[Theorem 7.1 part iii, Ferrand-Conducteur]
Proposition 37.67.3. In Situation 37.67.1 the pushout $Y \amalg _ Z X$ exists in the category of schemes. Picture The diagram is a fibre square, the morphism $a$ is integral, the morphism $b$ is a closed immersion, and as sheaves of rings where $c = a \circ i = b \circ j$.
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.
$\square$
\[ \xymatrix{ Z \ar[r]_ i \ar[d]_ j & X \ar[d]^ a \\ Y \ar[r]^-b & Y \amalg _ Z X } \]
\[ \mathcal{O}_{Y \amalg _ Z X} = b_*\mathcal{O}_ Y \times _{c_*\mathcal{O}_ Z} a_*\mathcal{O}_ X \]
\[ b_*\mathcal{O}_ Y \longrightarrow c_*\mathcal{O}_ Z \longleftarrow a_*\mathcal{O}_ X \]
\[ \mathcal{O}_{Y \amalg _ Z X} = b_*\mathcal{O}_ Y \times _{c_*\mathcal{O}_ Z} a_*\mathcal{O}_ X \]
\[ C_ i \cong B_ i \otimes _{B_ i \times _{C_ i} A_ i} A_ i \]
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