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
\[ f : X = \mathop{\mathrm{Spec}}(k[t, x_1, x_2, \ldots , y_1, y_2, \ldots ]) \longrightarrow \mathop{\mathrm{Spec}}(k[x_1, x_2, \ldots , y_1, y_2, \ldots ]) = Y \]
This is a morphism of finite presentation. Let $Z$ be the closed subset of $X$ defined by
\[ x_1(t - 1) = 0,\quad x_2(t - 1)(t - 2) = 0,\quad x_3(t - 1)(t - 2)(t - 3) = 0,\quad \ldots \]
Let $U = \bigcup _{j \geq 1} U_ j$ be the open of $X$ defined by
\[ U_ j = \text{points where }y_ j(t - 1)(t - 2) ... (t - j)\text{ is nonzero} \]
Then we have
\[ f(Z \cap U_ j) = \text{points where }x_1, \ldots , x_ j\text{ are zero and }y_ j\text{ is nonzero} \]
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. $\square$
Post a comment
Your email address will not be published. Required fields are marked.
In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).
All contributions are licensed under the GNU Free Documentation License.
Comments (0)