Behavior of duality under pull-back

Question

I have a technical question on commutative algebra. I am not an expert in the subject, and I would like to know if there are "typical conditions" making the following possible.

Let $\varphi:R\to S$ be a surjective ring homomorphisms (rings are commutative with identity). Let $X$ be an $S$-module, and regard it as an $R$-module as well via $\varphi$. I am interested in recovering the $S$-dual $\text{Hom}_S(X,S)$ of $X$ from the $R$-dual $\text{Hom}_R(X,R)$, in a natural way. Do you know of a way to do so, possibly under suitable supplementary hypothesis?

A naive guess that comes to mind is to try and consider $\text{Hom}_R(X,R)\otimes_R S$. How is this module related to $\text{Hom}_S(X,S)$?

I am interested in the very special case where $S$ (resp. $R$) is a order in a product $\prod K_i$ (resp. $\prod K_i\times\prod L_j$) of finitely many number fields, and where $X$ is a torsion free, finitely generated $S$-module such that $X\otimes\mathbf{Q}$ is isomorphic to $\prod K_i^{n_i}$, for some $n_i\geq 1$. Moreover both $S$ and $R$ are Gorenstein.

Answer 1

To to do this, one just has to replace the $\text{Hom}$ by an $\text{Ext}$. Let me explain. You already assumed that both $R$ and $S$ are Gorenstein, so that makes things easier.

Since $R$ and $S$ are Gorenstein, they are locally equidimensional, and so let us assume that $R$ is pure dimension $d$ and $S$ is pure dimension $e$ (by localizing if necessary). Then if $X$ is finitely generated,

$$ \begin{array}{rl} & \text{Hom}_S(X, S) \\ = & \text{Hom}_S(X, \omega_S) \\ = & h^{-e}(R\text{Hom}_S(X, \omega_S[e])) \\ = & h^{-e}(R\text{Hom}_R(X, \omega_R[d]) \\ = & h^{d-e}(R \text{Hom}_R(X, \omega_R) \\ = & \text{Ext}^{d-e}_R(X, \omega_R) \\ = & \text{Ext}^{d-e}_R(X, R). \end{array} $$ The fact that $\omega_R = R$ and $\omega_S = S$ comes from the fact that $R$ and $S$ are Gorenstein (localize further if necessary). The real content is the third equality which is a basic case of Grothendieck duality (really, it's a derived version of something like Hom-tensor adjointness, plus some formulas for the dualizing module).

Answer 2

Take $X=S$; write $S$ as the quotient of $R$ by an ideal $I$. Then $\text{Hom}_S(X,S)=S$, while $\text{Hom}_R(X,R)$ is the ideal of elements in $R$ annihilated by $I$. They have in general no relations — e.g. if $R$ is a domain and $I\neq (0)$, $\text{Hom}_R(X,R)=0$.

Answer 3

Small comment: certainly there's a natural map $\text{Hom}_R(X, R) \otimes_R S \to \text{Hom}_S(X, S)$ given by composing with $\varphi$ and extending by linearity. We can't hope for this map to be an isomorphism in general because the RHS sends colimits to limits but the LHS won't because if $\varphi$ is nontrivial then $S$ won't be flat over $R$. A typical example of what can go wrong here is to take $\varphi : \mathbb{Z} \to \mathbb{Z}/2\mathbb{Z}$ and $X = \mathbb{Z}/2\mathbb{Z}^n$.