The Stacks project

Proof. Let $y \in Y$. Choose an affine open $U \subset X$ such that $j^{-1}(\{ y\} ) \subset i^{-1}(U)$ (possible by assumption). Choose an affine open $V \subset Y$ neighbourhood of $y$ such that $j^{-1}(V) \subset i^{-1}(U)$. This is possible because $j : Z \to Y$ is a closed morphism (Morphisms, Lemma 29.44.7) and $i^{-1}(U)$ contains the fibre over $y$. Since $j$ is integral, the scheme theoretic fibre $Z_ y$ is the spectrum of an algebra integral over a field. By Limits, Lemma 32.11.6 we can find an $\overline{f} \in \Gamma (i^{-1}(U), \mathcal{O}_{i^{-1}(U)})$ such that $Z_ y \subset D(\overline{f}) \subset j^{-1}(V)$. Since $i|_{i^{-1}(U)} : i^{-1}(U) \to U$ is a closed immersion of affines, we can choose an $f \in \Gamma (U, \mathcal{O}_ U)$ whose restriction to $i^{-1}(U)$ is $\overline{f}$. After replacing $U$ by the principal open $D(f) \subset U$ we find affine opens $y \in V \subset Y$ and $U \subset X$ with

Now we (in some sense) repeat the argument. Namely, we choose $g \in \Gamma (V, \mathcal{O}_ V)$ such that $y \in D(g)$ and $j^{-1}(D(g)) \subset i^{-1}(U)$ (possible by the same argument as above). Then we can pick $f \in \Gamma (U, \mathcal{O}_ U)$ whose restriction to $i^{-1}(U)$ is the pullback of $g$ by $i^{-1}(U) \to V$ (again possible by the same reason as above). Then we finally have affine opens $y \in V' = D(g) \subset V \subset Y$ and $U' = D(f) \subset U \subset X$ with $j^{-1}(V') = i^{-1}(V')$. $\square$

Proof. As a topological space we set $Y \amalg _ Z X$ equal to the pushout of the diagram in the category of topological spaces (Topology, Section 5.29). This is just the pushout of the underlying sets (Topology, Lemma 5.29.1) endowed with the quotient topology. On $Y \amalg _ Z X$ we have the maps of sheaves of rings

and we can define

as the fibre product in the category of sheaves of rings. To prove that we obtain a scheme we have to show that every point has an affine open neighbourhood. This is clear for points not in the image of $c$ as the image of $c$ is a closed subset whose complement is isomorphic as a ringed space to $(Y \setminus j(Z)) \amalg (X \setminus i(Z))$.

A point in the image of $c$ corresponds to a unique $y \in Y$ in the image of $j$. By Lemma 37.67.2 we find affine opens $U \subset X$ and $V \subset Y$ with $y \in V$ and $i^{-1}(U) = j^{-1}(V)$. Since the construction of the first paragraph is clearly compatible with restriction to compatible open subschemes, to prove that it produces a scheme we may assume $X$, $Y$, and $Z$ are affine.

If $X = \mathop{\mathrm{Spec}}(A)$, $Y = \mathop{\mathrm{Spec}}(B)$, and $Z = \mathop{\mathrm{Spec}}(C)$ are affine, then More on Algebra, Lemma 15.6.2 shows that $Y \amalg _ Z X = \mathop{\mathrm{Spec}}(B \times _ C A)$ as topological spaces. To finish the proof that $Y \times _ Z X$ is a scheme, it suffices to show that on $\mathop{\mathrm{Spec}}(B \times _ C A)$ the structure sheaf is the fibre product of the pushforwards. This follows by applying More on Algebra, Lemma 15.5.3 to principal affine opens of $\mathop{\mathrm{Spec}}(B \times _ C A)$.

The discussion above shows the scheme $Y \amalg _ X Z$ has an affine open covering $Y \amalg _ X Z = \bigcup W_ i$ such that $U_ i = a^{-1}(W_ i)$, $V_ i = b^{-1}(W_ i)$, and $\Omega _ i = c^{-1}(W_ i)$ are affine open in $X$, $Y$, and $Z$. Thus $a$ and $b$ are affine. Moreover, if $A_ i$, $B_ i$, $C_ i$ are the rings corresponding to $U_ i$, $V_ i$, $\Omega _ i$, then $A_ i \to C_ i$ is surjective and $W_ i$ corresponds to $A_ i \times _{C_ i} B_ i$ which surjects onto $B_ i$. Hence $b$ is a closed immersion. The ring map $A_ i \times _{C_ i} B_ i \to A_ i$ is integral by More on Algebra, Lemma 15.6.3 hence $a$ is integral. The diagram is cartesian because

This follows as $B_ i \times _{C_ i} A_ i \to B_ i$ and $A_ i \to C_ i$ are surjective maps whose kernels are the same.

