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I'm reading a paper called Overrings of Prufer domains by Gilmer (available on MathSciNet), and other than this small point in the proof of (i)$\implies$(ii) in Theorem 2, everything else has become clear to me after some thought. The bit in question is on page 336. Would you please help me see why this is true?

The statement we want to show is in the title: If $D$ is a one-dimensional Prufer domain and if $\Delta$ is the set of maximal ideals of $D$, then given $M_{\alpha}\in\Delta$, we want to show $D_{M_{\alpha}}\nsupseteq\cap_{\beta\neq\alpha}D_{M_{\beta}}$ implies that $M_{\alpha}$ is the radical of an ideal with two generators.

Here is the proof I have, mostly from Gilmer except with some details clarified (apologies about the poor formatting and extraneous dollar signs; this site isn't cooperating). I have boldfaced/italicized my point of confusion: Gilmer comes up with this element $u$, and its reason for existence eludes me.

Let $\Delta ' =\{ > M_{\lambda}\}=\Delta-\{ M_{\alpha}\}$, and let $v_{\beta}$ be the valuation associated with $D_{M_{\beta}}$ for each $\beta$. Since $D_{M_{\alpha > }}\nsupseteq\cap D_{M_{\lambda}}$, there exist $a,b\in D$ such that $a/b\in \cap D_{M_{\lambda}} - > D_{M_{\alpha }}$, so $v_{\alpha}(a)< > v_{\alpha}(b)$ and $v_{\lambda}(a)\geq > v_{\lambda}(b)$ for each $\lambda$. Hence, $v_{\alpha}(b/a)>0$ (so, it's a nonunit in $D_{M_{\alpha}}$) and $b/a\in M_{\alpha}D_{M_{\alpha}}$; say $b/a=s/t$ where $s\in M_{\alpha}$, $t\in D-M_{\alpha}$. Then $v_{\alpha}(t)=0 < v_{\alpha}(s)$ and $v_{\lambda}(t)\geq v_{\lambda}(s)$ for each $\lambda$ (since that was the case for $a$ and $b$). We now let $\Omega '$ be the set of $M_{\lambda}$'s which contain $s$ and we let $\Omega = \Omega '\cup\{ > M_{\alpha}\}$. Thus, $\Omega$ is the set of maximal ideals of $D$ containing $s$. Claim: If $P\in\Delta$ and $P\subseteq\cup_{T\in\Omega}T$, then $P\in\Omega$. Proof of claim: Indeed, if $P\notin\Omega$ (i.e. $s\notin P$), then $p+ds=1$ for some $p\in P$, $d\in D$; it then follows that $p\notin T$ for $T\in\Omega$, so $p\in P-(\cup_{T\in\Omega}T)$. Thus, the claim is proved. This observation shows that if $N=D-(\cup_{T\in\Omega}T)$ and if $D'=D_N$, then $\{TD'\}_{T\in\Omega }$ is the set of maximal ideals of $D'$, and hence $D'$ is one-dimensional (by hypothesis) and Prufer (since overrings of Prufer domains are Prufer), so

$D'_{M_{\alpha}D'} = > D_{M_{\alpha}}\nsupseteq\cap_{T\in\Omega > '}D'_{TD'}=\cap_{T\in\Omega '}D_T$

by hypothesis. Also, if $M_{\beta}\in\Omega$, $D_{M_{\beta}}=D'_{M_{\beta}D'}$$ > implies that $v_{\beta}$ is the > valuation associated with > $D'{M{\beta}D'}$$. For each such $M_{\beta}\in\Omega '$ we then have $v_{\beta}(t)\geq v_{\beta}(s)>0$ while $v_{\alpha}(s)>v_{\alpha}(t)=0$, which implies by the proof of Lemma 7 ($s$ is in the Jacobson radical of $D'$) and by work done earlier in the proof, there exists $u\in M_{\alpha}D'-(\cup_{T\in\Omega '}TD')$. I do not understand where Gilmer got this u from. There is no loss of generality in assuming $u\in M_{\alpha} - \cup_{T\in\Omega '}T$. To show that the ideal $B=(s,u)$ in $D$ has radical $M_{\alpha}$, clearly $B\subseteq M_{\alpha}$, and we show it is not contained in any other prime/maximal ideals (recall $D$ is one-dimensional). If $M_{\lambda}\in\Delta ' -\Omega '$, then $s\notin M_{\lambda}$ so $B\nsubseteq M_{\lambda}$; moreover, $M_{\lambda}\in\Omega '$ implies that $u\notin M_{\lambda}D'$, implying $u\notin M_{\lambda}$, again implying $B\nsubseteq M_{\lambda}$. Therefore, $\hbox{rad}(B)=M_{\alpha}$, which is what we wanted to show.

Lemma 7 and its proof are as follows:

Lemma 7: Suppose $D$ is one-dimensional and that its Jacobson radical is nonzero. If > $\Delta=\{ M_{\beta}\}$ and if $M_{\alpha}\in\Delta$ is the radical of an ideal with two > generators, then there exists $m_{\alpha}\in M_{\alpha}$ such that $1-m_{\alpha}\in > M_{\beta}$ for each $\beta\neq\alpha$.

Proof: Let $x$ be a nonzero element of the Jacobson radical. If $v_{\beta}$ is the > valuation associated with $D_{M_{\beta}}$ for each $\beta$, then $v_{\beta}(x)>0$ for each $\beta$. By hypothesis, there exist $u, t\in M_{\alpha}$ such that $M_{\alpha}$ is the radical of $(u,t)$. Since $v_{\alpha}$ is rank one, there's some integer $n$ such that $v_{\alpha}(u^n)> v_{\alpha}(x)$ and $v_{\alpha}(t^n)>v_{\alpha}(x)$. If we let $B=(t^n, u^n, x)$, then $\sqrt B=M_{\alpha}$ and the minimal $v_{\alpha}$-value of an element of $B$ is $v_{\alpha}(x)$. Since $B$ is invertible ($D$ is Prufer), we have $(x)=AB$ for some ideal $A$ of $D$. So, since $A$ is invertible and $v_{\alpha}$ must attain a minimal value on $A$, the minimal value must be zero, so $A\nsubseteq M_{\alpha}$, but for $\beta\neq\alpha$, $AB=(x)\subseteq M_{\beta}$ while $B\nsubseteq M_{\beta}$, so (since the $M_{\beta}$ are prime), $A\subseteq\cap_{\beta\neq\alpha}M_{\beta}$. If we choose $a\in A-M_{\alpha}$, then for some $m_{\alpha}\in M_{\alpha}$ and $d\in D$, $m_{\alpha}+da=1$. So, for $\beta\neq\alpha$, $1-m_{\alpha}=da\in M_{\beta}$.

I find it bizarre that Gilmer uses the proof of Lemma 7, because it heavily uses its hypothesis that the maximal ideal it considers is the radical of an ideal with two generators, a hypothesis we don't have in the proof of Theorem 2. I can't see how anything can be extracted from that proof given we don't have that hypothesis. Do you see how we can use Lemma 7's proof?

Thanks so much! I realize this is long-winded and requires a lot of patience to read through. He whosoever delivers an explanation deserveth a cookie! :)

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