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

This article gives the statement, and possibly proof, of a fact about how a property of a homomorphism of commutative unital rings, forces a property for the induced map on spectra
View other facts about induced maps on spectra

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.