Finally, we can apply Lemmas 37.14.1 and 37.14.2 to conclude our construction is a pushout in the category of schemes. $\square$

Proof. The morphism $Y \amalg X \to Y \amalg _ Z X$ is surjective and universall closed. Thus we may apply Morphisms, Lemma 29.41.11. $\square$

Proof. Looking on affine opens we recover the result of More on Algebra, Lemma 15.5.1. $\square$

Proof. The pushout $Y \amalg _ Z X$ exists by Proposition 37.67.3. To see that the pushout $Y' \amalg _{Z'} X'$ exists, we check condition (3) of Situation 37.67.1 holds for $(X', Y', Z', i', j')$. Namely, let $y' \in Y'$ and denote $y \in Y$ the image. Choose $U \subset X$ affine open with $i(j^{-1}(y)) \subset U$. Choose a quasi-compact open $U' \subset X'$ contained in $f^{-1}(U)$ containing the quasi-compact subset $i'((j')^{-1}(\{ y'\} ))$. By Lemma 37.66.8 we see that $U'$ is quasi-affine. Since $Z'_{y'}$ is the spectrum of an algebra integral over a field, we can apply Limits, Lemma 32.11.6 and we find there exists an affine open subscheme of $U'$ containing $i'((j')^{-1}(\{ y'\} ))$ as desired.

Having verified existence we check the other assertions. Affine locally we are exactly in the situation of More on Algebra, Lemma 15.7.7 with $B \to D$ and $A' \to C'$ locally quasi-finite¹. In particular, the morphism $Y' \amalg _{Z'} X' \to Y \amalg _ Z X$ is locally of finite type. The squares in of the diagram are cartesian by More on Algebra, Lemma 15.6.4. Since being locally quasi-finite can be checked on fibres (Morphisms, Lemma 29.20.6) we conclude that $Y' \amalg _{Z'} X' \to Y \amalg _ Z X$ is locally quasi-finite.

We still have to check $Y' \amalg _{Z'} X' \to Y \amalg _ Z X$ is separated. Observe that $Y' \amalg X' \to Y' \amalg _{Z'} X'$ is universally closed and surjective by Proposition 37.67.3. Since also the morphism $Y' \amalg X' \to Y \amalg _ Z X$ is separated (as it factors as $Y' \amalg X' \to Y \amalg X \to Y \amalg _ Z X$) we conclude by Morphisms, Lemma 29.41.11. $\square$

Proof. If we start with $(X', Y', Z', i', j', f, g, h)$ as in Lemma 37.67.6 with $f, g, h$ flat or étale, then $Y' \amalg _{Z'} X' \to Y \amalg _ Z X$ is flat or étale by More on Algebra, Lemma 15.7.7.

For the converse, let $W \to Y \amalg _ Z X$ be a separated and locally quasi-finite morphism. Set $X' = W \times _{Y \amalg _ Z X} X$, $Y' = W \times _{Y \amalg _ Z X} Y$, and $Z' = W \times _{Y \amalg _ Z X} Z$ with obvious morphisms $i', j', f, g, h$. Form the pushout $Y' \amalg _{Z'} X'$. We obtain a morphism

of schemes over $Y \amalg _ X Z$ by the universal property of the pushout. If we do not assume that $W \to Y \amalg _ Z X$ is flat, then in general this morphism won't be an isomorphism. (In fact, More on Algebra, Lemma 15.6.5 shows the displayed arrow is a closed immersion but not an isomorphism in general.) However, if $W \to Y \times _ Z X$ is flat, then it is an isomorphism by More on Algebra, Lemma 15.7.7. $\square$

Proof. This is a special case of Proposition 37.67.3. Observe that hypothesis (3) in Situation 37.67.1 is immediate because the fibres of $j$ are singletons. Finally, reverse the roles of the arrows to conclude that both $a$ and $b$ are closed immersions. $\square$

Proof. We know that the pushouts exist by Lemma 37.67.8. In particular we get the morphism $h$. Hence we may replace all schemes in sight by affine schemes. In this case the assertions of the lemma are equivalent to the corresponding assertions of More on Algebra, Lemma 15.7.7. $\square$