Let $B$ be a $A$-algebra which is free of finite rank as $A$-module. Let $X$ be a finitely generated projective left $B$ module. (So $X$ is also a f.g. projective $A$ module.) Are these homomorphism groups naturally isomorphic as right $B$-modules? $$\mathrm{Hom}_A(X,A)\stackrel{?}{\cong} \mathrm{Hom}_B(X,B).$$

By dimension count (e.g. in special case of fields) one can see easily that these two are really isomorphic. After a lot of work I found a long proof of natural equivalence which only works for the case of division rings with an assumption on characteristic, which doesn't give an explicit isomorphism.

So my question is: Are these two $B$-modules always naturally isomorphic? (At least for the case of a division ring $B$ over a field $A$)

I also wonder if there is a simple explicit natural morphism in general (without above restrictions on $B$ and $X$ ) from one side to the other which gives a natural isomorphism in good cases.

Edit. I simplified my proof and found an explicit natural morphism from the right side to the left. Let $f \in \mathrm{Hom}_B(X,B)$, I define the corresponding element $t_f\in \mathrm{Hom}_A(X,A)$ as follows: For every $x\in X$ define a $A$-linear morphism $T_f(x):X\to X$ by $(T_f(x))(y) := f(y)x$. Then $t_f(x):= \mathrm{tr}_A(T_f(x)).$ I can show that $t:\mathrm{Hom}_B(X,B)\to \mathrm{Hom}_A(X,A)$ is an isomorphism for separable finite algebras over fields. But the question remains unsolved in the general case.

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    $\begingroup$ They are not always isomorphic as $B$-modules, even in a very simple situation : $A$ a field, $B$ commutative, $X=B$. Then it is well-known that the $B$-module $\mathrm{Hom}_A(B,A)$ is free of rank 1 if and only if $B$ is Gorenstein. $\endgroup$
    – abx
    Feb 1, 2015 at 21:23
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    $\begingroup$ @abx I supposed that B is a free A algebra. But in general one can ask of a natural morphism from one side to the other. $\endgroup$
    – Mostafa
    Feb 1, 2015 at 21:31

1 Answer 1


To make abx's comment more explicit, consider the following example:

Let $A=k$ be a field. Let $B=k[s,t]/(s^2,t^2,st)$, which is not Gorenstein. We see that $B$ is an $A$-algebra which is free of rank $3$ as an $A$-module. Set $_BX=\!_BB$. We compute that ${\rm Hom}_B(X,B)\cong B_B$.

Now, consider the hom set ${\rm Hom}_A(X,A)$. This hom set has the structure of a $B$-module via the following action: Given $\varphi\in {\rm Hom}_A(X,A)$, $b\in B$, and $x\in X$, we have $(\varphi\cdot b):x\mapsto \varphi(bx)$.

Now, suppose by way of contradiction, that there is an isomorphism of $B$-modules $\theta:B_B\cong {\rm Hom}_A(X,A)$. Let $\varphi:=\theta(1)$. Thus ${\rm Hom}_A(X,A)=\varphi\cdot B$.

Write $\varphi(1)=a_1,\varphi(s)=a_2,\varphi(t)=a_3$ with $a_1,a_2,a_3\in k$. Given any $b_1+b_2s+b_3t\in B$ (with $b_1,b_2,b_2\in k$) then

$$(\varphi\cdot b)(s)=\varphi(bs)=\varphi(b_1s)=a_2b_1$$ and $$(\varphi\cdot b)(t)=\varphi(bt)=\varphi(b_1t)=a_3b_1.$$

Thus, if $\psi\in {\rm Hom}_A(X,A)=\varphi\cdot B$, we must have $\begin{pmatrix}\psi(s)\\ \psi(t) \end{pmatrix}\in \begin{pmatrix}a_2\\ a_3 \end{pmatrix}k$, which contradicts the fact that elements in ${\rm Hom}_A(X,A)$ can be defined arbitrarily on the set $\{1,s,t\}$.


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