# Question about paragraph 1.J in Matsumura's "Commutative Algebra"

I started reading Matsumura's book "Commutative algebra", and I got stuck in a specific statement in paragraph 1.J.

The paragraph states that, starting with a ring $$A$$ and $$S \subset A$$ a multiplicatively closed subset of $$A$$, if we have some other ring $$B$$ and the sequence of arrows $$A \overset{f}{\to} B \overset{g}{\to} S^{-1}A$$ satisfying $$g \circ f = \pi$$, where $$\pi : A \to S^{-1}A$$ is the canonical map; and also satisfying that for all $$b \in B$$, there is $$r \in S$$ such that $$f(r)b \in f(A)$$, then $$S^{-1}B \cong f(S)^{-1}B \cong S^{-1}A$$ (this part I understand).

After this, the author states that a particular case worth noting is when $$A$$ is an integral domain, $$\mathfrak{p} = A-S \in \operatorname{Spec}(A)$$ and $$B$$ a ring such that $$A \subset B \subset A_{\mathfrak{p}}$$, then $$B_{\mathfrak{p}'} = A_{\mathfrak{p}} \cong B_{\mathscr{p}}$$, where $$\mathfrak{p}' = \mathfrak{p} A_{\mathfrak{p}} \cap B$$ and $$B_{\mathscr{p}}:= B \otimes_{A} A_{\mathfrak{p}}$$.

Since the second paragraph is a particular case of the first one, I assume $$f,g$$ are inclusions in the second paragraph and so $$f(S)=S$$ which would imply $$A_{\mathfrak{p}} = S^{-1}A = S^{-1}B = B_{\mathfrak{p}'}$$, and so it would suffice to show that $$S = B-\mathfrak{p}'$$, something in which I am not succeeding. Clearly $$S \subset S' := B - \mathfrak{p}'$$. Also, if $$\frac{a}{s} \in S'$$, then since $$\frac{a}{s} \notin \mathfrak{p} A_{\mathfrak{p}}$$, $$a \notin \mathfrak{p}$$ and so $$a \in S$$. But how do I get that $$a = rs$$ for some $$r \in S$$?

Also, about the isomorphism in the second paragraph, if we have that $$B = R^{-1}A$$, where $$R = U(B) \cap S$$ and $$U(B)$$ is the set of units in B, then I can easily get the isomorphism, but again, I can see clearly that $$R^{-1}A \subset B$$ but not the other inclusion. This would also mean that all subrings $$B\subset F(A)$$ with $$A \subset B$$ are localizations of A with respect to a multiplicative subset of $$A$$, and I don't know if that's true.

Are my guesses right? And if so, how do I prove them?

Thank you for all the help in advance :)

• Why do you say that $S^{-1}A=S^{-1}B$? May 9, 2021 at 19:15
• Because in this case the isomorphism $S^{-1}A \cong S^{-1}B$ is actually the identity map May 9, 2021 at 19:18