2
$\begingroup$

Let $R$ be a Prufer domain. If $0 \ne a \in R$ is such that $Ra \cap Rb$ is principal ideal for every $b \in R$, then is it true that $Ra+Rb$ is also principal for every $b\in R$ ?

Over Prufer domains, torsion-free modules are flat , so if $K$ is the fraction field of $R$ then any subring $S$ of $K$ containing $R$, is a flat $R$-module, hence $S$ over $R$ has Going Down property . So this On GCD and LCM of elements in integral domains which has the property that any over ring has Going Down is related.

$\endgroup$
4
+50
$\begingroup$

In fact, for any given nonzero $a$ and $b$, if $Ra\cap Rb$ is principal so is $Ra+Rb$. Here is one way to see it (surely there must be a more down-to-earth proof). Without assuming $Ra\cap Rb$ principal, we have an exact sequence of $R$-modules $$\begin{array}{ccccccccc} 0&\longrightarrow&Ra\cap Rb&\longrightarrow& Ra\oplus Rb&\longrightarrow&Ra+Rb&\longrightarrow&0\\ &&x&\longmapsto &(x,-x)\\ &&&&(x,y)&\longmapsto & x+y \end{array}$$ where, because $R$ is a Prüfer domain, both $Ra\cap Rb$ and $Ra+ Rb$ are locally free of rank 1 (while $Ra\oplus Rb$ is of course free). So the sequence splits and, taking determinants, we conclude that $Ra\cap Rb$ and $Ra+ Rb$ are inverse to each other in $\mathrm{Pic}(R)$. In particular, $Ra+ Rb$ is free if $Ra\cap Rb$ is. QED

$\endgroup$
  • 1
    $\begingroup$ You don't need to go to Picard group ... $M \oplus R \cong R^2$ already implies $M$ is free of rank $1$. Also, that the sequence splits can be seen from the fact that $Ra+Rb$ is a finitely generated ideal , hence projective because the domain is Prufer. $\endgroup$ – user111524 Mar 30 '18 at 17:40
2
$\begingroup$

Here is a constructive proof (though I am not saying that the proofs given so far have been non-constructive; I have not tracked down their dependencies).

I will rely on the book Henri Lombardi, Claude Quitté, Commutative algebra: Constructive methods. Finite projective modules, arXiv:1605.04832v2 (translated by Tania K. Roblot, published by Springer 2015, but the link is to an updated version gracefully posted on the arXiv). Specifically, I will need the following fact from that book:

(1) If $R$ is a Prüfer ring (not necessarily a Prüfer domain!), and if $\mathfrak{a}$ and $\mathfrak{b}$ are two finitely generated ideals of $R$, then $\mathfrak{a} \mathfrak{b} = \left(\mathfrak{a} \cap \mathfrak{b}\right) \left(\mathfrak{a} + \mathfrak{b}\right)$.

This is part of Exercise 10 point 3 in Chapter VIII of the above book.

Here is your claim, slightly extended (as Laurent pointed out, it is not necessary to all-quantify over $b$ in the assumption):

Proposition 1. Let $R$ be a Prüfer domain. Let $a \in R$ and $b \in R$ be such that the ideal $Ra \cap Rb$ of $R$ is principal. Then, the ideal $Ra + Rb$ of $R$ also is principal.

Proof of Proposition 1. We WLOG assume that $a \neq 0$, since otherwise the claim is obvious (because $a = 0$ entails $Ra + Rb = Rb$). For similar reasons, we WLOG assume that $b \neq 0$. The ideal $Ra \cap Rb$ of $R$ is principal; in other words, $Ra \cap Rb = Rc$ for some $c \in R$. Consider this $c$. The element $ab$ of $R$ belongs to both $Ra$ and $Rb$, and thus belongs to $Ra \cap Rb = Rc$. In other words, $ab = cx$ for some $x \in R$. Consider this $x$. We shall show that $Ra + Rb = Rx$.

Indeed, (1) (applied to $\mathfrak{a} = Ra$ and $\mathfrak{b} = Rb$) yields $\left(Ra\right)\left(Rb\right) = \underbrace{\left(Ra \cap Rb\right)}_{=Rc} \left(Ra + Rb\right) = Rc \left(Ra + Rb\right) = c\left(Ra + Rb\right)$. Thus, $c\left(Ra + Rb\right) = \left(Ra\right)\left(Rb\right) = R\underbrace{ab}_{=cx} = Rcx = c\left(Rx\right)$.

But $R$ is an integral domain. Hence, from $a \neq 0$ and $b \neq 0$, we obtain $ab \neq 0$, so that $cx = ab \neq 0$ and thus $c \neq 0$. Therefore, if two ideals $I$ and $J$ of $R$ satisfy $cI = cJ$, then $I = J$ (to prove this, just argue elementwise, again using the fact that $R$ is an integral domain). Applying this to $I = Ra + Rb$ and $J = Rx$, we conclude that $Ra + Rb = Rx$ (since $c\left(Ra + Rb\right) = c\left(Rx\right)$). Hence, the ideal $Ra + Rb$ is principal. $\square$

Remark. The above proof still works if we replace the condition that $R$ be a domain by the condition that $a$ and $b$ be non-zero-divisors.

$\endgroup$
1
$\begingroup$

Here is an alternative proof of

Claim. Let $R$ be a Prüfer domain and let $a, b \in R$. Then $Ra + Rb$ is a principal ideal if and only if $Ra \cap Rb$ is.

If $R$ is any commutative domain with identity, it is easily checked that $Ra \cap Rb$ is principal if $Ra + Rb$ is. Note that the localization of a Prüfer domain at a prime ideal is a valuation domain, hence a Bézout domain.

Proof. It suffices to show that $(Ra + Rb)(Ra \cap Rb) = Rab$. By localisation at a maximal ideal, we can assume that $R$ is a Bézout domain. In this case, the identity is immediate.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy