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Let $R_1 \to T$ and $R_2 \to T$ be homomorphisms of commutative rings. Consider the fiber product $R=R_1 \times_T R_2$. Let $R \to R'$ be a homomorphism of commutative rings, and define $R'_i$ to be the pushout

$\begin{array}{c} R_1 \times_T R_2 & \rightarrow & R' \\ \downarrow && \downarrow \\ R_i & \rightarrow & R'_i \end{array}$

Similarily, define $T'$ as the pushout of $T \leftarrow R \to R'$. Then we have a canonical homomorphism $R' \to R'_1 \times_{T'} R'_2$. Observe that the composition $R_1 \times_T R_2 \to R' \to R'_1 \times_{T'} R'_2$ is "diagonal", it is induced by $R_i \to R'_i$ and $T \to T'$.

Question. Under which conditions on $R_i \to T$ is $R' \to R'_1 \times_{T'} R'_2$ an isomorphism for all choices of homomorphisms $R \to R'$? In other words, which fiber products of commutative rings are stable?

Remark. The question is clearer when one states it in the dual category of affine schemes (not schemes!): Let $S=\mathrm{Spec}(T)$, $X_i=\mathrm{Spec}(R_i)$. We have a morphism $X' \to X_1 \cup_S X_2$. Let $X'_i \to X_i$ be the pullback along $X_i \to X_1 \cup_S X_2$, and $S' \to S$ be the pullback along $S \to X_1 \cup_S X_2$. The question is, when is the canonical morphism $X'_1 \cup_{S'} X'_2 \to X'$ an isomorphism?

Example. Let $T=0$. If $f : R_1 \times R_2 \to R'$ is a homomorphism, then $S'=0$, $R'_1 = R'/ \langle f(0,1) \rangle = f(1,0) R'$ with the idempotent $f(1,0)$ as a unit, and similarily $R'_2 = f(0,1) R'$ with $f(0,1)$ as a unit, and it is clear that $R' = R'_1 \times R'_2$.

More generally, if $R_1 \to T$ and $R_2 \to T$ are surjective, say with kernels $I_1$ and $I_2$, then the condition holds for $\alpha : R_1 \times_T R_2 \to R'$ if and only if $\langle g(I_1 \times 0) \rangle \cap \langle g(0 \times I_2) \rangle = 0$. For which $I_1,I_2$ does this hold?

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