Serge Lang, Algebra: Chapter 2, Exercise 1.
Let $S$ be a multiplicative subset of a commutative ring $A$, containing 1 but not 0. Let $\mathfrak{p}$ be a maximal element in the set of ideals of $A$, whose intersection with $S$ is empty. Show that $\mathfrak{p}$ is prime.
My approach to this problem is the following: Consider the ring of fractions $S^{-1}A$. We will denote the set of ideals whose intersection with $S$ is empty by $X$ for convenience. Now consider the mapping
$$\psi: J(A) \rightarrow J(S^{-1}A)$$
where $J(A)$ is the set of ideals of $A$. Then $\psi(X)$ is the set of proper ideals of $S^{-1}A$, since it cannot contain $(1) = S^{-1}A$ (If it did, the intersection of $S$ with some ideal in $X$ would not be empty). Since $\mathfrak{p}$ is maximal in $X$ (I assume that this maximality is meant with respect to set inclusions), and $\psi$ preserves inclusions, $\psi(\mathfrak{p})$ is maximal in $\psi(X)$. But this means it is a maximal, hence prime, ideal of $S^{-1}A$.
Now my question is: Can i conclude from this that $\mathfrak{p}$ must be a maximal/prime ideal in $A$? If so, how?
The second comma in the question confused me at first, but to be clear, we assume $\frak{p}$ is maximal in $X$ (not in $J(A)$), and you show that this implies $\psi(\frak{p})$ is prime in $S^{-1}A$.
You are close to proving what you want. It is important to realise exactly what $\psi$ does; if $\phi:A\to S^{-1}A$ is the canonical map $a\mapsto\frac{a}{1}$, then $\psi$ is the map $\psi(I)=\phi(I)(S^{-1}A)$, i.e. it is the extension of the ideal. You probably know this, it is just important to be clear about.
Now, you are missing two ingredients: First, show that the contraction of $\psi(\frak{p})$, i.e. $\phi^{-1}(\psi(\frak{p}))$, is in fact $\frak{p}$, and secondly, show that the contraction of a prime ideal is always a prime ideal. This second step you can do completely elementarily, or you can do it as a one-liner with a small trick using a characterisation of prime ideals.
For reference, what you are proving is essentially the following important theorem:
For a commutative ring $A$ with multiplicative subset $S$ and $\phi:A\to S^{-1}A$ the canonical map, there is a natural (surjective) map from the ideals not intersecting $S$ in $A$ to the ideals of $S^{-1}A$, given by extension of ideals. Furthermore, this correspondence preserves prime ideals, and is bijective when restricted to this subset, the inverse given by contraction.