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Let $\phi:X\to Y$ be a morphism between irreducible quasiprojective varieties. If $\phi$ has a dense image in $Y$ can we conclude that its image has an interior? It really feels like it, but I couldn't show it. Maybe there is a counterexample? I guess for specific type of quasi-projective varieties resp. morphisms the image is a constructible set. A dense constructible set must surely have an interior, no?

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  • $\begingroup$ You mean interior in the sense of "union of all open subsets of $Y$ which contain the image"? This exists by definition, but I don't know how useful it is. $\endgroup$ May 26 at 5:49
  • $\begingroup$ No, I want to know (equivalently) that there is a nonempty Zariki open set that is contained in the image. Obviously if the image has an interior, then it is dense but is the reverse also true. I have sufficient example of sets that are dense but have no interior in the Zariski topology, but not one for which it is also an image of a morphism. $\endgroup$
    – quantum
    May 26 at 6:05
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    $\begingroup$ @quantum By Chevalley's theorem $\phi(X)$ always contains a non-empty open subset of $\overline{\phi(X)}$ -- therefore if $\phi(X)$ is dense then $\phi(X)$ has non-empty interior. $\endgroup$ May 26 at 6:43

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The result you are asking for is essentially equivalent* to Chevalley's theorem on the constructibility of the image of a morphism between Noetherian schemes.

More precisely, one may make the following statement (the name is ad hoc):

Theorem (weak Chevalley's theorem, [GW, Theorem 10.19]): Let $f\colon X\to Y$ be a dominant (i.e. having dense image) finite type morphism of Noetherian schemes. Then, $f(X)$ contains a dense open subset of $Y$.

Of course, for $X$ and $Y$ irreducible quasi-projective varieties this reduces to your precise question.

You should then rest at ease knowing:

a) your intuition is good to suspect this is true,

b) you couldn't prove it (this is a semi-tricky result).

References:

[GW] Görtz, U. and Wedhorn, T., 2010. Algebraic Geometry I: Schemes. Vieweg+ Teubner.

*I mean that many proofs of Chevalley's theorem prove the claim you're asking about as a lemma, and then deduce Chevalley's therorem fairly easily from this lemma.

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  • $\begingroup$ Thank you Alex. It is indeed reassuring if an expert tells me to "rest at ease" giving the reasons you wrote. Yes, now I remember solving Chevalley Theorem in Hartshorne (exercise II.3.19 ). Thanks for the reference indeed! $\endgroup$
    – quantum
    May 26 at 8:51

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