Why is $ \mathrm{Frac} ( A / \mathfrak{p} ) = A_{\mathfrak{p}} / \mathfrak{p} A_{\mathfrak{p}} $?

Question

$ A $ is a commutative ring, $ \mathfrak{p} \in \mathrm{Spec} A $, $ A_{\mathfrak{p}} = ( A \backslash \mathfrak{p} )^{-1} A $, $ \mathrm{Frac} ( A / \mathfrak{p} )$ is the field of fractions of $ A / \mathfrak{p} $.

I would like to know why is $ \mathrm{Frac} ( A / \mathfrak{p} ) = A_{\mathfrak{p}} / \mathfrak{p} A_{\mathfrak{p}} $ ?

Thanks a lot.

Answer 1

The fast answer is that localization is an exact functor, so that the exact sequence $$0 \to {\mathfrak p} \to A \to A/{\mathfrak p} \to 0$$ is taken by localization to an exact sequence $$0 \to {\mathfrak p}A_{\mathfrak p} \to A_{\mathfrak p} \to (A/{\mathfrak p})_{\bar{\mathfrak p}} = (A/{\mathfrak p})_{(0)} = \mbox{Frac}(A/{\mathfrak p}) \to 0,$$

showing what you want.

But we can be lower-tech. First off, $A_{\mathfrak p}$ is a local ring, with maximal ideal ${\mathfrak p} A_{\mathfrak p}$, so you know at least that $k({\mathfrak p}) = A_{\mathfrak p}/{\mathfrak p}A_{\mathfrak p}$ is a field.

Consider the map $a \mapsto a/1\colon A \to A_{\mathfrak p}$ followed by the quotient map $A_{\mathfrak p} \to k({\mathfrak p})$.

I claim the kernel of the composition $A \to k({\mathfrak p})$ is ${\mathfrak p}$. Indeed, the kernel is made up of those elements $a \in A$ such that $a/1 = p/s \in {\mathfrak p} A_{\mathfrak p}$, where $p \in {\mathfrak p}$ and $s \in S = A \setminus {\mathfrak p}$, meaning there is $t \in S$ such that $t(sa-p) = 0$ in $A$, or $tsa = tp \in {\mathfrak p}$. Since $ts \in S$, it follows that $a \in {\mathfrak p}$.

So the map descends to an injection $A/{\mathfrak p} \to k({\mathfrak p})$. To see this is the field of fractions, it suffices to show every nonzero element of $A/{\mathfrak p}$ is invertible in $k({\mathfrak p})$ and every element of ${\mathfrak p}$ is a fraction of elements in the image of $A/{\mathfrak p} \to k({\mathfrak p})$.

If $\bar s \in A/{\mathfrak p}$ is nonzero, then it has a preimage $s \in S = A \setminus {\mathfrak p}$, so there is an $s^{-1} \in A_{\mathfrak p}$ and hence a $(s + {\mathfrak p}A_{\mathfrak p})^{-1}$ in the quotient $k({\mathfrak p})$.

As for fractions generating, everything in $k({\mathfrak p})$ is of the form $(a/s) + {\mathfrak p}A_{\mathfrak p}$ for $a \in A$ and $s \in S$. But this is $(a + {\mathfrak p}A_{\mathfrak p})(s + {\mathfrak p}A_{\mathfrak p})^{-1}$.

Answer 2

I think as a first approximation you should try to build maps both ways and show that they are inverse to one another. If you get stuck, you can ask us for help. But morally I think it's also good to note that both rings do the same sort of universal job, so of course they are canonically isomorphic.

Let's state what the ring homomorphisms $\operatorname{Frac}(A/\mathfrak{p}) \to B$ are by unwinding universal properties. First, they are the $A/\mathfrak{p} \to B$ that send all nonzero elements to units. Now, to give a morphism $A/\mathfrak{p} \to B$ is to give a morphism $\phi\colon A \to B$ that takes $\mathfrak{p}$ to $0$, and the unit condition translates into requiring that $\phi(A - \mathfrak{p})$ consist of units. I think this is the right characterizing property:

A map of rings $A \to \tilde{A}$ such that any map $A \to B$ that sends $\mathfrak{p}$ to $0$ and $A -\mathfrak{p}$ to units factors uniquely through $\tilde{A}$.

I leave it to you to wind up the universal properties in the opposite order, to get $A_\mathfrak{p}/\mathfrak{p}A_\mathfrak{p}$. When making arguments like this one must mention the Yoneda lemma even though I wouldn't bother learning about it now.