# $R$ be a commutative ring, $x \in R$, $I$ an ideal such that $I+\langle x \rangle$ and $(I:x)$ are principal ideals, then is $I$ a principal ideal?

Let $R$ be a commutative ring, $x \in R$, $I$ be an ideal such that $I+\langle x \rangle$ and $(I:x):=\{r \in R : rx \in I\}$ are both principal ideals. Then is $I$ also a principal ideal ?

Let $J:=(I,x)=I+\langle x \rangle =(c)$ and $(I:x)=(d)$. Let $i\in I \subseteq J$, so that $i=uc$ for some $u \in R$, then $urc \in I, \forall r \in R$, so $u(c)=uJ \subseteq I$, so $ux \in I$, so $u \in (I:x)=(d)$, so $u=vd$, thus $i=uc=vdc \in (cd)$, so $I \subseteq (cd)$. For the reverse inclusion, $d \in (I:x) \implies dx \in I$ and also $dI \subseteq I$, so $dJ=d(I+\langle x \rangle)\subseteq I$, so that $d(c) \subseteq I$, so that $dc=cd \in I$, thus $(cd) \subseteq I$. Hence we conclude $(cd)=I$.
Say, we write $I+(x)=(a)$ and that $(I:x)=(b)$ for suitable $a, b \in R$. Since $(I: I + (x)) = (I:x)$, we have that $(I:a) = (b)$ as well. We claim that $I=(ab)$.
Proof of Claim: Let $i \in I$. Then, we have that $i=ra$ for some $r \in R$. Since $ra \in I$, we note that $r \in (I:a)=(b)$. In particular, $r=sb$ for some $s \in R$. We now note that $i=sba=sab$, as $R$ is commutative. Thus, $I \subseteq (ab)$. Now, via the trivial inclusion, $(I+(x)) \cdot (I:(I+(x)) \subseteq I$, we note that $ab \in I$, thus, $(ab) \subseteq I$. We are now through!