Property of quotient ideals in a number Field

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Let $mathbbK$ be an algebraic number field.



Let $I$ and $J$ be two fractional ideals of $mathbbK$ and $q in mathbbN$ a positive integer.



Is it true that $I/qI $ isomorphic to $J/qJ$? If yes, what is the isomorphism?



If no, is there a condition on $q$ that make it True?




linear-algebra ring-theory field-theory algebraic-number-theory ideals




share|cite|improve this question



















  • Isomorphic in what sense? As abelian groups? As $mathcalO_mathbbK$-modules?
    – Gal Porat
    Jul 20 at 15:21











  • Besides, it will certainly not be true in any reasonable sense if you take $mathbbK = mathbbQ$, $I=(0), J=(1)$
    – Gal Porat
    Jul 20 at 15:22










  • What happen If $mathbbK$ is a cyclotomic number field?, and the ideals are not trivials.
    – C.S.
    Jul 20 at 16:37










  • But you have still not specified what you mean by your question...
    – Gal Porat
    Jul 20 at 17:48










  • No condition depending only on $q$ seems to exist as both $I/qI$ and $J/qJ$ are annihilated by $q$. There must be conditions on $I$ and $J$ to have isomorphism in certain circumstances.
    – joy
    Jul 23 at 8:23














up vote
0
down vote

favorite












Let $mathbbK$ be an algebraic number field.



Let $I$ and $J$ be two fractional ideals of $mathbbK$ and $q in mathbbN$ a positive integer.



Is it true that $I/qI $ isomorphic to $J/qJ$? If yes, what is the isomorphism?



If no, is there a condition on $q$ that make it True?




linear-algebra ring-theory field-theory algebraic-number-theory ideals




share|cite|improve this question



















  • Isomorphic in what sense? As abelian groups? As $mathcalO_mathbbK$-modules?
    – Gal Porat
    Jul 20 at 15:21











  • Besides, it will certainly not be true in any reasonable sense if you take $mathbbK = mathbbQ$, $I=(0), J=(1)$
    – Gal Porat
    Jul 20 at 15:22










  • What happen If $mathbbK$ is a cyclotomic number field?, and the ideals are not trivials.
    – C.S.
    Jul 20 at 16:37










  • But you have still not specified what you mean by your question...
    – Gal Porat
    Jul 20 at 17:48










  • No condition depending only on $q$ seems to exist as both $I/qI$ and $J/qJ$ are annihilated by $q$. There must be conditions on $I$ and $J$ to have isomorphism in certain circumstances.
    – joy
    Jul 23 at 8:23












up vote
0
down vote

favorite









up vote
0
down vote

favorite











Let $mathbbK$ be an algebraic number field.



Let $I$ and $J$ be two fractional ideals of $mathbbK$ and $q in mathbbN$ a positive integer.



Is it true that $I/qI $ isomorphic to $J/qJ$? If yes, what is the isomorphism?



If no, is there a condition on $q$ that make it True?




linear-algebra ring-theory field-theory algebraic-number-theory ideals




share|cite|improve this question











Let $mathbbK$ be an algebraic number field.



Let $I$ and $J$ be two fractional ideals of $mathbbK$ and $q in mathbbN$ a positive integer.



Is it true that $I/qI $ isomorphic to $J/qJ$? If yes, what is the isomorphism?



If no, is there a condition on $q$ that make it True?





linear-algebra ring-theory field-theory algebraic-number-theory ideals





share|cite|improve this question










share|cite|improve this question




share|cite|improve this question









asked Jul 20 at 13:47









C.S.

134




134











  • Isomorphic in what sense? As abelian groups? As $mathcalO_mathbbK$-modules?
    – Gal Porat
    Jul 20 at 15:21











  • Besides, it will certainly not be true in any reasonable sense if you take $mathbbK = mathbbQ$, $I=(0), J=(1)$
    – Gal Porat
    Jul 20 at 15:22










  • What happen If $mathbbK$ is a cyclotomic number field?, and the ideals are not trivials.
    – C.S.
    Jul 20 at 16:37










  • But you have still not specified what you mean by your question...
    – Gal Porat
    Jul 20 at 17:48










  • No condition depending only on $q$ seems to exist as both $I/qI$ and $J/qJ$ are annihilated by $q$. There must be conditions on $I$ and $J$ to have isomorphism in certain circumstances.
    – joy
    Jul 23 at 8:23
















  • Isomorphic in what sense? As abelian groups? As $mathcalO_mathbbK$-modules?
    – Gal Porat
    Jul 20 at 15:21











  • Besides, it will certainly not be true in any reasonable sense if you take $mathbbK = mathbbQ$, $I=(0), J=(1)$
    – Gal Porat
    Jul 20 at 15:22










  • What happen If $mathbbK$ is a cyclotomic number field?, and the ideals are not trivials.
    – C.S.
    Jul 20 at 16:37










  • But you have still not specified what you mean by your question...
    – Gal Porat
    Jul 20 at 17:48










  • No condition depending only on $q$ seems to exist as both $I/qI$ and $J/qJ$ are annihilated by $q$. There must be conditions on $I$ and $J$ to have isomorphism in certain circumstances.
    – joy
    Jul 23 at 8:23















Isomorphic in what sense? As abelian groups? As $mathcalO_mathbbK$-modules?
– Gal Porat
Jul 20 at 15:21





Isomorphic in what sense? As abelian groups? As $mathcalO_mathbbK$-modules?
– Gal Porat
Jul 20 at 15:21













Besides, it will certainly not be true in any reasonable sense if you take $mathbbK = mathbbQ$, $I=(0), J=(1)$
– Gal Porat
Jul 20 at 15:22




Besides, it will certainly not be true in any reasonable sense if you take $mathbbK = mathbbQ$, $I=(0), J=(1)$
– Gal Porat
Jul 20 at 15:22












What happen If $mathbbK$ is a cyclotomic number field?, and the ideals are not trivials.
– C.S.
Jul 20 at 16:37




What happen If $mathbbK$ is a cyclotomic number field?, and the ideals are not trivials.
– C.S.
Jul 20 at 16:37












But you have still not specified what you mean by your question...
– Gal Porat
Jul 20 at 17:48




But you have still not specified what you mean by your question...
– Gal Porat
Jul 20 at 17:48












No condition depending only on $q$ seems to exist as both $I/qI$ and $J/qJ$ are annihilated by $q$. There must be conditions on $I$ and $J$ to have isomorphism in certain circumstances.
– joy
Jul 23 at 8:23




No condition depending only on $q$ seems to exist as both $I/qI$ and $J/qJ$ are annihilated by $q$. There must be conditions on $I$ and $J$ to have isomorphism in certain circumstances.
– joy
Jul 23 at 8:23















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