## 30.18 Chow's Lemma

In this section we prove Chow's lemma in the Noetherian case (Lemma 30.18.1). In Limits, Section 32.12 we prove some variants for the non-Noetherian case.

Lemma 30.18.1. Let $S$ be a Noetherian scheme. Let $f : X \to S$ be a separated morphism of finite type. Then there exist an $n \geq 0$ and a diagram

$\xymatrix{ X \ar[rd] & X' \ar[d] \ar[l]^\pi \ar[r] & \mathbf{P}^ n_ S \ar[dl] \\ & S & }$

where $X' \to \mathbf{P}^ n_ S$ is an immersion, and $\pi : X' \to X$ is proper and surjective. Moreover, we may arrange it such that there exists a dense open subscheme $U \subset X$ such that $\pi ^{-1}(U) \to U$ is an isomorphism.

Proof. All of the schemes we will encounter during the rest of the proof are going to be of finite type over the Noetherian scheme $S$ and hence Noetherian (see Morphisms, Lemma 29.15.6). All morphisms between them will automatically be quasi-compact, locally of finite type and quasi-separated, see Morphisms, Lemma 29.15.8 and Properties, Lemmas 28.5.4 and 28.5.8.

The scheme $X$ has only finitely many irreducible components (Properties, Lemma 28.5.7). Say $X = X_1 \cup \ldots \cup X_ r$ is the decomposition of $X$ into irreducible components. Let $\eta _ i \in X_ i$ be the generic point. For every point $x \in X$ there exists an affine open $U_ x \subset X$ which contains $x$ and each of the generic points $\eta _ i$. See Properties, Lemma 28.29.4. Since $X$ is quasi-compact, we can find a finite affine open covering $X = U_1 \cup \ldots \cup U_ m$ such that each $U_ i$ contains $\eta _1, \ldots , \eta _ r$. In particular we conclude that the open $U = U_1 \cap \ldots \cap U_ m \subset X$ is a dense open. This and the fact that the $U_ i$ are affine opens covering $X$ are all that we will use below.

Let $X^* \subset X$ be the scheme theoretic closure of $U \to X$, see Morphisms, Definition 29.6.2. Let $U_ i^* = X^* \cap U_ i$. Note that $U_ i^*$ is a closed subscheme of $U_ i$. Hence $U_ i^*$ is affine. Since $U$ is dense in $X$ the morphism $X^* \to X$ is a surjective closed immersion. It is an isomorphism over $U$. Hence we may replace $X$ by $X^*$ and $U_ i$ by $U_ i^*$ and assume that $U$ is scheme theoretically dense in $X$, see Morphisms, Definition 29.7.1.

By Morphisms, Lemma 29.39.3 we can find an immersion $j_ i : U_ i \to \mathbf{P}_ S^{n_ i}$ for each $i$. By Morphisms, Lemma 29.7.7 we can find closed subschemes $Z_ i \subset \mathbf{P}_ S^{n_ i}$ such that $j_ i : U_ i \to Z_ i$ is a scheme theoretically dense open immersion. Note that $Z_ i \to S$ is proper, see Morphisms, Lemma 29.43.5. Consider the morphism

$j = (j_1|_ U, \ldots , j_ m|_ U) : U \longrightarrow \mathbf{P}_ S^{n_1} \times _ S \ldots \times _ S \mathbf{P}_ S^{n_ m}.$

By the lemma cited above we can find a closed subscheme $Z$ of $\mathbf{P}_ S^{n_1} \times _ S \ldots \times _ S \mathbf{P}_ S^{n_ m}$ such that $j : U \to Z$ is an open immersion and such that $U$ is scheme theoretically dense in $Z$. The morphism $Z \to S$ is proper. Consider the $i$th projection

$\text{pr}_ i|_ Z : Z \longrightarrow \mathbf{P}^{n_ i}_ S.$

This morphism factors through $Z_ i$ (see Morphisms, Lemma 29.6.6). Denote $p_ i : Z \to Z_ i$ the induced morphism. This is a proper morphism, see Morphisms, Lemma 29.41.7 for example. At this point we have that $U \subset U_ i \subset Z_ i$ are scheme theoretically dense open immersions. Moreover, we can think of $Z$ as the scheme theoretic image of the “diagonal” morphism $U \to Z_1 \times _ S \ldots \times _ S Z_ m$.

