Frac$(R/p)=(R/p)_p$?

Question

$R$ is a commutative ring with unity and $p$ is a prime ideal.

Then Frac$(R/p)=(R/p)_p$?

Both Frac$(R/p)$ and $(R/p)_p$ contain $R/p$ but how do we know $(R/p)_p$ is the smallest field containing $R/p$? I guess the canonical homomorphism is the isomorphism between them.

Answer 1

It's very simple: passing from $\;R/\mathfrak p\;$ to $\;(R/\mathfrak p)_{\mathfrak p}\simeq R_{\mathfrak p}\big/\mathfrak pR_{\mathfrak p}\;$ consists in making invertible the nonzero congruence classes in $R/\mathfrak p$. Isn't that the fraction field of $R/\mathfrak p\mkern1mu$?