2
$\begingroup$

Let $f:X\rightarrow Y$ be a morphism with connected fibers between projective varieties (not necessarily flat). Let $D\subset Y$ be an irreducible divisor. Let us look at the cycle $f^{-1}(D)\subset X$.

What can we say about the dimension of the irreducible components of $f^{-1}(D)$?

$\endgroup$

1 Answer 1

4
$\begingroup$

I don't know what you mean by "the cycle $f^{-1}(D)\subset Y$" (and not just because $f^{-1}(D)\subset X$), but for your question you don't need it to be a "cycle", so let's just assume you want the set.

The answer depends on $f:X\to Y$. For instance, if $X=X_1\cup X_2$ where $X_i$ are irreducible and closed in $X$, $Y=Y_1\cup Y_2$ where $Y_i$ are irreducible and closed in $Y$, $Y_1\cap Y_2=\{P\}$ is a single point, $f_1=f|_{X_1}: X_1\to Y_1$ is the blowing-up of $Y_1$ at $P$, $f_2=f|_{X_2}: X_2\to Y_2$ is the blowing-up of $Y_2$ at a subvariety of dimension $d$ contained in $Y_2$ and containing $P$, and then if $D\subset Y_1$ is a divisor which contains $P$, then $f^{-1}(D)$ will be a divisor on $X_1$, but will have dimension $\dim Y_2-d$ on $X_2$. So, even assuming that $Y$ is equidimensional $f^{-1}(D)$ might have different dimensional irreducible components.

So perhaps you'd want to assume that $X$ and $Y$ are irreducible, but I think even then you can get funny behaviour. Let $f:X\to Y$ be a small morphism, for example contracting a single curve on a threefold. Since your $X$ and $Y$ are projective, so is $f$ and let $H$ be a relatively very ample divisor that does not contain the entire special fiber (for simplicity let's assume that $f$ has only one non-trivial fiber) and let $D=f(H)$. Then $f^{-1}(D)=H\cup \text{special fiber}$, so again you can have irreducible components of (almost) arbitrary dimension.

One case when $f^{-1}(D)$ has only codimension $1$ components is if $D$ is a $\mathbb Q$-Cartier divisor.

$\endgroup$
2
  • $\begingroup$ Thanks a lot. Of course I was assuming $X$ irreducible. Then, in particular if $Y$ is smooth any component of $f^{-1}(D)$ has dimension $dim(X)-1$? $\endgroup$
    – user82886
    Feb 23, 2016 at 22:31
  • 3
    $\begingroup$ If $Y$ is smooth, then $D$ is Cartier in which case $f^*D$ is a divisor whose support equals to $f^{-1}(D)$. $\endgroup$ Feb 23, 2016 at 23:34

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.