The Stacks project

Proof. Denote $h : R \to X$ denote the composition of either $s$ or $t$ with $g$. Then $h$ is proper by Morphisms of Spaces, Lemmas 67.40.3 and 67.40.4. The sheaves

are coherent $\mathcal{O}_ X$-algebras by Cohomology of Spaces, Lemma 69.20.2. The $X$-morphisms $s$, $t$ induce $\mathcal{O}_ X$-algebra maps $s^\sharp , t^\sharp $ from the first to the second. Set

Then $\mathcal{A}$ is a coherent $\mathcal{O}_ X$-algebra and we can define

as in Morphisms of Spaces, Definition 67.20.8. By Morphisms of Spaces, Remark 67.20.9 and functoriality of the $\underline{\mathop{\mathrm{Spec}}}$ construction there is a factorization

and the morphism $g' : Y \to X'$ equalizes $s$ and $t$.

Before we show that $X'$ is the coequalizer of $s$ and $t$, we show that $Y \to X'$ and $X' \to X$ have the desired properties. Since $\mathcal{A}$ is a coherent $\mathcal{O}_ X$-module it is clear that $X' \to X$ is a finite morphism of algebraic spaces. This proves (1). The morphism $Y \to X'$ is proper by Morphisms of Spaces, Lemma 67.40.6. This proves (2). Denote $Y \to Y' \to X$ with $Y' = \underline{\mathop{\mathrm{Spec}}}_ X(g_*\mathcal{O}_ Y)$ the Stein factorization of $g$, see More on Morphisms of Spaces, Theorem 76.36.4. Of course we obtain morphisms $Y \to Y' \to X' \to X$ fitting with the morphisms studied above. Since $\mathcal{O}_{X'} \subset g_*\mathcal{O}_ Y$ is a finite extension we see that $Y' \to X'$ is finite and surjective. Some details omitted; hint: use Algebra, Lemma 10.36.17 and reduce to the affine case by étale localization. Since $Y \to Y'$ is surjective (with geometrically connected fibres) we conclude that $Y \to X'$ is surjective. This proves (3). To show that $X' \to X$ is universally injective, we have to show that $X' \to X' \times _ X X'$ is surjective, see Morphisms of Spaces, Definition 67.19.3 and Lemma 67.19.2. Since $Y \to X'$ is surjective (see above) and since base changes and compositions of surjective morphisms are surjective by Morphisms of Spaces, Lemmas 67.5.5 and 67.5.4 we see that $Y \times _ X Y \to X' \times _ X X'$ is surjective. However, since $Y \to X'$ equalizes $s$ and $t$, we see that $Y \times _ X Y \to X' \times _ X X'$ factors through $X' \to X' \times _ X X'$ and we conclude this latter map is surjective. This proves (4). Finally, if $g$ is surjective, then since $g$ factors through $X' \to X$ we see that $X' \to X$ is surjective. Since a surjective, universally injective, finite morphism is a universal homeomorphism (because it is universally bijective and universally closed), this proves (5).

In the rest of the proof we show that $Y \to X'$ is the coequalizer of $s$ and $t$ in the category of algebraic spaces over $S$. Observe that $X'$ is locally Noetherian (Morphisms of Spaces, Lemma 67.23.5). Moreover, observe that $Y \times _{X'} Y \to Y \times _ X Y$ is an isomorphism as $Y \to X'$ equalizes $s$ and $t$ (this is a categorical statement). Hence in order to prove the statement that $Y \to X'$ is the coequalizer of $s$ and $t$, we may and do assume $X = X'$. In other words, $\mathcal{O}_ X$ is the equalizer of the maps $s^\sharp , t^\sharp : g_*\mathcal{O}_ Y \to h_*\mathcal{O}_ R$.

Let $X_1 \to X$ be a flat morphism of algebraic spaces over $S$ with $X_1$ locally Noetherian. Denote $g_1 : Y_1 \to X_1$, $h_1 : R_1 \to X_1$ and $s_1, t_1 : R_1 \to Y_1$ the base changes of $g, h, s, t$ to $X_1$. Of course $g_1$ is proper and $R_1 = Y_1 \times _{X_1} Y_1$. Since we have flat base change for pushforward of quasi-coherent modules, Cohomology of Spaces, Lemma 69.11.2, we see that $\mathcal{O}_{X_1}$ is the equalizer of the maps $s_1^\sharp , t_1^\sharp : g_{1, *}\mathcal{O}_{Y_1} \to h_{1, *}\mathcal{O}_{R_1}$. Hence all the assumptions we have are preserved by this base change.

At this point we are going to check conditions (1) and (2) of Lemma 81.3.3. Condition (1) follows from Lemma 81.5.1 and the fact that $g$ is proper and surjective (because $X = X'$). To check condition (2), by the remarks on base change above, we reduce to the statement discussed and proved in the next paragraph.