Set $V_ i = p_ i^{-1}(U_ i)$. Note that $p_ i|_{V_ i} : V_ i \to U_ i$ is proper. Set $X' = V_1 \cup \ldots \cup V_ m$. By construction $X'$ has an immersion into the scheme $\mathbf{P}^{n_1}_ S \times _ S \ldots \times _ S \mathbf{P}^{n_ m}_ S$. Thus by the Segre embedding (see Constructions, Lemma 27.13.6) we see that $X'$ has an immersion into a projective space over $S$.

We claim that the morphisms $p_ i|_{V_ i}: V_ i \to U_ i$ glue to a morphism $X' \to X$. Namely, it is clear that $p_ i|_ U$ is the identity map from $U$ to $U$. Since $U \subset X'$ is scheme theoretically dense by construction, it is also scheme theoretically dense in the open subscheme $V_ i \cap V_ j$. Thus we see that $p_ i|_{V_ i \cap V_ j} = p_ j|_{V_ i \cap V_ j}$ as morphisms into the separated $S$-scheme $X$, see Morphisms, Lemma 29.7.10. We denote the resulting morphism $\pi : X' \to X$.

We claim that $\pi ^{-1}(U_ i) = V_ i$. Since $\pi |_{V_ i} = p_ i|_{V_ i}$ it follows that $V_ i \subset \pi ^{-1}(U_ i)$. Consider the diagram

$\xymatrix{ V_ i \ar[r] \ar[rd]_{p_ i|_{V_ i}} & \pi ^{-1}(U_ i) \ar[d] \\ & U_ i }$

Since $V_ i \to U_ i$ is proper we see that the image of the horizontal arrow is closed, see Morphisms, Lemma 29.41.7. Since $V_ i \subset \pi ^{-1}(U_ i)$ is scheme theoretically dense (as it contains $U$) we conclude that $V_ i = \pi ^{-1}(U_ i)$ as claimed.

This shows that $\pi ^{-1}(U_ i) \to U_ i$ is identified with the proper morphism $p_ i|_{V_ i} : V_ i \to U_ i$. Hence we see that $X$ has a finite affine covering $X = \bigcup U_ i$ such that the restriction of $\pi$ is proper on each member of the covering. Thus by Morphisms, Lemma 29.41.3 we see that $\pi$ is proper.

Finally we have to show that $\pi ^{-1}(U) = U$. To see this we argue in the same way as above using the diagram

$\xymatrix{ U \ar[r] \ar[rd] & \pi ^{-1}(U) \ar[d] \\ & U }$

and using that $\text{id}_ U : U \to U$ is proper and that $U$ is scheme theoretically dense in $\pi ^{-1}(U)$. $\square$

Remark 30.18.2. In the situation of Chow's Lemma 30.18.1:

1. The morphism $\pi$ is actually H-projective (hence projective, see Morphisms, Lemma 29.43.3) since the morphism $X' \to \mathbf{P}^ n_ S \times _ S X = \mathbf{P}^ n_ X$ is a closed immersion (use the fact that $\pi$ is proper, see Morphisms, Lemma 29.41.7).

2. We may assume that $\pi ^{-1}(U)$ is scheme theoretically dense in $X'$. Namely, we can simply replace $X'$ by the scheme theoretic closure of $\pi ^{-1}(U)$. In this case we can think of $U$ as a scheme theoretically dense open subscheme of $X'$. See Morphisms, Section 29.6.

3. If $X$ is reduced then we may choose $X'$ reduced. This is clear from (2).

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