I tried to prove this theorem :
All prime ideals of a Dedekind domain is invertible. i.e, For every prime ideal $\mathfrak{p}$ of Dedekind domain $R$, there exists $\mathfrak{p}^{-1} \subseteq Frac(R)$ such that $\mathfrak{p}\mathfrak{p}^{-1} = R$ where $Frac(R)$ is a field of fractions of $R$.
An article which I'm reading says that $\mathfrak{p}\mathfrak{p}^{-1} = R$ when $\mathfrak{p}^{-1} = \{x \in Frac(R) : x\mathfrak{p} \subseteq R\}$. The first step of the article is to prove $\mathfrak{p} \subseteq \mathfrak{p}\mathfrak{p}^{-1} \subseteq R$, and the second step is to prove $\mathfrak{p}^{-1} \subsetneq R$. This two steps proves the theorem since $\mathfrak{p}$ is a maximal ideal of $R$.
My questions are
(1) Is $\mathfrak{p}\mathfrak{p}^{-1}$ an ideal of $R$? If then, how can I prove it? If not, why does $\mathfrak{p} \subseteq \mathfrak{p}\mathfrak{p}^{-1} \subseteq R$ implies $\mathfrak{p}\mathfrak{p}^{-1} = \mathfrak{p}$ or $\mathfrak{p}\mathfrak{p}^{-1} = R$?
(2) Why does $\mathfrak{p}^{-1} \subsetneq R$ implies $\mathfrak{p}\mathfrak{p}^{-1} \neq \mathfrak{p}$?
I kept thinking about this questions for several days but I don't have any idea. Can anyone help me?
(Just to be clear - I am assuming that $\mathfrak{p}$ is a nonzero prime ideal, and I would guess this is what you meant as well.)
(1) Yes, it is an ideal.
It is not difficult to see that $\mathfrak{p}^{-1}$ is an $R$-submodule of $\mathrm{Frac}(R)$ and that such submodules are closed under multiplication - hence $\mathfrak{p}\mathfrak{p}^{-1}$ is is an $R$-submodule of $\mathrm{Frac}(R)$. It is also not difficult to see that the inclusion $\mathfrak{p}\mathfrak{p}^{-1} \subseteq R$ holds. Thus, it is an $R$-submodule of $R$, i.e. an ideal.
The inclusion $\mathfrak{p} \subseteq \mathfrak{p}\mathfrak{p}^{-1}$ comes from the fact that $1 \in \mathfrak{p}^{-1}$ (which is true since $1\mathfrak{p}=\mathfrak{p} \subseteq R$). Then the fact $\mathfrak{p}\mathfrak{p}^{-1}=\mathfrak{p}$ or $\mathfrak{p}\mathfrak{p}^{-1}=R$ follows from the above inclusion and the fact that $\mathfrak{p}$ is a maximal ideal of $R$ (recall that Dedekind domains have Krull dimension $1$).
Regarding (2):
Firstly, I would guess you meant $\mathfrak{p}^{-1}\not \subseteq R$ instead of $\mathfrak{p}^{-1} \subsetneq R$ in this context: The latter is certainly not true for any ideal $\mathfrak{p}$, so the implication is trivially true (or do you assume in this part of your question that $\mathfrak{p}$ is a general fractional ideal of $R$?). Moreover, since $\mathfrak{p}$ is an ideal, we have $1 \in \mathfrak{p}^{-1}$ and thus, $R \subseteq \mathfrak{p}^{-1}$. So we may assume even $R \subsetneq \mathfrak{p}^{-1}$.
So assume for contradiction that we have $\mathfrak{p}\mathfrak{p}^{-1}=\mathfrak{p}$. Notice that this implies that for any $a,b \in \mathfrak{p}^{-1},$ $ab \in \mathfrak{p}^{-1}$. Thus, $\mathfrak{p}^{-1}$ forms a ring containing $R$. Also, since $R$ is Noetherian and ${p}^{-1}$ its fractional ideal, it is finitely generated $R$-module.
Take any $a \in \mathfrak{p}^{-1} \setminus R$. Then we have $R \subseteq R[a] \subseteq \mathfrak{p}^{-1}$ and $\mathfrak{p}^{-1}$ is a ring finitely generated as an $R$-module, hence $a$ is integral over $R$ (by the characterization of an integral element). This is a contradiction, since $R$ is integrally closed.