The Stacks project

Proof. Let $S \subset A$ be a multiplicative subset. Then $S^{-1}A \to S^{-1}B$ is a ring map which induces a dominant morphism $\mathop{\mathrm{Spec}}(S^{-1}B) \to \mathop{\mathrm{Spec}}(S^{-1}A)$ as well (see Lemmas 29.8.4 and 29.25.9). Hence Lemma 29.55.1 produces an $S^{-1}A$-subalgebra $(S^{-1}B)' \subset S^{-1}B$. The statement means that $S^{-1}B' = (S^{-1}B)'$ as $S^{-1}A$-subalgebras of $S^{-1}B$.

To see this is true, we will use the construction of $B'$ and $(S^{-1}B)'$ in the proof of Lemma 29.55.1. In the first step, we see that $B'$ is the inverse image of the $A_{red}$-subalgebra $B'_{red} \subset B_{red}$ constructed for the ring map $A_{red} \to B_{red}$ and similarly for $(S^{-1}B)'$. Noting that $S^{-1}B_{red} = (S^{-1}B)_{red}$ this reduces us to the case discussed in the next paragraph.

If $A$ and $B$ are reduced, we have constructed $B'$ as the union of the subalgebras $A[b_1, \ldots , b_ n]$ such that for $i = 1, \ldots , n$ we have

1. $b_ i^2, b_ i^3 \in A[b_1, \ldots , b_{i - 1}]$, or

2. there exists a prime number $p$ with $pb_ i, b_ i^ p \in A[b_1, \ldots , b_{i - 1}]$.

Similarly for $(S^{-1}B)' \subset S^{-1}B$. Thus it is clear that the image of $B' \to B \to S^{-1}B$ is contained in $(S^{-1}B)'$. To show that the corresponding map $S^{-1}B' \to (S^{-1}B)'$ is surjective, one uses Lemma 29.46.3 to clear denominators successively; we omit the details. $\square$