Quotient field of a certain quotient ring

Question

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 $A/\mathfrak p$.

Answer 1

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 $a+\mathfrak{p}/s+\mathfrak{p}\in\mathrm{Frac}(A/\mathfrak{p})$, 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.

Answer 2

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 $A/\mathfrak p$.