Decomposition of a not f.g. module over PID

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For a module $M$ over PID which is not finitely generated, do we still have the isomorphism $Mcong Tor(M)oplus M/Tor(M)$? If not can you give a counter-example?




abstract-algebra modules principal-ideal-domains




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    @Bernard: That presumably means the torsion submodule, the set of $min M$ that are annihilated by some nonzero element of the ring.
    – Eric Wofsey
    Jul 26 at 22:25










  • @EricWofsey; Why, yes of course!
    – Bernard
    Jul 26 at 22:28














up vote
5
down vote

favorite












For a module $M$ over PID which is not finitely generated, do we still have the isomorphism $Mcong Tor(M)oplus M/Tor(M)$? If not can you give a counter-example?




abstract-algebra modules principal-ideal-domains




share|cite|improve this question

















  • 1




    @Bernard: That presumably means the torsion submodule, the set of $min M$ that are annihilated by some nonzero element of the ring.
    – Eric Wofsey
    Jul 26 at 22:25










  • @EricWofsey; Why, yes of course!
    – Bernard
    Jul 26 at 22:28












up vote
5
down vote

favorite









up vote
5
down vote

favorite











For a module $M$ over PID which is not finitely generated, do we still have the isomorphism $Mcong Tor(M)oplus M/Tor(M)$? If not can you give a counter-example?




abstract-algebra modules principal-ideal-domains




share|cite|improve this question













For a module $M$ over PID which is not finitely generated, do we still have the isomorphism $Mcong Tor(M)oplus M/Tor(M)$? If not can you give a counter-example?





abstract-algebra modules principal-ideal-domains





share|cite|improve this question












share|cite|improve this question




share|cite|improve this question








edited Jul 26 at 22:25









Eric Wofsey

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asked Jul 26 at 21:49









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  • 1




    @Bernard: That presumably means the torsion submodule, the set of $min M$ that are annihilated by some nonzero element of the ring.
    – Eric Wofsey
    Jul 26 at 22:25










  • @EricWofsey; Why, yes of course!
    – Bernard
    Jul 26 at 22:28












  • 1




    @Bernard: That presumably means the torsion submodule, the set of $min M$ that are annihilated by some nonzero element of the ring.
    – Eric Wofsey
    Jul 26 at 22:25










  • @EricWofsey; Why, yes of course!
    – Bernard
    Jul 26 at 22:28







1




1




@Bernard: That presumably means the torsion submodule, the set of $min M$ that are annihilated by some nonzero element of the ring.
– Eric Wofsey
Jul 26 at 22:25




@Bernard: That presumably means the torsion submodule, the set of $min M$ that are annihilated by some nonzero element of the ring.
– Eric Wofsey
Jul 26 at 22:25












@EricWofsey; Why, yes of course!
– Bernard
Jul 26 at 22:28




@EricWofsey; Why, yes of course!
– Bernard
Jul 26 at 22:28










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No. For instance, over the ring $mathbbZ$, let $M=prodmathbbZ/(p)$ where the product ranges over all primes $p$. Then the torsion subgroup $T$ of $M$ is just the direct sum $bigoplus mathbbZ/(p)$, set set of elements of $M$ which have only finitely many nonzero coordinates. The quotient $M/T$ is then divisible, since for any $min M$ and any nonzero $ainmathbbZ$, we can divide $m$ by $a$ after modifying only finitely many coordinates of $m$ (namely, those corresponding to primes that divide $a$).



However, the only element of $M$ which is divisible by every nonzero integer is $0$, since if $min M$ is divisible by a prime $p$ then its $p$-coordinate must be $0$. So $M$ has no nontrivial divisible subgroups, and in particular it does not have a subgroup isomorphic to $M/T$.






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    No. For instance, over the ring $mathbbZ$, let $M=prodmathbbZ/(p)$ where the product ranges over all primes $p$. Then the torsion subgroup $T$ of $M$ is just the direct sum $bigoplus mathbbZ/(p)$, set set of elements of $M$ which have only finitely many nonzero coordinates. The quotient $M/T$ is then divisible, since for any $min M$ and any nonzero $ainmathbbZ$, we can divide $m$ by $a$ after modifying only finitely many coordinates of $m$ (namely, those corresponding to primes that divide $a$).



    However, the only element of $M$ which is divisible by every nonzero integer is $0$, since if $min M$ is divisible by a prime $p$ then its $p$-coordinate must be $0$. So $M$ has no nontrivial divisible subgroups, and in particular it does not have a subgroup isomorphic to $M/T$.






    share|cite|improve this answer

























      up vote
      9
      down vote



      accepted










      No. For instance, over the ring $mathbbZ$, let $M=prodmathbbZ/(p)$ where the product ranges over all primes $p$. Then the torsion subgroup $T$ of $M$ is just the direct sum $bigoplus mathbbZ/(p)$, set set of elements of $M$ which have only finitely many nonzero coordinates. The quotient $M/T$ is then divisible, since for any $min M$ and any nonzero $ainmathbbZ$, we can divide $m$ by $a$ after modifying only finitely many coordinates of $m$ (namely, those corresponding to primes that divide $a$).



      However, the only element of $M$ which is divisible by every nonzero integer is $0$, since if $min M$ is divisible by a prime $p$ then its $p$-coordinate must be $0$. So $M$ has no nontrivial divisible subgroups, and in particular it does not have a subgroup isomorphic to $M/T$.






      share|cite|improve this answer























        up vote
        9
        down vote



        accepted







        up vote
        9
        down vote



        accepted






        No. For instance, over the ring $mathbbZ$, let $M=prodmathbbZ/(p)$ where the product ranges over all primes $p$. Then the torsion subgroup $T$ of $M$ is just the direct sum $bigoplus mathbbZ/(p)$, set set of elements of $M$ which have only finitely many nonzero coordinates. The quotient $M/T$ is then divisible, since for any $min M$ and any nonzero $ainmathbbZ$, we can divide $m$ by $a$ after modifying only finitely many coordinates of $m$ (namely, those corresponding to primes that divide $a$).



        However, the only element of $M$ which is divisible by every nonzero integer is $0$, since if $min M$ is divisible by a prime $p$ then its $p$-coordinate must be $0$. So $M$ has no nontrivial divisible subgroups, and in particular it does not have a subgroup isomorphic to $M/T$.






        share|cite|improve this answer













        No. For instance, over the ring $mathbbZ$, let $M=prodmathbbZ/(p)$ where the product ranges over all primes $p$. Then the torsion subgroup $T$ of $M$ is just the direct sum $bigoplus mathbbZ/(p)$, set set of elements of $M$ which have only finitely many nonzero coordinates. The quotient $M/T$ is then divisible, since for any $min M$ and any nonzero $ainmathbbZ$, we can divide $m$ by $a$ after modifying only finitely many coordinates of $m$ (namely, those corresponding to primes that divide $a$).



        However, the only element of $M$ which is divisible by every nonzero integer is $0$, since if $min M$ is divisible by a prime $p$ then its $p$-coordinate must be $0$. So $M$ has no nontrivial divisible subgroups, and in particular it does not have a subgroup isomorphic to $M/T$.







        share|cite|improve this answer













        share|cite|improve this answer



        share|cite|improve this answer











        answered Jul 26 at 22:03









        Eric Wofsey

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