Interior of the image of a morphism

Question

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?

Answer 1

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.