On domain in which all right finitely generated ideals are principal

Question

Let $R$ - a domain in which all right finitely generated ideals are principal. I want to prove that $aR \cap bR \ne 0$ for all $a, b \ne 0$.

My idea is the following. Let $(a)$ and $(b)$ - two right principal ideals. We consider $(a) + (b) = \{ \, ar + bs \mid r,s \in R \, \}$. From the condition "in which all right finitely generated ideals are principal" it follows that $(a) + (b) = (c)$, for some $c \in R$, i.e. $ar + bs = c t$, $t \in R$. How to prove from here using about domain $R$ that $aR \cap bR \ne 0$?

Answer 1

Suupose that $c \in aR \cap bR$, then $c= as = br$, for some $r,s \in R$. Moreover, since finite right ideals are principal, $(a,b) = (a)+(b) = (d)$ for some $ d \in R$, hence we may assume that $a = ds',b=dr'$ and $c = dss'=drr' \implies s's=r'r$ since $R$ is a domain. In fact, then converse is also true: for any $\rho,\sigma \in R$, if they satisfy $r'\rho = s'\sigma $, then $a \rho = b \sigma \in aR \cap bR$.

Assume that $aR\cap bR =0$, then the only possible pair $(\rho,\sigma)$ is $(0,0)$, as $R$ is domain. However, it can be shown that there is some $\alpha, \beta \in R-\{0\}$, such that $d = a \alpha +b \beta = d(s'\alpha+r'\beta) \implies s'\alpha + r'\beta = 0$, which is a contradiction.

Answer 2

It seems like you aren't using the precise results of what it means for the sum of two principal right ideals to be principal.

If you know $aR + bR=cR$, then right away you know two things:

• $c=ar+bs$ for some $r,s\in R$, because $cR\subseteq aR + bR$
• $cr'=a$, $b=cs'$ for some nonzero $r',s'\in R$, because $aR+bR\subseteq cR$

Note that in the first bullet, if $bs=0$, then $bR\subseteq aR=cR$ and the intersection is clearly nonzero. Without loss of generality we may suppose $bs$ is nonzero.

Combining the two bullet points, $a=cr'=arr'+bsr'$, and rewriting, $a(1-rr')=bsr'$. We're in a domain, so the product of $bs$ and $r'$ is nonzero, and this witnesses $aR\cap bR\neq \{0\}$.

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For future readers and searchability, this problem is, in other words, showing that

a right Bezout domain is a right Ore domain.