# Thread: Quasi-local domain is a PID if its localizations at maximal ideals are PIDS

1. ## Quasi-local domain is a PID if its localizations at maximal ideals are PIDS

Let $\displaystyle A$ be a quasi-local domain (i.e. a commutative domain with finitely many maximal ideals). Suppose that for every maximal ideal $\displaystyle M$ of $\displaystyle A$, the localization $\displaystyle A_M$ is a P.I.D. Show that $\displaystyle A$ is a P.I.D.

2. I think most texts define quasi-local to mean that there's only one maximal ideal. Do you mean semi-quasilocal? What's the reference for this problem?

3. Thank you for your interest. A local ring is a ring having a single maximal ideal. A quasi-local ring is a ring having a finite number of maximal ideals.

The reference is Lorenzini's "An introduction to arithmetic geometry", AMS.

4. Specifically, where in Lorenzini's book is this problem?

5. Chapter III, problem 3 on page 126.

6. I'm still thinking about this but here's my approach. First consider only local domains for which there's only one maximal ideal. Let $\displaystyle I\subset A$ be an ideal. Then the expansion $\displaystyle I^E$ of $\displaystyle I$ to $\displaystyle A_M$ is principal, by assumption. Let $\displaystyle a/s$ be its generator, where $\displaystyle a\in I$ and $\displaystyle s\in S=A\setminus M$. Now $\displaystyle a$ is a candidate for the generator of $\displaystyle I$ but now to prove it. Obviously the maximality of $\displaystyle M$ is critical.

7. OK, I think I have a proof. It may not be the most direct, however. We know that any local domain who's maximal ideal is principal must be a discrete valuation ring. Now, let $X$ be the maximal spectrum of $A$. Then,

$\displaystyle A=\cap_{M\in X}A_M.$

Since each $\displaystyle A_M$ is a DVR and the intersection is finite, we have that $\displaystyle A$ is a PID.

8. Originally Posted by ojones
OK, I think I have a proof. It may not be the most direct, however. We know that any local domain who's maximal ideal is principal must be a discrete valuation ring. Now, let $X$ be the maximal spectrum of $A$. Then,

$\displaystyle A=\cap_{M\in X}A_M.$

Since each $\displaystyle A_M$ is a DVR and the intersection is finite, we have that $\displaystyle A$ is a PID.
Why is the statement $\displaystyle A=\cap_{M\in X}A_M.$ true?

9. This is proposition 9.7 (page 74) in Lorenzini's book.