# Pulling Back Cohen-Macaulay Sheaves

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?

• The only reasonable condition is if $f$ is flat. – 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).

• 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? – Lucas Mason-Brown Dec 15 '17 at 13:03
• Yes, you can, I will edit. – Hailong Dao Dec 15 '17 at 16:31