# Finite flat pullback of the diagonal

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 $$\Delta_X$$.

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.
• In the separable case (e.g. characteristic $0$), the multiplicity will always be $1$; I added some details. Commented Mar 19, 2023 at 14:23
• P.S. Another notation for $(f \times f)^{-1}(\Delta_Y)$ is $X \times_Y X$, so for instance that explains why $f$ is étale if and only if $\Delta_X \hookrightarrow (f\times f)^{-1}(\Delta_Y)$ is an open immersion [Tag 02GE]. Commented Mar 19, 2023 at 16:31