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Let $f: X \rightarrow Y$ be a morphism of schemes of finite type. $X$ is reduced, $Y$ is an integral scheme with generic point $\eta$ and $k(\eta)$, the residue field of $\eta$, has characteristic $0$. Let $W$ be a closed subset of $X$, viewed as a reduced closed subscheme of $X$. In an article I am reading it then states "Since $k(\eta)$ is a field of characteristic zero and $X_{\eta}$ is reduced, we have $\dim W_{\eta} < \dim X_{\eta}$." I was wondering how this follows. Any explanation would be appreciated. Thank you.

PS Here $W$ is a proper closed subset of $X$

PPS We assume $X_{\eta} \not = \emptyset$

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    $\begingroup$ Yea I'm confused too. What if we take $W = X$? $\endgroup$ Aug 5, 2019 at 18:17
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    $\begingroup$ If one takes $W$ to be a proper closed subset of $X$, then the reason this should hold is that the condition "$W$ intersect a fiber is the whole fiber" is a closed subset of $Y$, which we may avoid when calculating at the generic point. $\endgroup$
    – KReiser
    Aug 5, 2019 at 19:47
  • $\begingroup$ @MinseonShin let me fix that.. thank u. $\endgroup$
    – Johnny T.
    Aug 5, 2019 at 20:16
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    $\begingroup$ There's yet another problem if the morphism is not dominant, i.e., if $X_\eta=\emptyset$. $\endgroup$
    – Ben
    Aug 8, 2019 at 21:13
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    $\begingroup$ Should $X$ be irreducible (or maybe $W$ does not contain the generic point of any irreducible component of $X$)? Otherwise take $Y = \operatorname{Spec} k$, let $X = \operatorname{Spec} k[x,y]/(xy)$ be the union of the two axes in the plane, and let $W = \operatorname{Spec} k[x,y]/(y)$ be the $x$-axis. $\endgroup$ Aug 8, 2019 at 21:57

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As pointed out in the comments, things can get a little strange when $X$ is reducible or $f$ isn't dominant. We'll treat the case where $X$ is irreducible and $f$ is dominant while trusting the reader to make the correct generalization of this to the case of $X$ reducible (treat each component $X_{\alpha}$ individually via the composite of the closed immersion $X_\alpha\to X$ and $f$, etc).


Here's a few results we'll need:

Theorem (Generic flatness; EGA IV2 Theoreme 6.9.1): Let $Y$ be a locally noetherian integral scheme with $f:X\to Y$ a finite type morphism and $\mathcal{F}$ a coherent $\mathcal{O}_X$ module. Then there's a nonempty open subscheme $U\subset Y$ so that $\mathcal{F}$ restricted to $f^{-1}(U)$ is flat over $U$.

Lemma (EGA IV2 Corollaire 6.1.2): Let $X,Y$ be locally noetherian schemes with $f:X\to Y$ flat. Then for all $x\in X$, we have $\dim_x X = \dim_{f(x)} Y + \dim_x X_{f(x)}$.

Fact: If $X$ is a scheme, $\dim X = \sup_{x\in X} \dim_x X$.

Fact: For an irreducible scheme $X$, the local dimension at any point is equal to the global dimension of $X$.

In addition to the hypotheses in the problem statement, we assume $X$ irreducible, $f$ dominant, and $W$ a proper closed subscheme.

By two applications of generic flatness (to $\mathcal{F}=\mathcal{O}_X$ and $\mathcal{F}=\mathcal{O}_W$), we may assume that both $f:X\to Y$ and $f|_W:W\to Y$ are flat by replacing $Y$ with a slightly smaller open subscheme which doesn't affect our calculations at the generic point*. Let $n=\dim X$ and $m=\dim Y$. If there exists a $w\in W$ so that $\dim_w W_{f(w)} \geq n-m$, then we have $\dim_w W \geq n-m+m = n$, which contradicts the fact that $W$ is a proper closed subscheme. So for all $w\in W$, $\dim_w W_{f(w)} < n-m$, so in particular $\dim W_\eta < n-m = \dim X_\eta$ and we're done.

If you're looking for more intuition about what should happen with respect to fibers of a morphism, Stacks has a good section on it here.

*: If this makes $W=\emptyset$, we're already done and we can stop right here - h/t Ben in the comments pointing out I didn't write this down.

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    $\begingroup$ Actually, for this argument to work (the dimension of $W$ has to stay the same upon replacing $Y$ with a suitable open subset), we have to assume that $f|_W$ is dominant as well, but that’s no serious restriction: Otherwise, $W_\eta=\emptyset$ and there is nothing to show. $\endgroup$
    – Ben
    Aug 12, 2019 at 12:14
  • $\begingroup$ Thank you very much!! $\endgroup$
    – Johnny T.
    Aug 12, 2019 at 19:01
  • $\begingroup$ @Ben correct. I definitely remember thinking this as I wrote the post but then didn't write it down. $\endgroup$
    – KReiser
    Aug 12, 2019 at 19:29
  • $\begingroup$ @KReiser In the paper they somehow use the fact that $k(\eta)$ has characteristic $0$. I guess it's actually not necessary? Would you happen to have any idea what they may be talking about? (or is the argument above using this fact somewhere?) $\endgroup$
    – Johnny T.
    Aug 16, 2019 at 15:35
  • $\begingroup$ I'm not using characteristic zero anywhere - generic flatness is true in all characteristics. I'm not sure what they would need characteristic zero for, but maybe I'm too settled on this argument and not thinking well about other ones. $\endgroup$
    – KReiser
    Aug 16, 2019 at 18:12

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