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Let $R$ be an integral domain. Consider the set $$S := \big\{a \in R\smallsetminus \{0\} : Ra+Rx \text{ is a principal ideal } \forall x \in R \big \}.$$ Is $S$ a saturated multiplicative closed subset of $R$? If in general $S$ is not saturated or multiplicative closed, what if we assume $R$ is a GCD domain? Is the claim true then?

Assume $R$ is local and let $x \in R$ and $a |in S$. Then there exists $d \in R $ such that $Ra+Rx=Rd$, where $d | a$ and $d|x$. Let $a'=a/d, x '= x/d$. We have then $Ra'+Rx '= R$. As $R$ is a local ring, one of $a'$ or $x'$ must be a unit, i.e., we have $a|x$ or $x|a$. Thus in a local ring $R$, the following holds $$S = \{a \in R \smallsetminus \{0\}: \text{ for every } x \in R,\, x|a \text{ or } a|x\}.$$ I can show that in any integral domain $R$, the set $\{a \in R \smallsetminus \{0\} : \text{ for every } x \in R,\, x|a \text{ or } a|x\}$ is a saturated multiplicative closed set.

I have no idea what happens if the ring is not local.

Let $R$ be an integral domain. Consider the set $$S := \big\{a \in R\smallsetminus \{0\} : Ra+Rx \text{ is a principal ideal } \forall x \in R \big \}.$$ Is $S$ a saturated multiplicative closed subset of $R$? If in general $S$ is not saturated or multiplicative closed, what if we assume $R$ is a GCD domain? Is the claim true then  ?

If we assume $R$ is local , let $x \in R$ , then $a \in S \implies \exists d \in R $ such that $Ra+Rx=Rd$  , where $d | a , d|x$ , then let $a'=a/d , x '=x/d$ , then $Ra'+Rx '=R$ , but $R$ is a local ring  , hence one of $a' $ and $x'$ must be a unit i.e. either $a|x$ or $x|a$  . Thus in a local ring $R$  ,

$\big\{a \in R\smallsetminus \{0\} : Ra+Rx \text{ is a principal ideal } \forall x \in R \big \} $

$=\{a \in R \smallsetminus \{0\} : $ for every $x \in R$ , either $x|a$ or $a|x\}$ ; and I can show that in any integral 

domain , $\{a \in R \smallsetminus \{0\} : $ for every $x \in R$ , either $x|a$ or $a|x\}$ is a saturated multiplicative closed set  .

I have no idea what happens if the ring is not local  .

Let $R$ be an integral domain. Consider the set $$S := \big\{a \in R\smallsetminus \{0\} : Ra+Rx \text{ is a principal ideal } \forall x \in R \big \}.$$ Is $S$ a saturated multiplicative closed subset of $R$? If in general $S$ is not saturated or multiplicative closed, what if we assume $R$ is a GCD domain? Is the claim true then  ?

If we assume $R$ is local , let $x \in R$ , then $a \in S \implies \exists d \in R $ such that $Ra+Rx=Rd$  , where $d | a , d|x$ , then let $a'=a/d , x '=x/d$ , then $Ra'+Rx '=R$ , but $R$ is a local ring  , hence one of $a' $ and $x'$ must be a unit i.e. either $a|x$ or $x|a$  . Thus in a local ring $R$  ,

$\big\{a \in R\smallsetminus \{0\} : Ra+Rx \text{ is a principal ideal } \forall x \in R \big \} $

$=\{a \in R \smallsetminus \{0\} : $ for every $x \in R$ , either $x|a$ or $a|x\}$ ; and I can show that in any integral

  domain , $\{a \in R \smallsetminus \{0\} : $ for every $x \in R$ , either $x|a$ or $a|x\}$ is a saturated multiplicative closed set  .

I have no idea what happens if the ring is not local  .

Let $R$ be an integral domain. Consider the set $$S := \big\{a \in R\smallsetminus \{0\} : Ra+Rx \text{ is a principal ideal } \forall x \in R \big \}.$$ Is $S$ a saturated multiplicative closed subset of $R$? If in general $S$ is not saturated or multiplicative closed, what if we assume $R$ is a GCD domain? Is the claim true then?

Assume $R$ is local and let $x \in R$ and $a |in S$. Then there exists $d \in R $ such that $Ra+Rx=Rd$, where $d | a$ and $d|x$. Let $a'=a/d, x '= x/d$. We have then $Ra'+Rx '= R$. As $R$ is a local ring, one of $a'$ or $x'$ must be a unit, i.e., we have $a|x$ or $x|a$. Thus in a local ring $R$, the following holds $$S = \{a \in R \smallsetminus \{0\}: \text{ for every } x \in R,\, x|a \text{ or } a|x\}.$$ I can show that in any integral domain $R$, the set $\{a \in R \smallsetminus \{0\} : \text{ for every } x \in R,\, x|a \text{ or } a|x\}$ is a saturated multiplicative closed set.

I have no idea what happens if the ring is not local.

Let $R$ be an integral domain. Consider the set $$S := \big\{a \in R\smallsetminus \{0\} : Ra+Rx \text{ is a principal ideal } \forall x \in R \big \}.$$ Is $S$ a saturated multiplicative closed subset of $R$? If in general $S$ is not saturated or multiplicative closed, what if we assume $R$ is a GCD domain? Is the claim true then?

Let $R$ be an integral domain. Consider the set $$S := \big\{a \in R\smallsetminus \{0\} : Ra+Rx \text{ is a principal ideal } \forall x \in R \big \}.$$ Is $S$ a saturated multiplicative closed subset of $R$? If in general $S$ is not saturated or multiplicative closed, what if we assume $R$ is a GCD domain? Is the claim true then  ?

If we assume $R$ is local , let $x \in R$ , then $a \in S \implies \exists d \in R $ such that $Ra+Rx=Rd$ , where $d | a , d|x$ , then let $a'=a/d , x '=x/d$ , then $Ra'+Rx '=R$ , but $R$ is a local ring , hence one of $a' $ and $x'$ must be a unit i.e. either $a|x$ or $x|a$ . Thus in a local ring $R$ ,

$\big\{a \in R\smallsetminus \{0\} : Ra+Rx \text{ is a principal ideal } \forall x \in R \big \} $

$=\{a \in R \smallsetminus \{0\} : $ for every $x \in R$ , either $x|a$ or $a|x\}$ ; and I can show that in any integral

domain , $\{a \in R \smallsetminus \{0\} : $ for every $x \in R$ , either $x|a$ or $a|x\}$ is a saturated multiplicative closed set .

I have no idea what happens if the ring is not local .