr)$. One may prove that ${\rm Hom}_S(R,S) \cong R$ as $S$-modules.

For example, if $S=k$ is a field and $R$ is a finite-dimensional $k$-algebra, the condition is equivalent to $R$ being a Frobenius algebra. Any references are welcome.

  • $\begingroup$ Are they not always isomorphic? $\endgroup$ – Anonymous Jun 4 at 23:16
  • $\begingroup$ @Anonymous, they are always isomorphic as $S$-modules, but since not every finite-dimensional $k$-algebra is Frobenius they cannot be always isomorphic. $\endgroup$ – C.Niculescu Jun 5 at 10:06
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    $\begingroup$ Surely you mean $r\cdot\phi(x)=\phi(xr)$. What you wrote is not a left action. $\endgroup$ – tkf Jun 5 at 12:39
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    $\begingroup$ Also, they are not always isomorphic as $S$ modules. In the case $R=\mathbb{R}[a,b]/\langle a^2,b^2,ab\rangle$ and $S=\mathbb{R}[b]/\langle b^2\rangle$ both $R$ and ${\rm Hom}_S(R,S)$ are isomorphic to $S\oplus S$ as vector spaces over $\mathbb{R}$. However as $S$ modules they are quite different. In the latter case $$(\mu_1+\mu_2b)(s_1,s_2)=((\mu_1+\mu_2b)s_1,\mu_1 s_2).$$ Hence the image of multiplication by $b$ is only 1-dimensional over $\mathbb{R}$. On the other hand, if $S$ is central in $R$ then they are isomorphic as $S$-modules, by @Anonymous now deleted answer. $\endgroup$ – tkf Jun 5 at 13:45
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    $\begingroup$ So should your question read "Let $S\subseteq R$ with $R$ a not necessarily commutative ring, and $S$ a ring contained in the centre of $R$, with $R$ free and finite dimensional as a module over $S$. Then $R$ acts on ${\rm Hom}_S(R,S)$ via $(r\cdot\phi)x=\phi(xr)$. With these definitions $R$ and ${\rm Hom}_S(R,S)$ are necesarrily isomorphic as left $S$ modules. What further conditions would guarantee they are isomorphic as $R$ modules?"? $\endgroup$ – tkf Jun 5 at 15:07

Not sure if this is what you are looking for, but certainly this is a necessary condition for your condition to hold.

Let $S$ be a central subring of a (not necessarily commutative) ring $R$, with $R$ free and finite dimensional as a module over $S$. Your condition is that $R$ is isomorphic to ${\rm Hom}_S(R,S)$ as left $R$ modules. Equivalently, there exists a map $\epsilon\colon R\to S$ such that every $S$-linear homomorphism $R \to S$ may be written in the form $\epsilon(\_a)$ for a unique $a\in R$.

A necessary condition for such an $\epsilon$ to exist is that finitely generated projective $R$-modules are injective relative to $S$. That is given an $R$-linear map of left $R$ modules $f\colon A \to M$ such that $f$ has a left inverse as a map of $S$ modules, any $R$-linear map $h\colon A \to P$ (for $P$ a finitely generated projective module) may be extended to an $R$-linear map $M \to P$.

$$ A\stackrel f\to M $$ $$h\downarrow \,\,\,\swarrow\quad$$ $$P\quad\quad$$

Proof: Suppose $\epsilon$ exists as above. It is sufficient to consider the case $P=R$, as the property of being relatively injective extends in an obvious way to (finite) direct sums and summands.

Given $m\in M$ we have an element of ${\rm Hom}_S(R,S)$ given by $$\lambda\mapsto \epsilon(hg(\lambda m))$$ where $g$ is the $S$-linear left inverse to $f$.

Thus we have $\hat h(m)\in R$ such that $$\epsilon(hg(\lambda m))=\epsilon(\lambda \hat h(m),$$ for all $\lambda\in R$. Then $\hat h$ is $R$-linear as for all $\lambda\in R$ we have $$\epsilon (\lambda \hat h(\mu m))=\epsilon(hg(\lambda\mu m))=\epsilon(\lambda\mu\hat h(m)).$$

Finally we note that $\hat hf=h$: $$\epsilon(\lambda\hat hf(a))=\epsilon(hg(\lambda f(a)))=\epsilon(hgf(\lambda a))=\epsilon(h(\lambda a))=\epsilon(\lambda h(a)),$$ for all $\lambda \in R, \,\, a \in A$.

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  • $\begingroup$ Thanks for the beautiful answer. $\endgroup$ – C.Niculescu Jun 8 at 11:39
  • $\begingroup$ Thanks. If you replace "$S$-modules" with "sets" in the definition of "injective relative to $S$" then you get the usual definition of injective. In particular when $S$ is a field "injective relative to $S$" is just the same as "injective". e.g. finitely generated projective modules over a Frobenius algebra are injective in the usual sense. However when doing homological algebra with coefficients in a ring (e.g. $\mathbb Z$), this relative result is really useful - e.g. $\mathbb Z[G]$ has your property for a finite group $G$, so the result is useful for group cohomology with integer coeffs. $\endgroup$ – tkf Jun 8 at 19:43
  • $\begingroup$ Interesting, I did not know that. Also, since last week, I have managed to dig some literature about this condition. As you mentioned in a previous past, this is very close to what people call a Frobenius extensions, in fact my condition will immediately imply that $R$ is a Frobenius extension of $S$. $\endgroup$ – C.Niculescu Jun 10 at 15:25

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