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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

Andrews
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user
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2 Answers2

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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.

Alex Kruckman
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    +1 I also think that OP will be benefited from "working through the details of the more "concrete" proof.". – Krish Sep 04 '17 at 15:53
  • @Alex Kruckman : why $a=pa'$ ? – user Sep 04 '17 at 16:25
  • @users Because that's how you multiply in the localization! If $a/b = (p/1)(a'/b')$, then $a = pa'$ and $b = 1b'$. – Alex Kruckman Sep 04 '17 at 17:32
  • @AlexKruckman : But $a/b=pa'/b'$ implies $s(b' a - bpa')=0 $ for some $s \notin P$ ... – user Sep 04 '17 at 17:35
  • So $sb'a = sbpa'\in P$. Since $s\notin P$, and $b'\notin P$, and $P$ is prime, $a\in P$. But ok, maybe I've been a little sloppy in what I wrote above, I"ll rewrite that paragraph. – Alex Kruckman Sep 04 '17 at 17:44
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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).

SC30
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    This proof gives an isomorphism of modules - is there a way to see automatically that it gives an isomorphism of rings as well? – Alex Kruckman Sep 04 '17 at 15:48
  • @SC30 but you still need to explain $(R/p)_p$ is isomorphic to the fraction field!better we know this localization. – Jian Sep 04 '17 at 15:51
  • @SC30 : Like Alex Kruckman , I am also tempted to comment that this only shows module isomorphism , which I already know as stated in the question – user Sep 04 '17 at 15:51
  • I would say that that given the ring map $R\rightarrow R/\mathfrak p$ we apply the localization functor obtaining a canonical map $R_{\mathfrak p}\rightarrow (R/\mathfrak p)_{\mathfrak p}$ (if you draw a diagram using the naturality of the functor you should get that those two things are canically identified as rings). – SC30 Sep 04 '17 at 16:09
  • @Sky I think it is obvious. There is a general property saying that if S contains all the nonzero divisors we have that $S^{-1}R$ is the fraction field of R. In our case it is much easier and immediate since we are dealing with an integral domain and we are inverting everything.... – SC30 Sep 04 '17 at 16:12