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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 ?

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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 $(cd)=I$.

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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!

(This can be used to deduce that an ideal maximal with respect to not being principal is prime. In particular, if every prime ideal in a commutative ring is principal, then, so is every ideal.)

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