Finite flat pullback of the diagonal

Question

Let $X, Y$ be smooth projective connected complex varieties of the same pure dimension $d$ and $f : X\to Y$ a finite flat surjective morphism.

Let $\Delta_X$ be the closed subscheme of $X\times X$ that is the scheme-theoretic image of the diagonal morphism $X\to X\times X$, and $\Delta_Y$ the scheme-theoretic image of the diagonal morphism $Y\to Y\times Y$.

Do we have $f^{-1}\Delta_Y=\Delta_X$ scheme-theoretically? (i.e. $\Delta_Y\times_{Y\times Y}(X\times X)=\Delta_X$)

I'd expect $\Delta_X$ to be one of the irreducible components of $f^{-1}\Delta_Y$, but was wondering about an explicit example and whether one could say something about the discrepancy between $f^{-1}\Delta_Y$ and $\Delta_X$.

Answer 1

If $f$ is finite flat of degree $d$, then $f \times f \colon X \times X \to Y \times Y$ has degree $d^2$, but $\Delta_f \colon \Delta_X \to \Delta_Y$ has degree $d$. So equality cannot hold scheme-theoretically unless $f$ is an isomorphism.

A fairly explicit case is the finite étale Galois case, where $(f \times f)^{-1}(\Delta_Y)$ is the disjoint union of the graphs $\Gamma_\sigma$ of deck transformations $\sigma \colon X \to X$.

If $f$ is only étale but not Galois, then $(f\times f)^{-1}(\Delta_Y)$ will be smooth and $\Delta_X$ is still a connected component, but the other components will not map isomorphically onto $X$ under their projections.

If $f$ is generically étale (i.e. separable), then there is a dense open $U \subseteq Y$ above which the above holds, so $(f \times f)^{-1}(\Delta_Y)$ is still generically smooth (i.e. geometrically reduced). But for each component of the branch divisor $D \subseteq Y$, there will be another irreducible component of $(f \times f)^{-1}(\Delta_Y)$ intersecting $\Delta_X$ above $D$.

In the inseparable case, you get situations where $\Delta_X$ occurs with multiplicity $>1$, but that only happens in positive characteristic.