Lemma 110.25.1. There exists a morphism $f : X \to Y$ of finite presentation between affine schemes and a locally closed subset $T$ of $X$ such that $f(T)$ is not a finite union of locally closed subsets of $Y$.

## 110.25 Images of locally closed subsets

Chevalley's theorem says that the image of a constructible set by a finitely presented morphism of affine schemes is constructible, see Algebra, Theorem 10.29.10 and Morphisms, Section 29.22. We will see the same thing does not hold for images of locally closed subsets.

Let $k$ be a field of characteristic $0$. Consider the projection morphism

This is a morphism of finite presentation. Let $Z$ be the closed subset of $X$ defined by

Let $U = \bigcup _{j \geq 1} U_ j$ be the open of $X$ defined by

Then we have

We claim that $B = f(Z \cap U) = \bigcup _{j \geq 1} f(Z \cap U_ j)$ is not a finite union of locally closed subsets of $Y$.

Proof of the claim. Suppose that $B = A_1 \cup \cdots \cup A_ m$ is a finite cover of $B$ by locally closed subsets of $Y$. We will show by induction on $n$ that $m \geq n$. The base case $n = 1$ is OK as $B$ is nonempty. Assume $n > 1$ and that the induction hypothesis holds for $n - 1$. Since the closure of $B$ is $(x_1 = 0)$, one of the $A_ i$ must contain some nonempty open subset of $(x_1 = 0)$. Then $A_ i$ must be open in $(x_1 = 0)$. But any such open subset cannot contain a point with $y_1 = 0$; indeed, for points of $B$, $y_1 = 0$ forces $x_2 = 0$, and this shows $B$ contains no neighborhood of $(x, y)$ inside $(x_1 = 0)$. Therefore, the remaining $m - 1$ elements restrict to a constructible cover of $B \cap (y_1 = 0)$. However, observe that the right shift map $x_ i \mapsto x_{i + 1}$, $y_ i \mapsto y_{i + 1}$ identifies $B$ with $B \cap (y_1 = 0)$! Thus by induction hypothesis, we see that $m - 1 \geq n - 1$ and we conclde $m \geq n$. This finishes the proof of the induction step and thereby establishes the claim.

**Proof.**
See discussion above.
$\square$

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