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.)