If a principal ideal is a multiple of another ideal, is it necessarily a principal multiple?

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Question. Let $R$ denote a commutative ring. Suppose $fraka$ and $frakb$ are principal ideals of $R$. Suppose there exists a (not-necessarily principal) ideal $frakj$ of $R$ satisfying $frakja=b.$ Does there necessarily exist a principal ideal $frakj$ such that $frakja=b$? If the answer is "no", are there assumptions on $R$ weaker than being a principal ideal ring that make the answer "yes"?




ring-theory commutative-algebra ideals




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    Question. Let $R$ denote a commutative ring. Suppose $fraka$ and $frakb$ are principal ideals of $R$. Suppose there exists a (not-necessarily principal) ideal $frakj$ of $R$ satisfying $frakja=b.$ Does there necessarily exist a principal ideal $frakj$ such that $frakja=b$? If the answer is "no", are there assumptions on $R$ weaker than being a principal ideal ring that make the answer "yes"?




    ring-theory commutative-algebra ideals




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      up vote
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      favorite









      up vote
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      down vote

      favorite











      Question. Let $R$ denote a commutative ring. Suppose $fraka$ and $frakb$ are principal ideals of $R$. Suppose there exists a (not-necessarily principal) ideal $frakj$ of $R$ satisfying $frakja=b.$ Does there necessarily exist a principal ideal $frakj$ such that $frakja=b$? If the answer is "no", are there assumptions on $R$ weaker than being a principal ideal ring that make the answer "yes"?




      ring-theory commutative-algebra ideals




      share|cite|improve this question











      Question. Let $R$ denote a commutative ring. Suppose $fraka$ and $frakb$ are principal ideals of $R$. Suppose there exists a (not-necessarily principal) ideal $frakj$ of $R$ satisfying $frakja=b.$ Does there necessarily exist a principal ideal $frakj$ such that $frakja=b$? If the answer is "no", are there assumptions on $R$ weaker than being a principal ideal ring that make the answer "yes"?





      ring-theory commutative-algebra ideals





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      asked Jul 31 at 5:30









      goblin

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          Let $mathfrak a = (a)$ and $mathfrak b = (b)$.



          $mathfrak b = mathfrak j mathfrak a subseteq mathfrak a$, so $b = ka$ for some $k$.



          Therefore, $mathfrak b = (b) = (ka) = (k) (a) = (k) mathfrak a$.






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            up vote
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            accepted










            Let $mathfrak a = (a)$ and $mathfrak b = (b)$.



            $mathfrak b = mathfrak j mathfrak a subseteq mathfrak a$, so $b = ka$ for some $k$.



            Therefore, $mathfrak b = (b) = (ka) = (k) (a) = (k) mathfrak a$.






            share|cite|improve this answer

























              up vote
              1
              down vote



              accepted










              Let $mathfrak a = (a)$ and $mathfrak b = (b)$.



              $mathfrak b = mathfrak j mathfrak a subseteq mathfrak a$, so $b = ka$ for some $k$.



              Therefore, $mathfrak b = (b) = (ka) = (k) (a) = (k) mathfrak a$.






              share|cite|improve this answer























                up vote
                1
                down vote



                accepted







                up vote
                1
                down vote



                accepted






                Let $mathfrak a = (a)$ and $mathfrak b = (b)$.



                $mathfrak b = mathfrak j mathfrak a subseteq mathfrak a$, so $b = ka$ for some $k$.



                Therefore, $mathfrak b = (b) = (ka) = (k) (a) = (k) mathfrak a$.






                share|cite|improve this answer













                Let $mathfrak a = (a)$ and $mathfrak b = (b)$.



                $mathfrak b = mathfrak j mathfrak a subseteq mathfrak a$, so $b = ka$ for some $k$.



                Therefore, $mathfrak b = (b) = (ka) = (k) (a) = (k) mathfrak a$.







                share|cite|improve this answer













                share|cite|improve this answer



                share|cite|improve this answer











                answered Jul 31 at 5:41









                Kenny Lau

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