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This question may be too vague, so feel free to specialize to particular examples.

Given a morphism of schemes $f:X\to Y$, I want to know what conditions one can impose on $f,X$ or $Y$ such that a generic point of $X$ will map to a generic point of $Y$. For example, if $X$ and $Y$ are irreducible etc.

Tian An
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2 Answers2

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Let $f:X\rightarrow Y$ be a map between irreducible schemes with generic points $\eta$ and $\theta$ respectively. It's possible to prove that $f(\eta)=\theta$ is equivalent to $f$ being dominant without assuming that the schemes are reduced. Here is the proof:

If $f(\eta)=\theta$ we have $\overline{f(X)}\supseteq \overline{f(\eta)}=\bar{\theta}=Y$. On the other hand, if $f$ is dominant then, as $f(\overline{A})\subseteq \overline {f(A)}$ for every continuous function and every set $A$, we have $$f(X)=f(\bar{\eta})\subseteq \overline{f(\eta)}.$$ So $f(\eta)$ is a point whose closure is $Y$, hence $f(\eta)=\theta$.

nowhere dense
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  • What is the definition of dominant? Is it $\overline{f(X)} = Y$? – Johnny T. Jul 17 '20 at 09:00
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    yes, it is just that – nowhere dense Jul 17 '20 at 11:41
  • I have a follow-up question. This same statement should hold in the case that $f$ is not a morphism $X \to Y$ but a rational map $X \to Y$ (i.e., exercise 7.5.A of Vakil's Foundations of AG: a rational map $\pi: X \to Y$ of irr. schemes is dominant iff the generic point is sent to the generic point). In this case, you only have some morphism $f: U \to Y$, for $U$ a dense open subset. given this, is it valid to say that $f(U) \subseteq f(\overline{\eta}) \subseteq \overline{f(\eta)}$? can we take $\overline{\eta}\subseteq U$ as the closure of $\eta$ in the subspace topology $U \subseteq X$ ? – pedroelpanda Nov 02 '22 at 23:10
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Only irreducible sets have a unique generic point. Since you mention 'the' generic point I will assume $X,Y$ are irreducible from now on. Let $\eta$ be the generic point of $Y$ and $\mu$ of $X$. Then $\eta$ is contained in each affine open of $Y$ and the same goes for $\mu$ in $X$. So we can restrict our attention to some affines $U\subset X$ and $V \subset Y$. The map becomes

$$f: U = \mbox{Spec } R \rightarrow \mbox{Spec } S = V,$$ $$\varphi: S \rightarrow R$$

and your question reduces to $\varphi^{-1}(\sqrt{(0)}) = \sqrt{(0)}$. Being irreducible gives that this nilradicals are prime. If you further assume that the schemes are reduced, then the nilradicals will be zero. Then we demand injectivity of $\varphi$. This must hold for every open affine $V$ and $U$, which is equivalent to the sheaf component $f^{\#}$ being injective or $f$ being dominant, i.e. having dense image.

bbnkttp
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    It seems to me that you are assuming $X$ and $Y$ are not merely irreducible but also integral? – Zhen Lin Feb 11 '15 at 20:31
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    This injectivity has a more geometric interpretation: the morphism $X \to Y$ should be dominant, i.e. have dense image. – tracing Feb 12 '15 at 02:16
  • @ZhenLin: You are right. Integrality comes in when I assume that the generic point is the zero ideal in the affine opens. Is this the only place?

    I only knew of the implication injective -> dominant. That is why I avoided using this, but after giving it a second thought the converse is also true. Thanks!

     – bbnkttp Feb 12 '15 at 13:45