I'm trying to prove that if $R$ is a local domain, and $S$ is a multiplicative subset of $R$ then the ring of fractions $S^{-1}R=\{\frac{a}{b}:a\in R,b\in S\}$ is also local.
I was trying to characterize the maximal ideals of $S^{-1}R$ but I had no success. I would appreciate some hints.
If you take two prime ideals $p_1, p_2$ so that neither is a subset of the other, and let $S = R\setminus (p_1\cup p_2)$, then $S^{-1}R$ has both $p_1$ and $p_2$ as ideals, but there is no proper ideal that contains both of them. Therefore $S^{-1}R$ isn't local.
For a concrete example, take $R = \Bbb C[x, y]_m$ with $m = (x,y)$. This is the ring of rational functions in two variables over $\Bbb C$ where the denominator must have a non-zero constant term. This ring is local because we localised at a maximal ideal, and i hope you will agree that it's a domain.
Next, let $p_1 = (x)$ and $p_2 = (y)$ and let $S = R\setminus (p_1 \cup p_2)$. In other words, $S$ consists of those rational functions in $R$ such that the numerator is neither divisible by $x$ nor by $y$ (when written in simplest terms). So $\frac{x + y}{x^2 - xy + 3} \in S$, for instance.
This gives us $S^{-1}R$, the ring of rational functions over $\Bbb C$ in two variables that can be written such that the denominator is divisible by neither $x$ nor $y$. In this ring, $x$ and $y$ are not invertible, thus they both belong to some maximal ideal. Yet any ideal that contains both of them must necessarily be the whole ring, since $x + y$ would be in such an ideal, and $x + y$ is invertible in $S^{-1}R$. This means that there must be at least two maximal ideals (I would bet large money on those two being $(x)$ and $(y)$), so $S^{-1}R$ is not local.