# Base change of trace for Gorenstein or Cohen-Macaulay morphisms

This is basically a question of functoriality for base change of CM morphisms.

EDIT: $\text{ }$ Brian Conrad sent me an email explaining the that this is indeed true, and follows from his book. I'll explain this in my case a little later. First I will state the question.

Suppose that $f : X \to V$ is an equidimensional (dimension $d$) finite type (reduced, if it helps) Cohen-Macaulay morphism (flat with Cohen-Macaulay fibers). I'm also happy to assume that $V$ is integral, excellent and has a dualizing complex. Additionally suppose that we have $f' : X' \to V$ another equidimensional (dimension $d$) finite-type (reduced) Cohen-Macaulay morphism that factors through $f$ as below.

$$f' : X' \xrightarrow{\phi} X \to V.$$

Further suppose that $\phi$ is finite (although the question could be asked more generally for proper $\phi$, I'll phrase it for finite $\phi$). If it helps at any point, please feel free to assume that $f$ and $f'$ are Gorenstein morphisms.

Recall that $\omega_{f}[d] = f^! \mathcal{O}_V$ and that by Brian Conrad's book [LINK: Google books] we know that both $\omega_f$ and $\omega_{f'}$ are compatible with base change.

I'd like to conclude that the following natural map is also compatible with base change:

$$\phi_* \omega_{f'} \cong R \mathcal{H}om_{O_X}(\phi_* O_{X'}, \omega_f ) \cong \mathcal{H}om_{O_X}(\phi_* O_{X'}, \omega_f ) \to \omega_f$$.
The map can be interpreted as evaluation at 1.

In other words, I'd like to know that the trace map of $\phi$ is compatible with base change. Furthermore, it would be even good enough to prove this in the $f, f'$ Gorenstein morphisms case.

One way to do this would be as follows. If $g : T \to V$ is any other morphism and $f_T : X_T \to T$ and $f_T': X_T' \to T$ are the base changes and $g_X : X_T \to X$ is the projection, is it true that the natural map (denoted [*] below) $$g_X^* \phi_* \omega_{f'} \cong g_X^* \mathcal{H}om_{O_X}(\phi_* O_{X'}, \omega_f ) \to \mathcal{H}om_{O_{X_T}}(\phi_* O_{X_T'}, \omega_{f_T} ) \cong (\phi_T)_* \omega_{f_T'}$$ between abstractly isomorphic sheaves is an isomorphism?

Edit: Brian Conrad pointed out to me that this is already a special case of what is in his Theorem 3.6.1 (assuming I understand everything right). Essentially the point is that in my case everything is finite type, which makes it all much easier.

A short answer, if you are willing to upgrade your discussion to a derived category situation. The natural base-change isomorphism of duality $${g'}^* f^! \cong {f'}^! g^*$$ holds under the hypothesis of a "Tor-independent" square. In other words, you may not have $g$ flat, but if $f$ is, it holds. This is the case always if you work with varieties over a field and the map $f$ is the structural morphism. All this is explained in Lipman's Notes on derived functors and Grothendieck duality in SLN 1960. See Theorem 4.4.1 and its corollaries.
• Hi Leo, Thanks, I'm more or less willing to upgrade to derived categories (since the things I'm pulling back actually have no interesting cohomology, even after upper shriek), but I thought I'd need some properness for $f$ as well there? This led us to Sasty's paper (which does away with the properness). $$\text{ }$$ But then I was also worried that I actually needed something a little more (ie, factoring that isomorphism through something) which seemed to require unraveling that isomorphism, and I was hoping that there was an easier way. Brian then pointed out that indeed there was! – Karl Schwede Feb 7 '13 at 14:45
• Roughly, everything works under the hypothesis that $f$ is proper. But then, using the localization property of "upper shriek" (Verdier: Base change for twisted inverse image of coherent sheaves) by Nagata localization it works on a separated situation. Of course, Sastry has the most general factorization theorem in this setting. – Leo Alonso Feb 7 '13 at 15:55
• Hi Leo, I guess I was a little worried about preserving the flatness of $f$ when compactifying (ie, can I compactify while keeping $f$ flat?). Is that known? Or is there some other way to do this? – Karl Schwede Feb 7 '13 at 18:40
• @Leo, Karl: Let $f:X\rightarrow Y$ be a finite surjection between 2-dimensional normal local noetherian schemes, with respective closed points $x$ and $y$. Then $U=X-x$ is finite flat over $V=Y-y$ since $U$ and $V$ are Dedekind, so $U\rightarrow Y$ is flat. I claim there is no proper flat $f':X'\rightarrow Y$ containing $U$ as dense open if $f$ isn't flat. Suppose $f'$ exists, so it is finite and ${f'}^{-1}(V)=U$. The normalization of $X'$ is clearly $X$. Clearly $X'$ is R$_1$, and it inherits $S_2$ from $Y$, so $X'$ is normal (i.e., $X'=X$) and hence $X$ is $Y$-flat. QED – user30379 Feb 15 '13 at 3:40