# When does ${\rm Hom}_S(R,S) \cong R$?

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.

• Are they not always isomorphic? – Anonymous Jun 4 at 23:16
• @Anonymous, they are always isomorphic as $S$-modules, but since not every finite-dimensional $k$-algebra is Frobenius they cannot be always isomorphic. – C.Niculescu Jun 5 at 10:06
• Surely you mean $r\cdot\phi(x)=\phi(xr)$. What you wrote is not a left action. – tkf Jun 5 at 12:39
• 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. – tkf Jun 5 at 13:45
• 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?"? – 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 $$\lambda \in R, \,\, a \in A$$.

• Thanks for the beautiful answer. – C.Niculescu Jun 8 at 11:39
• 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. – tkf Jun 8 at 19:43
• 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 $S$. – C.Niculescu Jun 10 at 15:25