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 $X$.

## 68.13 Embedding into affine space

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

**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

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

(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. $\square$

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