$Msubset N$ over Dedekind domain $O$. If module index $[M:N]=0$, then $M=N$?

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Let $M,N$ be two torsion free modules over Dedekind domain $O$ with same rank. Suppose $Msubset N$. Then one can talk about module index $[M:N]_O$. Module index is defined in terms of local field automorphism $l_p:M_pto N_p$ which exists by PID. Then $[M_p:N_p]_O_p=det(l_p)O_p=p^k_pO_p$. Then $[M:N]_O=prod_pin Spec(O)p^k_p$.



$textbfQ:$ If $[M:N]_O=O$ for $Nsubset M$, do I know $M=N$? It is clear that $Mcong N$. Do I know this morphism is $M=N$ morphism?




abstract-algebra number-theory




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    Try showing that $M_mathfrak p = N_mathfrak p$ for all primes $mathfrak p$ of $O$.
    – D_S
    Jul 19 at 22:34










  • @D_S Indeed, $M_psubset N_p$ and use PID classification theorem yields the result.
    – user45765
    Jul 19 at 22:38














up vote
1
down vote

favorite












Let $M,N$ be two torsion free modules over Dedekind domain $O$ with same rank. Suppose $Msubset N$. Then one can talk about module index $[M:N]_O$. Module index is defined in terms of local field automorphism $l_p:M_pto N_p$ which exists by PID. Then $[M_p:N_p]_O_p=det(l_p)O_p=p^k_pO_p$. Then $[M:N]_O=prod_pin Spec(O)p^k_p$.



$textbfQ:$ If $[M:N]_O=O$ for $Nsubset M$, do I know $M=N$? It is clear that $Mcong N$. Do I know this morphism is $M=N$ morphism?




abstract-algebra number-theory




share|cite|improve this question

















  • 1




    Try showing that $M_mathfrak p = N_mathfrak p$ for all primes $mathfrak p$ of $O$.
    – D_S
    Jul 19 at 22:34










  • @D_S Indeed, $M_psubset N_p$ and use PID classification theorem yields the result.
    – user45765
    Jul 19 at 22:38












up vote
1
down vote

favorite









up vote
1
down vote

favorite











Let $M,N$ be two torsion free modules over Dedekind domain $O$ with same rank. Suppose $Msubset N$. Then one can talk about module index $[M:N]_O$. Module index is defined in terms of local field automorphism $l_p:M_pto N_p$ which exists by PID. Then $[M_p:N_p]_O_p=det(l_p)O_p=p^k_pO_p$. Then $[M:N]_O=prod_pin Spec(O)p^k_p$.



$textbfQ:$ If $[M:N]_O=O$ for $Nsubset M$, do I know $M=N$? It is clear that $Mcong N$. Do I know this morphism is $M=N$ morphism?




abstract-algebra number-theory




share|cite|improve this question













Let $M,N$ be two torsion free modules over Dedekind domain $O$ with same rank. Suppose $Msubset N$. Then one can talk about module index $[M:N]_O$. Module index is defined in terms of local field automorphism $l_p:M_pto N_p$ which exists by PID. Then $[M_p:N_p]_O_p=det(l_p)O_p=p^k_pO_p$. Then $[M:N]_O=prod_pin Spec(O)p^k_p$.



$textbfQ:$ If $[M:N]_O=O$ for $Nsubset M$, do I know $M=N$? It is clear that $Mcong N$. Do I know this morphism is $M=N$ morphism?





abstract-algebra number-theory





share|cite|improve this question












share|cite|improve this question




share|cite|improve this question








edited Jul 19 at 22:48









Bernard

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asked Jul 19 at 22:25









user45765

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  • 1




    Try showing that $M_mathfrak p = N_mathfrak p$ for all primes $mathfrak p$ of $O$.
    – D_S
    Jul 19 at 22:34










  • @D_S Indeed, $M_psubset N_p$ and use PID classification theorem yields the result.
    – user45765
    Jul 19 at 22:38












  • 1




    Try showing that $M_mathfrak p = N_mathfrak p$ for all primes $mathfrak p$ of $O$.
    – D_S
    Jul 19 at 22:34










  • @D_S Indeed, $M_psubset N_p$ and use PID classification theorem yields the result.
    – user45765
    Jul 19 at 22:38







1




1




Try showing that $M_mathfrak p = N_mathfrak p$ for all primes $mathfrak p$ of $O$.
– D_S
Jul 19 at 22:34




Try showing that $M_mathfrak p = N_mathfrak p$ for all primes $mathfrak p$ of $O$.
– D_S
Jul 19 at 22:34












@D_S Indeed, $M_psubset N_p$ and use PID classification theorem yields the result.
– user45765
Jul 19 at 22:38




@D_S Indeed, $M_psubset N_p$ and use PID classification theorem yields the result.
– user45765
Jul 19 at 22:38















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