Having trouble with just one line in a proof on why nonzero prime ideals are maximal in a Dedekind domain

Solution 1:

If $\mathfrak{m}$ is any invertible fractional ideal in a domain $R$, then its inverse is the fractional ideal $(R:\mathfrak{m}) = \{x \in K \ | \ x \mathfrak{m} \subset R\}$.

Here the proof is saying that if $\mathfrak{p} \subset \mathfrak{m}$, we define

$\mathfrak{a} = \mathfrak{p} (R:\mathfrak{m})$.

Any element of $\mathfrak{a}$ is thus a finite sum of elements $xy$ with $x \in \mathfrak{p}$ and $y \mathfrak{m} \subset R$. Since $\mathfrak{p} \subset \mathfrak{m}$, this implies $y \mathfrak{p} \subset R$ and thus $yx \in R$. So $\mathfrak{a} \subset R$.

Note that we have not used the primality of $\mathfrak{p}$ or the maximality of $\mathfrak{m}$. This is a basic fact about invertible fractional ideals, often stated in mantra form as to contain is to divide.

[To address the OP's other question: the theorem in $\S 20.1$ of these notes shows that a domain has the property that all fractional ideals are invertible iff it is Noetherian, integrally closed of dimension one. The proof of the half you're asking about doesn't seem to use anything fancy. But, somewhat embarrassingly, in the proof I give the mantra "to contain is to divide" as though I've talked about it before, which an automated search reveals is not the case. :( So the simple argument given above seems to be missing from my notes as well. On the other hand, I do state and prove the characterization of the inverse ideal I used in my answer: see $\S 19.3$.]

Solution 2:

Maybe I'm just too sleepy this morning, but if $a$ is not a proper ideal, then $1\in a$ making $p=ma$ impossible for $p\subset m$ ......

Or I'm missing something?