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\begin{equation*} \DeclareMathOperator\Coim{Coim} \DeclareMathOperator\Coker{Coker} \DeclareMathOperator\Ext{Ext} \DeclareMathOperator\Hom{Hom} \DeclareMathOperator\Im{Im} \DeclareMathOperator\Ker{Ker} \DeclareMathOperator\Mor{Mor} \DeclareMathOperator\Ob{Ob} \DeclareMathOperator\Sh{Sh} \DeclareMathOperator\SheafExt{\mathcal{E}\mathit{xt}} \DeclareMathOperator\SheafHom{\mathcal{H}\mathit{om}} \DeclareMathOperator\Spec{Spec} \newcommand\colim{\mathop{\mathrm{colim}}\nolimits} \newcommand\lim{\mathop{\mathrm{lim}}\nolimits} \newcommand\Qcoh{\mathit{Qcoh}} \newcommand\Sch{\mathit{Sch}} \newcommand\QCohstack{\mathcal{QC}\!\mathit{oh}} \newcommand\Cohstack{\mathcal{C}\!\mathit{oh}} \newcommand\Spacesstack{\mathcal{S}\!\mathit{paces}} \newcommand\Quotfunctor{\mathrm{Quot}} \newcommand\Hilbfunctor{\mathrm{Hilb}} \newcommand\Curvesstack{\mathcal{C}\!\mathit{urves}} \newcommand\Polarizedstack{\mathcal{P}\!\mathit{olarized}} \newcommand\Complexesstack{\mathcal{C}\!\mathit{omplexes}} \newcommand\Pic{\mathop{\mathrm{Pic}}\nolimits} \newcommand\Picardstack{\mathcal{P}\!\mathit{ic}} \newcommand\Picardfunctor{\mathrm{Pic}} \newcommand\Deformationcategory{\mathcal{D}\!\mathit{ef}} \end{equation*}

Lemma 36.14.3. Let $X \to X'$ be a thickening of schemes and let $X \to Y$ be an affine morphism of schemes. Then there exists a pushout

\[ \xymatrix{ X \ar[r] \ar[d]_ f & X' \ar[d]^{f'} \\ Y \ar[r] & Y' } \]

in the category of schemes. Moreover, $Y \subset Y'$ is a thickening, $X = Y \times _{Y'} X'$, and

\[ \mathcal{O}_{Y'} = \mathcal{O}_ Y \times _{f_*\mathcal{O}_ X} f'_*\mathcal{O}_{X'} \]

as sheaves on $|Y| = |Y'|$.

Proof. We first construct $Y'$ as a ringed space. Namely, as topological space we take $Y' = Y$. Denote $f' : X' \to Y'$ the map of topological spaces which equals $f$. As structure sheaf $\mathcal{O}_{Y'}$ we take the right hand side of the equation of the lemma. To see that $Y'$ is a scheme, we have to show that any point has an affine neighbourhood. Since the formation of the fibre product of sheaves commutes with restricting to opens, we may assume $Y$ is affine. Then $X$ is affine (as $f$ is affine) and $X'$ is affine as well (see Lemma 36.2.3). Say $Y \leftarrow X \rightarrow X'$ corresponds to $B \rightarrow A \leftarrow A'$. Set $B' = B \times _ A A'$; this is the global sections of $\mathcal{O}_{Y'}$. As $A' \to A$ is surjective with locally nilpotent kernel we see that $B' \to B$ is surjective with locally nilpotent kernel. Hence $\mathop{\mathrm{Spec}}(B') = \mathop{\mathrm{Spec}}(B)$ (as topological spaces). We claim that $Y' = \mathop{\mathrm{Spec}}(B')$. To see this we will show for $g' \in B'$ with image $g \in B$ that $\mathcal{O}_{Y'}(D(g)) = B'_{g'}$. Namely, by More on Algebra, Lemma 15.5.3 we see that

\[ (B')_{g'} = B_ g \times _{A_ h} A'_{h'} \]

where $h \in A$, $h' \in A'$ are the images of $g'$. Since $B_ g$, resp. $A_ h$, resp. $A'_{h'}$ is equal to $\mathcal{O}_ Y(D(g))$, resp. $f_*\mathcal{O}_ X(D(g))$, resp. $f'_*\mathcal{O}_{X'}(D(g))$ the claim follows.

It remains to show that $Y'$ is the pushout. The discussion above shows the scheme $Y'$ has an affine open covering $Y' = \bigcup W'_ i$ such that the corresponding opens $U'_ i \subset X'$, $W_ i \subset Y$, and $U_ i \subset X$ are affine open. Moreover, if $A'_ i$, $B_ i$, $A_ i$ are the rings corresponding to $U'_ i$, $W_ i$, $U_ i$, then $W'_ i$ corresponds to $B_ i \times _{A_ i} A'_ i$. Thus we can apply Lemmas 36.14.1 and 36.14.2 to conclude our construction is a pushout in the category of schemes. $\square$


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