The Stacks project

Lemma 76.32.1. Let $S$ be a scheme. Let $f : X \to Y$ be an integral morphism of algebraic spaces over $S$. Let $y \in |Y|$ be a point which can be represented by a closed immersion $y : \mathop{\mathrm{Spec}}(k) \to Y$. Then there exists a factorization $X \to X' \to Y$ of $f$ such that

  1. $X' \to Y$ is integral,

  2. $X \to X'$ is an isomorphism over $X' \setminus X'_ y$,

  3. $X'_ y$ has a unique point $x'$ with $\kappa (x') = k$.

Moreover, if $f$ is finite and $Y$ is locally Noetherian, then $X' \to Y$ is finite.

Proof. By Morphisms of Spaces, Lemma 67.11.2 the sheaves $f_*\mathcal{O}_ X$, $(X_ y \to Y)_*\mathcal{O}_{X_ y}$, and $y_*\mathcal{O}_{\mathop{\mathrm{Spec}}(k)}$ are quasi-coherent sheaves of $\mathcal{O}_ Y$-algebras. Consider the maps

\[ f_*\mathcal{O}_ Y \longrightarrow (X_ y \to Y)_*\mathcal{O}_{X_ y} \longleftarrow y_*\mathcal{O}_{\mathop{\mathrm{Spec}}(k)} \]

The fibre product is a quasi-coherent sheaf of $\mathcal{O}_ Y$-algebras $\mathcal{A}'$ and we can define $X' \to Y$ as the relative spectrum of $\mathcal{A}'$ over $Y$, see Morphisms, Lemma 29.11.5. This construction commutes with arbitrary change of base. In particular, it is clear that over the open subspace $|Y| \setminus \{ y\} $ the morphism $X \to X'$ is an isomorphism and over $|Y| \setminus \{ y\} $ the morphism $X' \to Y$ is integral. It remains to prove the statements in a small neighbourhood of $y$. Choose an affine scheme $V = \mathop{\mathrm{Spec}}(R)$ and an ├ętale morphism $\varphi : V \to Y$ such that $y$ is in the image of $\varphi $. Then $V_ y$ is a closed subscheme of $V$ ├ętale over $k$, whence consists of finitely many points each with residue field separable over $k$ (see Decent Spaces, Remark 68.4.1). After shrinking $V$ we may assume there is a unique closed point $v = \mathop{\mathrm{Spec}}(l) \to V$ mapping to $y$ with $l/k$ finite separable. We may write $V \times _ Y X = \mathop{\mathrm{Spec}}(C)$ with $R \to C$ an integral ring map. The stated compatibility with base change gives us that $U \times _ X Y' = \mathop{\mathrm{Spec}}(C')$ where

\[ C' = C \times _{C \otimes _ R l} l \]

Since $R \to l$ is surjective, also $C \to C \otimes _ R l$ is surjective and we see that this is a fibre product of the kind studied in More on Algebra, Situation 15.6.1 (with $A, A', B, B'$ corresponding to $C \otimes _ R l, C, l, C'$). Observe that $C'$ is an $R$-subalgebra of $C$ and hence is integral over $R$; this proves (1). Finally, More on Algebra, Lemma 15.6.2 shows that $V \times _ X Y' = \mathop{\mathrm{Spec}}(C')$ has a unique point $y''$ lying over $v$ with residue $l$ (this corresponds with the obvious surjective map $C' \to l$). Thus $X_ y \times _{\mathop{\mathrm{Spec}}(k)} \mathop{\mathrm{Spec}}(l)$ has a unique point with residue field $l$. Since $l/k$ is finite separable, this implies $X'_ y$ has a unique point with residue field $k$, i.e., (3) holds.

To prove the final statement, observe that if $Y$ is locally Noetherian, then $R$ is a Noetherian ring and if $f$ is finite, then $R \to C$ is finite. Then $C'$ is a finite type $R$-algebra by More on Algebra, Lemma 15.5.1. This proves that $X' \to Y$ is finite. $\square$


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