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Consider a morphism of quasi-projective algebraic sets $Xoversetvarphirightarrow Y$. What can we say about $varphi(X)$ in $overlinevarphi(X)$? Is $varphi(X)$ open in its closure? Does $varphi(X)$ contain an open set of $overlinevarphi(X)$?

$varphi(X)$ is not necessarily open in it's closure: consider $Bbb A^2to Bbb A^2$ by $(x,y)mapsto (x,xy)$. The image of this morphism is $Bbb A^2setminus (0,c) $, which is not open in $Bbb A^2$ but does have $Bbb A^2$ as its closure.

The chief result in this area is Chevalley's theorem on constructible sets. The following is from Hartshorne's book Algebraic Geometry, exercises II.3.18 and II.3.19:

II.3.18. Let $X$ be a Zariski topological space. A constructible subset of $X$ is a subset which belongs to the smallest family $mathfrakF$ of subsets of $X$ so that $mathfrakF$ contains all opens of $X$, is closed under finite intersections, and contains all complements of members of $mathfrakF$.

a) A subset of $X$ is locally closed if it's the intersection of an open subset with a closed subset. Show that a subset of $X$ is constructible if and only if it can be written as a finite disjoint union of locally closed subsets.

b) Show that a constructible subset of an irreducible Zariski space is dense iff it contains the generic point. Furthermore, in that case, it contains a nonempty open subset.

(snip)

II.3.19. The real importance of the notion of constructible sets derives from the following theorem of Chevalley - see Cartan and Chevalley [Geometrie Algebrique, Seminar Cartan-Chevalley, expose 7] and see also Matsumura [Commutative Algebra, Ch. 2, S6]: let $f: Xto Y$ be a morphism of finite type noetherian schemes. Then the image of any constructible set in $X$ is a constructible set in $Y$. In particular, $f(X)$ is constructible in $Y$. Prove the theorem in the following steps:

a) Reduce to showing that $f(X)$ is itself constructible, in the case where $X,Y$ are affine, integral noetherian schemes and $f$ is dominant.

b)* In that case, show that $f(X)$ contains a nonempty open subset of $Y$ by using the following result from commutative algebra: let $Asubset B$ be an inclusion of noetherian integral domains, such that $B$ is a finitely generated $A$-algebra. Then given a nonzero element $bin B$, there is a nonzero element $ain A$ with the following property: if $phi:Ato K$ is any homomorphism of $A$ to an algebraically closed field $K$ such that $phi(a)neq 0$, then $phi$ extends to a morphism $phi':Bto K$ with $phi'(b)neq 0$. [Hint: Prove this by induction on the number of generators of $B$ over $A$. For the case of one generator, prove it directly. In the application, take $b=1$.]

c) Now use noetherian induction on $Y$ to complete the proof.

(snip)

Whether or not you go and actually prove this result yourself, it's true, and the description of constructible as a finite disjoint union of locally closed should make understanding the question of $overlinevarphi(X)$ containing an open quite straightforwards.