Assume $S = \mathop{\mathrm{Spec}}(A)$ is an affine scheme, $X = X'$ is an affine scheme, and $Z$ is an affine scheme over $S$. We have to show that

is bijective. However, this is clear from the fact that $X = X'$ which implies $\mathcal{O}_ X$ is the equalizer of the maps $s^\sharp , t^\sharp : g_*\mathcal{O}_ Y \to h_*\mathcal{O}_ R$ which in turn implies

Namely, we have

and similarly for $Y$ and $R$, see Properties of Spaces, Lemma 66.33.1. $\square$

Proof. Existence of $X_1$ and the fact that $X_1 \to X$ is a finite universal homeomorphism is a special case of Lemma 81.13.1. The formation of $X_1$ commutes with étale localization on $X$ (see proof of Lemma 81.13.1). Thus the morphism $X_1 \to X$ is an isomorphism over $U$. It is immediate from the construction that $Z \to X$ lifts to $X_1$. $\square$

Proof. Let's look a bit more closely at the construction of $X_ n$ and how it changes as we increase $n$. We have $X_ n = \underline{\mathop{\mathrm{Spec}}}(\mathcal{A}_ n)$ where $\mathcal{A}_ n$ is the equalizer of $s_ n^\sharp $ and $t_ n^\sharp $ going from $g_{n , *}\mathcal{O}_{Y_ n}$ to $h_{n, *}\mathcal{O}_{R_ n}$. Here $g_ n : Y_ n = X' \amalg Z_ n \to X$ and $h_ n : R_ n = Y_ n \times _ X Y_ n \to X$ are the given morphisms. Let $\mathcal{I} \subset \mathcal{O}_ X$ be the coherent sheaf of ideals corresponding to $Z$. Then

Similarly, we have a decomposition

As $Z_ n \to X$ is a monomorphism, we see that $X' \times _ X Z_ n = Z_ n \times _ X X'$ and that this identification is compatible with the two morphisms to $X$, with the two morphisms to $X'$, and with the two morphisms to $Z_ n$. Denote $f_ n : X' \times _ X Z_ n \to X$ the morphism to $X$. Denote

By the remarks above we find that

We have canonical maps

of coherent $\mathcal{O}_ X$-algebras. The statement of the lemma means that for $n$ large enough there exists an $m \geq 0$ such that the image of $\mathcal{A}_{n + m} \to \mathcal{A}_ n$ is isomorphic to $\mathcal{O}_ X$. This we may check étale locally on $X$. Hence by Properties of Spaces, Lemma 66.6.3 we may assume $X$ is an affine Noetherian scheme.

Since $X_ n \to X$ is an isomorphism over $U$ we see that the kernel of $\mathcal{O}_ X \to \mathcal{A}_ n$ is supported on $|Z|$. Since $X$ is Noetherian, the sequence of kernels $\mathcal{J}_ n = \mathop{\mathrm{Ker}}(\mathcal{O}_ X \to \mathcal{A}_ n)$ stabilizes (Cohomology of Spaces, Lemma 69.13.1). Say $\mathcal{J}_{n_0} = \mathcal{J}_{n_0 + 1} = \ldots = \mathcal{J}$. By Cohomology of Spaces, Lemma 69.13.2 we find that $\mathcal{I}^ t \mathcal{J} = 0$ for some $t \geq 0$. On the other hand, there is an $\mathcal{O}_ X$-algebra map $\mathcal{A}_ n \to \mathcal{O}_ X/\mathcal{I}^ n$ and hence $\mathcal{J} \subset \mathcal{I}^ n$ for all $n$. By Artin-Rees (Cohomology of Spaces, Lemma 69.13.3) we find that $\mathcal{J} \cap \mathcal{I}^ n \subset \mathcal{I}^{n - c}\mathcal{J}$ for some $c \geq 0$ and all $n \gg 0$. We conclude that $\mathcal{J} = 0$.

Pick $n \geq n_0$ as in the previous paragraph. Then $\mathcal{O}_ X \to \mathcal{A}_ n$ is injective. Hence it now suffices to find $m \geq 0$ such that the image of $\mathcal{A}_{n + m} \to \mathcal{A}_ n$ is equal to the image of $\mathcal{O}_ X$. Observe that $\mathcal{A}_ n$ sits in a short exact sequence

and similarly for $\mathcal{A}_{n + m}$. Hence it suffices to show

for some $m \geq 0$. To do this we may work étale locally on $X$ and since $X$ is Noetherian we may assume that $X$ is a Noetherian affine scheme. Say $X = \mathop{\mathrm{Spec}}(R)$ and $\mathcal{I}$ corresponds to the ideal $I \subset R$. Let $\mathcal{A} = \widetilde{A}$ for a finite $R$-algebra $A$. Let $f_*\mathcal{O}_{X'} = \widetilde{B}$ for a finite $R$-algebra $B$. Then $R \to A \subset B$ and these maps become isomorphisms on inverting any element of $I$.

Note that $f_{n, *}\mathcal{O}_{X' \times _ X Z_ n}$ is equal to $f_*(\mathcal{O}_{X'}/I^ n\mathcal{O}_{X'})$ in the notation used in Cohomology of Spaces, Section 69.22. By Cohomology of Spaces, Lemma 69.22.4 we see that there exists a $c \geq 0$ such that

is contained in $I^{n + m}B$. On the other hand, as $R \to B$ is finite and an isomorphism after inverting any element of $I$ we see that $I^{n + m}B \subset \mathop{\mathrm{Im}}(I^ n \to B)$ for $m$ large enough (can be chosen independent of $n$). This finishes the proof as $A \subset B$. $\square$