I'm studying Prufer domains and I found in the book "Multiplicative Ideal Theory" by R. Gilmer the following characterization given in theorem 25.2 (this is not exactly the way the theorem is stated).
Theorem: Let $D$ be an integral domain. TFAE:
i) $D$ is a Prufer domain.
ii) For every nonzero f.g. ideals $I,J$ of $D$, $(I:J)+(J:I)=D$.
iii) For every $x,y\in D\setminus \{0\}$, $(x:y)+(y:x)=D$.
(Here $(I:J)=\{r\in R: rJ\subseteq I\}$ and $(x:y)$ means $\bigl(\langle x\rangle :\langle y\rangle\bigr)$).
Gilmer uses some extra results involving localizations of rings in order to deduce that $D$ can be assumed to be a local ring, but he doesn't prove any of those results. I proved them and it seems that in the end the proof isn't really short. So I decided to avoid those extra results about localizations and I got this alternative proof.
Proof: (i)$\implies$(ii): Let be $I,J$ nonzero f.g ideals of $D$. Let's set $$K=(I:J)+(J:I)\;...(^*)$$ so $K$ is an ideal of $D$. Our goal is to show that $K=D$. In order to prove the above we consider a maximal ideal $M$ of $D$. Localizing at $M$ and using that $D$ is Prufer gives us the valuation ring $D_M$. Now if we apply extension of ideals in (*) we get $$K_M=((I:J)+(J:I))_M=(I:J)_M+(J:I)_M=(I_M:J_M)+(J_M:I_M)\;...(**)$$
(In general it isn't true that $S^{-1}(I:J)=(S^{-1}I:S^{-1}J)$, but when $J$ is f.g. the above result becomes true).
As $D_M$ is a valuation ring, then either $I_M\subseteq J_M$ or $J_M\subseteq I_M$. In any case we deduce that the RHS of (**) is equal to $D_M$, therefore $K_M=D_M$. Since $M$ is an arbitrary maximal ideal, we can conclude that $K=D$ (this is essentially the only result that I use of localizations).
ii)$\implies$(iii): This is obvious.
iii)$\implies$(i): Let $P$ be an arbitrary prime ideal of $D$. Using a well-known characterization of Prufer domains, we only need to prove that $D_P$ is a valuation ring. In order to do that, by assuming that $K=Frac(D)$ is also the field of fractions of $D_P$ (this is possible because $D_P$ is an overring of $D$), we need to show that if $k\in K\setminus \{0\}$, then either $k$ or $k^{-1}$ belongs to $D_P$.
Let's write $k=x/y$ with nonzero $x,y\in D$. By hypothesis $(x:y)+(y:x)=D$, so applying extension of ideals at $D_P$ we get $$(x_P:y_P)+(y_P:x_P)=D_P.$$
As $D_P$ is a local ring, from the above it follows that either $(x_P:y_P)=D_P$ or $(y_P:x_P)=D_P$. Let's suppose that $(x_P:y_P)=D_P$, so in particular $1\in (x_P:y_P)$ and from this we deduce that $y/1\in x_P$. This means that there is some $\alpha\in x_P$ such that $y/1=\alpha$, but we can write $\alpha=u/s$ with $u\in \langle x\rangle$ and $s\in D\setminus P$ for at least one representation of $\alpha$. Thus, if $u=rx$ with $r\in D$, we have $$\frac{y}{1}=\frac{rx}{s}$$ $$\implies \frac{y}{x}=\frac{r}{s}\in D_P.$$
Therefore, $k^{-1}\in D_P$. Analogously, if $(y_P:x_P)=D_P$ we will deduce that $k\in D_P$. Hence, $D_P$ is a valuation ring.
So I want to know if my approach is right. Thanks in advance for comments and/or answers.
