The Stacks project

68.13 Embedding into affine space

Some technical lemmas to be used in the proof of Chow's lemma later.

Lemma 68.13.1. Let $S$ be a scheme. Let $f : U \to X$ be a morphism of algebraic spaces over $S$. Assume $U$ is an affine scheme, $f$ is locally of finite type, and $X$ quasi-separated and locally separated. Then there exists an immersion $U \to \mathbf{A}^ n_ X$ over $X$.

Proof. Say $U = \mathop{\mathrm{Spec}}(A)$. Write $A = \mathop{\mathrm{colim}}\nolimits A_ i$ as a filtered colimit of finite type $\mathbf{Z}$-subalgebras. For each $i$ the morphism $U \to U_ i = \mathop{\mathrm{Spec}}(A_ i)$ induces a morphism

\[ U \longrightarrow X \times U_ i \]

over $X$. In the limit the morphism $U \to X \times U$ is an immersion as $X$ is locally separated, see Morphisms of Spaces, Lemma 65.4.6. By Lemma 68.5.12 we see that $U \to X \times U_ i$ is an immersion for some $i$. Since $U_ i$ is isomorphic to a closed subscheme of $\mathbf{A}^ n_{\mathbf{Z}}$ the lemma follows. $\square$

Remark 68.13.2. We have seen in Examples, Section 108.23 that Lemma 68.13.1 does not hold if we drop the assumption that $X$ be locally separated. This raises the question: Does Lemma 68.13.1 hold if we drop the assumption that $X$ be quasi-separated? If you know the answer, please email stacks.project@gmail.com.

Lemma 68.13.3. Let $S$ be a scheme. Let $f : Y \to X$ be a morphism of algebraic spaces over $S$. Assume $X$ Noetherian and $f$ of finite presentation. Then there exists a dense open $V \subset Y$ and an immersion $V \to \mathbf{A}^ n_ X$.

Proof. The assumptions imply that $Y$ is Noetherian (Morphisms of Spaces, Lemma 65.28.6). Then $Y$ is quasi-separated, hence has a dense open subscheme (Properties of Spaces, Proposition 64.13.3). Thus we may assume that $Y$ is a Noetherian scheme. By removing intersections of irreducible components of $Y$ (use Topology, Lemma 5.9.2 and Properties, Lemma 28.5.5) we may assume that $Y$ is a disjoint union of irreducible Noetherian schemes. Since there is an immersion

\[ \mathbf{A}^ n_ X \amalg \mathbf{A}^ m_ X \longrightarrow \mathbf{A}^{\max (n, m) + 1}_ X \]

(details omitted) we see that it suffices to prove the result in case $Y$ is irreducible.

Assume $Y$ is an irreducible scheme. Let $T \subset |X|$ be the closure of the image of $f : Y \to X$. Note that since $|Y|$ and $|X|$ are sober topological spaces (Properties of Spaces, Lemma 64.15.1) $T$ is irreducible with a unique generic point $\xi $ which is the image of the generic point $\eta $ of $Y$. Let $\mathcal{I} \subset X$ be a quasi-coherent sheaf of ideals cutting out the reduced induced space structure on $T$ (Properties of Spaces, Definition 64.12.5). Since $\mathcal{O}_{Y, \eta }$ is an Artinian local ring we see that for some $n > 0$ we have $f^{-1}\mathcal{I}^ n \mathcal{O}_{Y, \eta } = 0$. As $f^{-1}\mathcal{I}\mathcal{O}_ Y$ is a finite type quasi-coherent ideal we conclude that $f^{-1}\mathcal{I}^ n\mathcal{O}_ V = 0$ for some nonempty open $V \subset Y$. Let $Z \subset X$ be the closed subspace cut out by $\mathcal{I}^ n$. By construction $V \to Y \to X$ factors through $Z$. Because $\mathbf{A}^ n_ Z \to \mathbf{A}^ n_ X$ is an immersion, we may replace $X$ by $Z$ and $Y$ by $V$. Hence we reach the situation where $Y$ and $X$ are irreducible and $Y \to X$ maps the generic point of $Y$ onto the generic point of $X$.

Assume $Y$ and $X$ are irreducible, $Y$ is a scheme, and $Y \to X$ maps the generic point of $Y$ onto the generic point of $X$. By Properties of Spaces, Proposition 64.13.3 $X$ has a dense open subscheme $U \subset X$. Choose a nonempty affine open $V \subset Y$ whose image in $X$ is contained in $U$. By Morphisms, Lemma 29.38.2 we may factor $V \to U$ as $V \to \mathbf{A}^ n_ U \to U$. Composing with $\mathbf{A}^ n_ U \to \mathbf{A}^ n_ X$ we obtain the desired immersion. $\square$


Comments (0)


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).

Unfortunately JavaScript is disabled in your browser, so the comment preview function will not work.

All contributions are licensed under the GNU Free Documentation License.




In order to prevent bots from posting comments, we would like you to prove that you are human. You can do this by filling in the name of the current tag in the following input field. As a reminder, this is tag 088K. Beware of the difference between the letter 'O' and the digit '0'.