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In the PlanetMath article above, in the second paragraph of the proof of the first lemma, it says

[...] let $\mathfrak{p}$ be a prime ideal, and $\mathfrak{m}$ be a maximal ideal containing $\mathfrak{p}$. As $\mathfrak{m}$ is invertible, there exists an ideal $\mathfrak{a}$ such that $\mathfrak{p} = \mathfrak{m}\mathfrak{a}$.

My question is the following: why is $\mathfrak{a}$ an (integral) ideal of $R$ and not a fractional ideal of its field of fractions $k$? (So $\mathfrak{a}$ can potentially be not fully inside $R$.) This fact is crucial for the next line, in which primality of $\mathfrak{p}$ is used to establish that either $\mathfrak{a} \subseteq \mathfrak{p}$ or $\mathfrak{m} \subseteq \mathfrak{p}$.

Any help would be greatly appreciated, and the sooner the better. Many thanks in advance!

(Here is an alternative way to help: Present another proof of the fact that nonzero prime ideals are maximal in a Dedekind domain that does not assume material beyond a first course in abstract algebra. The definition we use for "Dedekind domain" is every nonzero fractional ideal is invertible, which we proved equivalent to the condition that every proper nonzero ideal factors uniquely into prime ideals; the proof given in Dummit and Foote also has a line that I don't see, so it is best to avoid that one.)

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+1 for providing the definition you're using :-) – joriki Aug 5 '11 at 9:02
@Pierre-Yves: good idea! – Andrea Mori Aug 5 '11 at 11:09

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?

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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 $\S 19.3$.]

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Thank you very much for your insightful answer. I now have much more intuition about ring ideals in general; spot-on explanations! – vwxf Aug 5 '11 at 17:38

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