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Let $A$ be a UFD, we know $A$ is a PID iff $A$ is a Bezout domain. Now I want to consider the case $A$ is not a PID.

So the question is that does every UFD $A$ which is not PID admit two distinct prime elements $a,b$ such that $(a,b)\neq A.$

qinxs
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1 Answers1

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I don't know the answer, but I will take a stab at the proof myself. Hope that this proof is either correct, or someone will be able to point to a mistake, in which case I will pull it out.

Lemma: If $p,q,r$ are three elements of a commutative ring $R$ with unity such that $(p,r)=(q,r)=R$, then $(pq,r)=R$ as well.

Proof: if $1=ap+br=cq+dr$ for some $a,b,c,d\in R$, then:

$$q=apq+brq$$ $$1=cq+dr=c(apq+brq)+dr=(ac)pq+(bq+d)r$$ $\blacksquare$

From there, using induction and symmetry, we can prove our statement.

Let $R$ be UFD which is not a Bezout domain. This means that there are two elements $p,q$ such that $(p,q)$ is not principal.

WLOG we can assume that $p,q$ are coprime. Otherwise, take $d=\gcd(p,q)$ and then $p/d, q/d$ are coprime and the ideal $(p/d, q/d)$ is already not principal.

Now, let $p=p_1\cdots p_n, q=q_1\cdots q_m$ where all $p_i, q_j$ are prime and none of the $p_i$'s is the same as any of the $q_j$'s. Now:

  • Either one of $(p_i, q_j)$ is a proper ideal, in which case we are done. (That ideal cannot be principal because $p_i,q_j$ are already coprime.)
  • Or, all $(p_i, q_j)=R$. Now, using the Lemma above multiple times we can conclude that $(p,q)=(p_1\cdots p_n,q_1\cdots q_m)=R$, which is a contradiction with the assumption that $(p,q)$ is not principal.

$\blacksquare$