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
Best Answer
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:
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.