Suppose $f:X\to Y$ is a finite morphism of varieties and $\mathcal{F}$ is a Cohen-Macaulay sheaf on $Y$. Under what conditions on $f$ is $f^*\mathcal{F}$ Cohen-Macaulay?

  • 3
    $\begingroup$ The only reasonable condition is if $f$ is flat. $\endgroup$ – Mohan Dec 15 '17 at 1:34

As far as being true for the most general situation, you need: $X,Y$ to be Cohen-Macaulay, $\dim \mathcal F_x =\dim Y_x$ ($\mathcal F_x$ is maximal Cohen-Macaulay) for all $x$ in the support of $\mathcal F$, and $f$ to have finite flat dimension. (more is true with a little extra technical assumption, you need the ``Cohen-Macaulay defect" to be constant along $f$, see the last paragraph).

Then what you need follows from the two local statements about f.g modules over a local ring $R$.

1) If M is maximal CM, and N has finite projective dimension, then $Tor_i(M,N)=0$ for all $i>0$ (M, N are Tor-independent). See: http://www.math.lsa.umich.edu/~hochster/711F06/L11.20.pdf

2) If $M,N$ are Tor-independent and $pd_RN<\infty$, we also have the depth formula: $depth(M\otimes N) + depth(R) = depth(M)+depth(N)$ See: https://www.math.unl.edu/~siyengar2/Papers/Abform.pdf

In particular, if N is CM then $M\otimes N$ is also CM of same depth. (here we use that $R$ is CM, so $depth(R) =depth(M)$).

If you drop any condition, it is not hard to find examples to show that the statement is no longer true. On the other hand, for special $X,Y,\mathcal F$, sometimes one can say a bit more. For example, if $R$ is a complete intersection, Tor-independence forces the depth formula to hold, without knowing that $N$ has finite flat dimension.

Added in response to OP's request: By looking at the depth formula in 2), the following more technical, but general statement is true: suppose $f: R\to S$ is a finite, local map of finite flat dimension, and $dim(R)-depth(R)=dim(S)-depth(S)$. Then for a maximal CM module $M$ over $R$, $M\otimes_R S$ is (maximal) CM over $S$. This cover both cases when $R,S$ are CM ( both sides of the equality is $0$), or if $f$ is flat (both dim and depth are preserved).

| cite | improve this answer | |
  • $\begingroup$ Thanks for the detailed answer! What can we say if $X$ is not CM? For example, if $f$ is flat, it seems $f^*F$ is automatically CM, without conditions on $X$ or $Y$. Can we do better? $\endgroup$ – Lucas Mason-Brown Dec 15 '17 at 13:03
  • 1
    $\begingroup$ Yes, you can, I will edit. $\endgroup$ – Hailong Dao Dec 15 '17 at 16:31

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.