an example of a Noetherian domain with finitely many non-principal maximal ideals [closed]

Let $F$ be an algebraically closed field, and consider the ring $F[X, Y]$ of polynomials over $F$ in two indeterminates $X$ and $Y$. Let $S$ be the multiplicatively closed set in $F[X, Y]$ generated by polynomials of the form $X−α$ and $Y−β$ for all non-zero elements $α, β \in F$, and set $R=S^{-1}F[X,Y]$ a Noetherian domain of Krull dimension two. The only non-principal maximal ideal of $R$ is $M=S^{-1}(X,Y)$, while $R$ has infinitely many principal (height one) maximal ideals.

Why The only non-principal maximal ideal of $R$ is $M=S^{-1} (X,Y)$?

why $R$ has infinitely many principal (height one) maximal ideals?

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closed as off-topic by Graham Leuschke, Daniel Moskovich, Andrés Caicedo, Andrey Rekalo, Theo Johnson-FreydJul 16 '13 at 3:08

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question does not appear to be about research level mathematics within the scope defined in the help center." – Graham Leuschke, Daniel Moskovich, Andrés Caicedo, Andrey Rekalo
If this question can be reworded to fit the rules in the help center, please edit the question.

What makes you think these things are true? – Steven Landsburg Jul 15 '13 at 6:17
This is a quote from the following paper: ams.org/journals/tran/2011-363-07/S0002-9947-2011-05249-9 When quoting works, it is very important to cite the source. – Karl Schwede Jul 15 '13 at 6:38
This question appears to be off-topic because neither cites its sources nor provides motivation for the question. – Theo Johnson-Freyd Jul 16 '13 at 3:08

I'm not sure if this question is appropriate for this site, but I'm going to give it the benefit of the doubt.

Look up an introductory text on commutative algebra and see what the prime ideals of $S^{-1} A$ correspond to in terms of $A$. ($A$ is a ring and $S$ is a multiplicative system).

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