# On domain in which all right finitely generated ideals are principal

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 $$aR \cap bR \ne 0$$?

• "finite" of "finitely generated" ? Commented May 31 at 12:04
• Finitely generated, of course. My typo. I fixed in post. Commented May 31 at 12:33
• Don't use mathjax to typeset italics. Use mark-up. Commented May 31 at 14:04

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.

• I think you meant $c = as = br$ and $c = d s's = d r' r$. But why $a = d s'$ and $b = d r'$? Commented May 31 at 14:12
• @Irene Yes I meant $c=as=br$. For your question, we must have that $a ,b \in aR+bR = (a,b) = dR$, hence $d |a,b$.
– Ubik
Commented May 31 at 14:17

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 $$aR\cap bR\neq \{0\}$$.

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

a right Bezout domain is a right Ore domain.