[Math] Preimage of codimension one subvarieties under a dominant map

algebraic-geometry

Let $f:X\to Y$ be a dominant morphism of projective varieties over an algebraically closed field. If $Z\subset Y$ is a codimension $1$ subvariety, do the irreducible components of $f^{-1}(Z)$ necessarily have codimension $1$ in $X$?

I am asking about codimension $1$ since I know codimension $1$ usually behaves the best.

If this is true, is there an easy explanation? If this is almost true, what is the correction?

Best Answer

[This answer has been edited to discuss the general case.]

I will assume that variety means irreducible (otherwise you could work on individual irreducible components). Then $f:X \to Y$ is dominant by assumption, and has closed image since its source is projective, thus it is surjective. Thus $f^{-1}(Z) \to Z$ is also surjective.

Now, as described in e.g. this MO answer (or in a Hartshorne exercise, maybe in Section 4 of Chapter II), for the proper map $f$, the function $y \mapsto \dim f^{-1}(y)$ is upper semicontinuous, so if $z \in Z$, then the dimensions of $f^{-1}(z)$ are at least $\dim X - \dim Y$. This implies that $f^{-1}(Z)$ contains components of codimension $1$. (The intuition is just that we can add the dimension of $Z$ and of a typical fibre. One way to make this precise is to note that if every component of $f^{-1}(Z)$ were of codimension at least $2$, then since it dominates $Z$, we would see that a generic fibre would be of dimension $\dim X - \dim Y -1$, whereas we already noted that every fibre has dimension at least $\dim X - \dim Y$.)


In general you can't do better than this, because there are morphisms of $3$-folds (just to take an example) which are birational, but in which the preimage of some particular point $y \in Y$ is a curve. Then if you take $Z$ to be a generic codim'n one subvariety passing through $y$, its preimage will be the union of a codim'n subvariety of $X$ (the proper transform of $Z$) and the curve $f^{-1}(P)$. So in general you can't expect $f^{-1}(Z)$ to be equidimensional.

Note also that $f^{-1}(Z)$ can also contain multiple components of codimension $1$. (E.g. let $X \to Y$ be the blow up of a surface at a point, and let $Z$ be a curve that passes through the blown up point.)


Rereading the question, I see that the point of the question might be the equidimensionality, and so you might be interested in the counterexample involving $3$-folds. This MO question and answers gives one such example.

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