$\space$Suppose $R$ is a subring of the ring $S$ with $1\in R$ and assume $S$ is integral and finitely generated (as a ring) over R. If $P$ is a maximal ideal in $R$ then there is a nonzero and finite number of maximal ideals $Q$ of $S$ with $Q\cap R = P$.
$\space$I'm working on the verification of above corollary (Dummit and Foote, Abstract Algebra, p. 695).
$\space$In the proof described in the text, authors say "To prove that there are only finitely many possible $Q$ it suffices to prove that there are only finitely many homomorphism from $S$ to a field containing $R/P$ that extend the homomorphism from $R$ to $R/P$.", however, I can't understand the reason why this approach works.
$\space$Of course, for each maximal ideal $Q$, we can construct homomorphism from $S$ to $S/Q$ ,which is a field containing $R/P$, by natural projection. And $S/Q$ is isomorphic to one of the field extension of the form $(R/P)[\overline s_1,\overline s_2....,\overline s_n]$, and possible extension of such form is finitlely many.
$\space$ If $Q_1 \neq Q_2$ implies $S/Q_1 \ncong S/Q_2$ we have done. However, I can't prove this, and intuitively, it possibly be false... Could anyone help? Thanks.