# Quotient field of a certain quotient ring

Let $A$ be a commutative integral domain and $\mathfrak p$ a prime ideal of $A$. Let $A_{\mathfrak p}$ be the localization of $A$ at $\mathfrak p$ and $\mathfrak{m}_{\mathfrak{p}}=\mathfrak pA_{\mathfrak p}$ its unique maximal ideal.

On page 10 of Serre's Local Fields, it states that $A_{\mathfrak p}/{\mathfrak{m}_{\mathfrak p}}$ is the field of fractions of the quotient ring $A/\mathfrak p$.

Could someone help me prove this?

In particular I'm unable to show that the residue field $A_{\mathfrak p}/{\mathfrak{m}_{\mathfrak{p}}}$ is contained in the quotient field of $A/\mathfrak p$.

• Dear @DonAntonio, Serre says "[$A_\mathfrak{p}$] is a local ring with maximal ideal $\mathfrak{p}A_\mathfrak{p}$ and residue field the field of fractions of $A/\mathfrak{p}$." The residue field of $A_\mathfrak{p}$ is its quotient modulo its unique maximal ideal, which is $A_\mathfrak{p}/\mathfrak{p}A_\mathfrak{p}$. In general $A_\mathfrak{p}$ won't be a domain. – Keenan Kidwell Aug 16 '13 at 19:12
• Oh, I think I see now my confussion: the OP is using $\,m_{\mathfrak p}\;$ instead of $\,\mathfrak pA_{\mathfrak p}\;$...and even YACP stressed this. – DonAntonio Aug 16 '13 at 19:26

## 2 Answers

The ring map $A/\mathfrak{p}\rightarrow A_\mathfrak{p}/\mathfrak{p}A_\mathfrak{p}$ is injective because the inverse image of $\mathfrak{p}A_\mathfrak{p}$ in $A$ (under the localization map $A\rightarrow A_\mathfrak{p}$) is equal to $\mathfrak{p}$ (one of the first things one proves about the prime ideal structure of a localization). So there is a unique $A/\mathfrak{p}$-algebra map $\mathrm{Frac}(A/\mathfrak{p})\rightarrow A_\mathfrak{p}/\mathfrak{p}A_\mathfrak{p}$, necessarily an injection since the source is a field. But given $a/s+\mathfrak{p}A_\mathfrak{p}$ in the target, $s\notin\mathfrak{p}$ means $s+\mathfrak{p}$ is non-zero in $A/\mathfrak{p}$, so $a+\mathfrak{p}/s+\mathfrak{p}\in\mathrm{Frac}(A)$, and this maps to $a/s+\mathfrak{p}A_\mathfrak{p}$ under the $A_\mathfrak{p}$-algebra map obtained above, so that map is an isomorphism.

Said somewhat differently, though more or less equivalently, if $I$ is an ideal of $A$, $S$ a multiplicative set, then the natural map $S^{-1}A\rightarrow\bar{S}^{-1}(A/I)$, where $\bar{S}$ is the image of $S$ in $A/I$, is surjective with kernel $S^{-1}I$, the ideal of $S^{-1}A$ generated by $I$, and so induces an isomorphism $S^{-1}A/S^{-1}I\cong\bar{S}(A/I)$ of $A/I$-algebras. Taking $S=A-\mathfrak{p}$, and $I=\mathfrak{p}$, you get an isomorphism $A_\mathfrak{p}/\mathfrak{p}A_\mathfrak{p}\cong(A/\mathfrak{p})_{\overline{A-\mathfrak{p}}}$. But $\overline{A-\mathfrak{p}}$ is exactly the set of non-zero elements of $A/\mathfrak{p}$, so the localization at $\overline{A-\mathfrak{p}}$ is the field of fractions of $A/\mathfrak{p}$. This isomorphism $A_\mathfrak{p}/\mathfrak{p}A_\mathfrak{p}\cong\mathrm{Frac}(A/\mathfrak{p})$ is the inverse to the one defined above.

• I see! I knew that by the universal property of the quotient field there is an injective ring homomorphism $Frac(A/{\mathfrak p}) \rightarrow A_{\mathfrak p}/{{\mathfrak p} A_{\mathfrak p}}$ but I wasn't sure that it was surjective. One question: doesn't the universal property state that this is only a homomorphism of fields and not a homomorphism of $A_{\mathfrak p}$-algebras? – Mark Rodriguez Aug 16 '13 at 20:09
• I actually meant to say $A/\mathfrak{p}$-algebra homomorphism," and this is definitely part of the universal property, i.e., the induced map $\mathrm{Frac}(A/\mathfrak{p})\rightarrow A_\mathfrak{p}/\mathfrak{p}A_\mathfrak{p}$ is the unique one which, when pre-composed with $A/\mathfrak{p}\rightarrow\mathrm{Frac}(A/\mathfrak{p})$, is equal to the given map. Another way of saying this is that the map $\mathrm{Frac}(A/\mathfrak{p})\rightarrow A_\mathfrak{p}/\mathfrak{p}A_\mathfrak{p}$ is an $A/\mathfrak{p}$-algebra homomorphism. Without this one might not have uniqueness. – Keenan Kidwell Aug 16 '13 at 20:13

Here is another way to describe the situation; of course it is not so different to Keenan's.

There is an exact sequence $0 \to \mathfrak p \to A \to A/\mathfrak p \to 0.$ Localization is exact, and so localizing at $\mathfrak p$ gives an exact sequence $$0 \to \mathfrak p A_{\mathfrak p} \to A_{\mathfrak p} \to (A/\mathfrak p)_{\mathfrak p} \to 0.$$ Thus it suffices to check that the localization of $A/\mathfrak p$ at $\mathfrak p$ is naturally identified with the fraction field of $A/\mathfrak p;$ but this is pretty clear, if you think about each in terms of adding certain denominators to the elements of $A/\mathfrak p$.

• How is the map $A_{\mathfrak p}\rightarrow (A/{\mathfrak p})_{\mathfrak p}$ in the second exact sequence defined? – Mark Rodriguez Aug 16 '13 at 21:02
• @YACP: Dear YACP, I don't really understand your comment. If $I$ is an ideal in the commutative ring $R$, then the third arrow exact sequence of modules $0 \to I \to R \to R/I \to 0$ is also the quotient map of rings $R \to R/I$, and the $R$-module structure on $R/I$ induces its ring structure (this module structure automatically factors through $R/I$). In short, it seems fine to me. Maybe you would want to spell out more details, but I think they are pretty straightforward. In any case, Keenan has already answered the question at a level that seems more appropriate for the OP; ... – Matt E Aug 17 '13 at 1:00
• ... I just wanted to add a slightly more streamlined perspective. Regards, – Matt E Aug 17 '13 at 1:01
• @MarkRodriguez: Dear Mark, It is the canonical map on localizations induced by the quotient map $A \to A/\mathfrak p.$ (Localization is a functor.) Regards, – Matt E Aug 17 '13 at 1:02
• @MarkRodriguez: Dear Mark, I literally mean the $A$-algebra $A/\mathfrak p$ localized at the multiplicative set $A \setminus \mathfrak p$. (The notation $M_{\mathfrak p}$ is the standard notation for the $A$-module $M$ localized at $A\setminus \mathfrak p$.) But this is canonically identifed with $(A/\mathfrak p)_{\overline{A \setminus \mathfrak p}}.$ Regards, – Matt E Aug 18 '13 at 1:46