6
$\begingroup$

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.

$\endgroup$
  • 3
    $\begingroup$ 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 '15 at 21:23
  • 1
    $\begingroup$ @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 '15 at 21:31
2
$\begingroup$

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 $\{1,s,t\}$.

$\endgroup$

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.