About the image of a morphism of algebraic sets.

Question

Consider a morphism of quasi-projective algebraic sets $X\overset{\varphi}{\rightarrow} Y$. What can we say about $\varphi(X)$ in $\overline{\varphi(X)}$? Is $\varphi(X)$ open in its closure? Does $\varphi(X)$ contain an open set of $\overline{\varphi(X)}$?

Answer 1

$\varphi(X)$ is not necessarily open in it's closure: consider $\Bbb A^2\to \Bbb A^2$ by $(x,y)\mapsto (x,xy)$. The image of this morphism is $\Bbb A^2\setminus \{(0,c) | c\neq 0\}$, 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 $\mathfrak{F}$ of subsets of $X$ so that $\mathfrak{F}$ contains all opens of $X$, is closed under finite intersections, and contains all complements of members of $\mathfrak{F}$.

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 $f: X\to Y$ 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 $A\subset B$ be an inclusion of noetherian integral domains, such that $B$ is a finitely generated $A$-algebra. Then given a nonzero element $b\in B$, there is a nonzero element $a\in A$ with the following property: if $\phi:A\to K$ is any homomorphism of $A$ to an algebraically closed field $K$ such that $\phi(a)\neq 0$, then $\phi$ extends to a morphism $\phi':B\to K$ 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 $\overline{\varphi(X)}$ containing an open quite straightforwards.