Lemma 37.14.1. Let $A' \to A$ be a surjection of rings and let $B \to A$ be a ring map. Let $B' = B \times _ A A'$ be the fibre product of rings. Set $S = \mathop{\mathrm{Spec}}(A)$, $S' = \mathop{\mathrm{Spec}}(A')$, $T = \mathop{\mathrm{Spec}}(B)$, and $T' = \mathop{\mathrm{Spec}}(B')$. Then

\[ \vcenter { \xymatrix{ S \ar[r]_ i \ar[d]_ f & S' \ar[d]^{f'} \\ T \ar[r]^{i'} & T' } } \quad \text{corresponding to}\quad \vcenter { \xymatrix{ A & A' \ar[l] \\ B \ar[u] & B' \ar[l] \ar[u] } } \]

is a pushout of schemes.

**Proof.**
By More on Algebra, Lemma 15.6.2 we have $T' = T \amalg _ S S'$ as topological spaces, i.e., the diagram is a pushout in the category of topological spaces. Next, consider the map

\[ ((i')^\sharp , (f')^\sharp ) : \mathcal{O}_{T'} \longrightarrow i'_*\mathcal{O}_ T \times _{g_*\mathcal{O}_ S} f'_*\mathcal{O}_{S'} \]

where $g = i' \circ f = f' \circ i$. We claim this map is an isomorphism of sheaves of rings. Namely, we can view both sides as quasi-coherent $\mathcal{O}_{T'}$-modules (use Schemes, Lemmas 26.24.1 for the right hand side) and the map is $\mathcal{O}_{T'}$-linear. Thus it suffices to show the map is an isomorphism on the level of global sections (Schemes, Lemma 26.7.5). On global sections we recover the identification $B' \to B \times _ A A'$ from statement of the lemma (this is how we chose $B'$).

Let $X$ be a scheme. Suppose we are given morphisms of schemes $m' : S' \to X$ and $n : T \to X$ such that $m' \circ i = n \circ f$ (call this $m$). We get a unique map of topological spaces $n' : T' \to X$ compatible with $m'$ and $n$ as $T' = T \amalg _ S S'$ (see above). By the description of $\mathcal{O}_{T'}$ in the previous paragraph we obtain a unique homomorphism of sheaves of rings

\[ (n')^\sharp : \mathcal{O}_ X \longrightarrow (n')_*\mathcal{O}_{T'} = m'_*\mathcal{O}_ T \times _{m_*\mathcal{O}_ T} n_*\mathcal{O}_ S \]

given by $(m')^\sharp $ and $n^\sharp $. Thus $(n', (n')^\sharp )$ is the unique morphism of ringed spaces $T' \to X$ compatible with $m'$ and $n$. To finish the proof it suffices to show that $n'$ is a morphism of schemes, i.e., a morphism of locally ringed spaces.

Let $t' \in T'$ with image $x \in X$. We have to show that $\mathcal{O}_{X, x} \to \mathcal{O}_{T', t'}$ is local. If $t' \not\in T$, then $t'$ is the image of a unique point $s' \in S'$ and $\mathcal{O}_{T', t'} = \mathcal{O}_{S', s'}$. Namely, $S' \setminus S \to T' \setminus T$ is an isomorphism of schemes as $B' \to A'$ induces an isomorphism $\mathop{\mathrm{Ker}}(B' \to B) = \mathop{\mathrm{Ker}}(A' \to A)$. If $t'$ is the image of $t \in T$, then we know that the composition $\mathcal{O}_{X, x} \to \mathcal{O}_{T', t'} \to \mathcal{O}_{T, t}$ is local and we conclude also.
$\square$

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