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Let $\pi : X \rightarrow Y$ be any morphism of projective varieties, $Z$ locally closed subset in $X$ and $p \in Y$ a general point. Denote by $Z_p$ the fiber $X_p=\pi^{-1}(p)$ restricted on $Z$. The goal is show that the closure $\bar{Z}_p$ is equal to $\bar{Z} \cap X_p$. This is what Harris argues at p. 141 (first course in AG):

First, we can restrict the map $\pi$ to the closure of $Z$, i.e. we can assume $X=\bar{Z}$, or in other words, that $Z$ is an open subset of $X$ (we can also assume that $X$ is irreducible). Next, we can assume the image of $\pi$ is dense, since otherwise both sides of the presumed inequality are empty. The result now follows immediately from applying the fiber dimension Theorem (11.2 in Harris, p. 60 in Shafarevich-1974) to the map $\pi$ on $X$ and on $W=X-Z$: if the dimensions of $Y,X$ and $W$ are $k,n$ and $m<n$ respectively, then the local dimension of $X_p$ is at least $n-k$ everywhere, while for general $p$ the dimension of $W_p$ must be $m-k<n-k$. It follows that $W_p$ cannot contain an irreducible component of $X_p$; thus $W_p$ lies in the closure of $Z_p$.

There are several things i don't understand about the above argument. But let me ask the first of them, since if i understand that, then i may understand the rest.

Question: Why does the local fiber dimension theorem applied on $W$ gives that $\dim W_p = m - k$? The first thing that this suggests is that $\pi(W)$ is dense and i can't see why is that.

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    $\begingroup$ I suspect that the dimension calculation for $W_p$ is implicitly assuming that $W$ dominates $Y$; otherwise, the general $W_p$ is empty and certainly does not contain a component of $X_p$. $\endgroup$ – John Brevik Aug 8 '15 at 12:14
  • $\begingroup$ @JohnBrevik: Your comment actually unlocked my understanding of all the issues that were bothering me. Thanks a lot! $\endgroup$ – Manos Aug 10 '15 at 2:27
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    $\begingroup$ Glad I could help! Thanks for the kind words :) $\endgroup$ – John Brevik Aug 10 '15 at 9:49

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