Integral extension implies inverse image of max-spectrum is max-spectrum

Statement
Suppose $$f:R \to S$$ is an integral extension of commutative unital rings. Then consider the induced map on spectra:

$$f^*: Spec(S) \to Spec(R)$$

that sends a prime ideal $$P$$ of $$S$$ to its contraction $$P^c = f^{-1}(P)$$ in $$R$$.

Then the following are true:


 * If $$\mathfrak{m}$$ is a maximal ideal of $$S$$, then $$f^*(\mathfrak{m})$$ is maximal in $$R$$
 * If $$f^*(P)$$ is maximal in $$R$$, then $$P$$ is maximal in $$S$$

Proof
The key idea here is the following fact: for a finite extension of integral domains, one of them is a field if and only if the other is.