Is $0 \to p \to R_p \to Frac(R/p) \to 0$ exact sequence?

Question

I have got the following task on my commutative algebra course:

Let us have a prime ideal $p$ in ring $R$. Consider the exact sequence $0 \to p \to R \to R/p \to 0$.

Now let's localize the sequence. It should stay exact. Which exact sequence we obtain.

I suspect $0 \to p \to R_p \to Frac(R/p) \to 0$

I denote by $Frac$ the ring obtained from the given by localization over all nonzerodivisors.

Is my suspicion true and why? I'm not sure with the result of localization in the last term.

Answer 1

When considering the exact sequence $0 \to p \to R \to R/p \to 0$ and

localizeing this sequence, we have the exact sequence $0 \to pR_p \to R_p \to (R/p)_p \to 0$. Thus we have $\frac{R_p}{pR_p}\cong (R/p)_p$ and since $\frac{R_p}{pR_p}$ is a field ($pR_p$ is a maximal ideal), $(R/p)_p$ is a field, and hence $(R/p)_p=Frac(R/p)$. Thus $$0 \to pR_p \to R_p \to Frac(R/p)\to 0$$ is an exact sequence.

Answer 2

The last term is correct. The Frac over a domain is the localisation on the prime ideal $0$. You can also verify that the localisation "commutes" with the quotient operation so $(\frac{R}{p})_{\bar{p}}=(\frac{R}{p})_{\bar{0}}$ is isomorphic to $\frac{(R)_p}{p(R)_p}$ but $(R)_p$ is a local ring so you will find your "Frac" and you should be able to write your exact sequence. What is $p$ localised as a $R$ module?...