I'm studying for the MGRE. The following was what I was originally going to put as my title, but it was too long and any part of it might be critical info: "Why is the intersection of all ideals, containing two elements (one of which is prime that isn't a divisor of the other), of an integral domain that is also a principal ideal also principal ideal ring?"
Prove: Let $a,b,p$ be elements of $D$, where $D$ is an integral domain which is also a principal ideal ring, and suppose $p$ is a divisor of $a b$. Then if $p$ is prime in $D$, $p$ is either a divisor of $a$ or $b$.
Solution: If either $a$ or $b$ is a unit or if $a$ or $b$ (or both) is an associate of $p$, then the theorem is trivial. Suppose the contrary and, moreover, suppose $p$ is not a divisor of $a$. Denote by $I$ the ideal in $D$ which is the intersection of all ideals in $D$ that contain $a$ and $p$. Since $I$ is a principal ideal, suppose it is generated by $c$ in $I$ so that $p = c x$ for some $x \in D$. Then either (i) $x$ is a unit in $D$ or (ii) $c$ is a unit in $D$.
(i) Suppose $x$ is a unit in $D$; then, by Theorem VIII, $p$ and its associate $c$ generate the same principle ideal $I$. Since $a$ is in $I$, we must have $a = g c = h p$ for some $g,h \in D$. But then $p$ is a divisor of $a$, a contradiction; hence $x$ is not a unit of $D$.
(ii) Suppose $c$ is a unit in $D$; then, $c^{-1} c = u$ is in $I$ and $I = D$. Now there exist $s,t \in D$ such that $u = p s + t a$, where $u$ is unity of $D$. Then $b = u b = (p s) b + (a t) b = p (s b) + a (t b)$, and, since $p$ is a divisor of $a b$, we have $p$ is a divisor of $b$ as required.
(Note: In this book, an integral domain is defined to have unity.)
It states in the solution of the problem that this intersection of ideals is a principal ideal but doesn't explain why. I don't see any theorems that say anything about this. Maybe it's straightforward and I'm too tired to see it. This is the first time I've been stumped in this book.
Also, at a later point, it says that the element that generates the intersection is a unit. And, this implies the intersection contains unity. And, that implies the intersection equals the integral domain. I understand why everything in this paragraph is true so far. But, next it says without explanation that the unity is a linear combo of the aforementioned prime and the element it is not a divisor of.
If any could please explain at least one of the two to me that would be greatly appreciated.
Edit: Sorry. The book never defined "principle ideal ring" and I assumed it meant a principle ideal endowed with D's other operator making it a ring. With this definition, it immediately cleared up my confusion for the first question.
For the second question, I just didn't think every element in D would be linear combo of a and p. It's actually still kind of messing with me conceptually. But, I am really tired. Thanks guys!
