$\mathfrak p$ be a prime ideal of a commutative ring $R$ , then $R_\mathfrak p/\mathfrak pR_\mathfrak p$ is the field of fractions of $R/\mathfrak p$?

Question

Let $\mathfrak p$ be a prime ideal of a commutative ring $R$ with unity , then how to show that the field $R_\mathfrak p/\mathfrak pR_\mathfrak p$ is isomorphic with the field of fractions of the integral domain $R/\mathfrak p$ ?

I can show that $R_\mathfrak p/\mathfrak pR_\mathfrak p \cong (R/\mathfrak p)_\mathfrak p$ as $R_\mathfrak p$ modules ... but I am not sure whether this implies the result I am asking or not

Answer 1

There's an obvious map $f\colon R_P\to \text{Frac}(R/P)$, given by $a/b\mapsto (a+P)/(b+P)$. You need to check:

• If $a/b\in R_P$, then $f(a/b)\in \text{Frac}(R/P)$, i.e., $b+P\neq 0$ in $R/P$.
• If $a/b = c/d$ in $R_P$, then $f(a/b) = f(c/d)$.
• $f$ is a homomorphism.

To show that $f$ induces an isomorphism $R_P/PR_P\cong \text{Frac}(R/P)$, you just need to show that the kernel of $f$ is $PR_P$. Well, suppose $a/b$ is in the kernel. Then $f(a/b) = (a+P)/(b+P)$ is equal to $0$ in $\text{Frac}(R/P)$, so $a+P$ is equal to $0$ in $R/P$. So $a\in P$, and $a/b = a(1/b) \in PR_P$. Conversely, if $a/b\in PR_P$, then it is equivalent to $p/b'$ for some $p\in P$ and $b'\notin P$. And $f(p/b') = (p+P)/(b'+P) = 0$, so $a/b$ is in the kernel (you've already checked that $f$ is well-defined, so it suffices to look at just one representative of its equivalence class).

There's a more category-theoretic way to do this exercise, where you check that $R_P/PR_P$ and $\text{Frac}(R/P)$ both satisfy the universal property that a ring homomorphism from this ring to $S$ is uniquely determined by a ring homomorphism $R\to S$ which maps every element of $P$ to $0$ and every element of $R\backslash P$ to a unit, and hence they must be isomorphic (by Yoneda's Lemma). But it sounds like you might benefit more at this stage from working through the details of the more "concrete" proof.

Answer 2

Since localization preserves exact sequences we have that the exact sequence
$$ 0 \to \mathfrak p \to R \to R/\mathfrak p \to 0$$ becomes
$$0 \to \mathfrak p R_{\mathfrak p} \to R_{\mathfrak p} \to (R/\mathfrak p)_{\mathfrak p} \to 0$$ Hence $(R/\mathfrak p)_{\mathfrak p}=R_{\mathfrak p}/\mathfrak pR_\mathfrak p$. From this you can conclude that it is the field of fractions since $R/\mathfrak p$ is a domain and localizing by $\mathfrak p$ means inverting all nonzero elements (there are no nonzero divisor in an integral domain).