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I'm stuck on the proof of this result:

Let $P \in \mathrm{Spec}(R)$ and $D = R \setminus P$. Show that $PD^{-1}$ is a maximal ideal in $RD^{-1}$.

Lemma. $D$ is a multiplicatively closed subset of $R$ with no zero divisors. So the ring of fractions $PD^{-1}$ exists.

Theorem. $PD^{-1}$ is a maximal ideal in $RD^{-1}$.

Proof. $PD^{-1}$ is a maximal ideal iff the only ideals containing $PD^{-1}$ are $PD^{-1}$ and $RD^{-1}$. Let $I$ be an ideal containing $PD^{-1}$ that is not equal to $PD^{-1}$.

I need to find an element $e \in I$ so that $e$ is a unit. That way, I can show that $I$ must be $RD^{-1}$. But I'm having trouble constructing the example element that is invertible. Could you give help?

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    What is $P$?... – 2017-02-08
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    It's in the title: P is a prime ideal of R. – 2017-02-08
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    This result follows easily from the *correspondence* betwen prime ideals of $R$ and prime ideals of $RD^{-1}$. Do you know that theorem? – 2017-02-08
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    Localization commutes with quotients $RD^{-1} / PD^{-1} \cong (R/P)D^{-1}$. Now show that this quotient is a field. – 2017-02-08
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    @Watson your "answer" didn't answer my question. Did you even read my question? I asked how could I construct an invertible element in a step of this proof. – 2017-02-08
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    @TomislavOstojich : 1) yes, I read your question 2) my comment was only here to provide a different approach 3) If $I$ contains strictly $PD^{-1}$, then you can find $x/y \in I$ such that $x \not \in P$, which is a unit in $RD^{-1}$. – 2017-02-08

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We have $P_D:=\{\frac{p}{s}; p\in P, s\in D \}$.

To see that $P_D$ is a maximal ideal, note that an element of $R_D$ not in $P_D$ is the form $\frac{s'}{s}; s'\in S$ and so has an inverse $\frac{s}{s'}$ and is a unit.