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 :)

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


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