Suppose $A$ is a commutative ring with $1$, $g\in A$, and $\mathfrak{p}$ is a prime ideal. We know that upon localising $\mathfrak{p}_g$ is maximal is $A_g$. (In particular it does not contain $g$.) Then is it true that $\mathfrak{p}$ is maximal in $A$?
We know that $A_g/\mathfrak{p}_g\cong (A/\mathfrak{p})_g$. If the LHS is a field so is RHS. Then it is forced that $A/\mathfrak{p}$ is also a field since we cannot get a field by inverting just one element (I know this is true for finitely generated $k$-algebras, but let me know if this is false in general). Thus, $\mathfrak{p}$ is a closed point.
Is my proof right or wrong. Meanwhile I am trying to show that $S^{-1}R$ is a field then $S$ is infinite (leaving the case when $R$ is already a field).
You can get a field by inverting one element. For instance let $A$ be the subring of $\Bbb Q$ consisting of the fractions with odd denominator. Then $\Bbb Q=A[1/2]$. A domain with $A[b]$ a field is called a G-domain. These are used in proofs of Hilbert's Nullstellensatz.