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

Question

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?

Answer 1

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.