Why is the decomposition of $operatorname{Tor}(M)$ into cyclic modules a “subset” of the decomposition of...












1














In my book it says that if $M$ is finitely generated over $R$, a P.I.D., then



$$M cong R^r oplus R/(a_1) oplus cdots oplus R/(a_n)$$



and that



$$operatorname{Tor}(M) cong R/(a_1) oplus cdots oplus R/(a_n)$$



I don't understand how the second part follows from the first. First, I don't see why $operatorname{Tor}(M)$ has to be finitely generated; I know that a submodule of a finitely generated module is not necessarily finitely generated.



Secondly, if we assume that $operatorname{Tor}(M)$ is finitely generated, then we can apply the theorem and get



$$operatorname{Tor}(M) cong R^k oplus R/(b_1) oplus cdots oplus R/(b_m)$$



I understand that since it's a torsion module, $k$ should equal zero. But I don't see why the $R/(b_1) oplus cdots oplus R/(b_m)$ part has to be the same as the $R/(a_1) oplus cdots oplus R/(a_n)$ part.






abstract-algebra modules finitely-generated







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    1














    In my book it says that if $M$ is finitely generated over $R$, a P.I.D., then



    $$M cong R^r oplus R/(a_1) oplus cdots oplus R/(a_n)$$



    and that



    $$operatorname{Tor}(M) cong R/(a_1) oplus cdots oplus R/(a_n)$$



    I don't understand how the second part follows from the first. First, I don't see why $operatorname{Tor}(M)$ has to be finitely generated; I know that a submodule of a finitely generated module is not necessarily finitely generated.



    Secondly, if we assume that $operatorname{Tor}(M)$ is finitely generated, then we can apply the theorem and get



    $$operatorname{Tor}(M) cong R^k oplus R/(b_1) oplus cdots oplus R/(b_m)$$



    I understand that since it's a torsion module, $k$ should equal zero. But I don't see why the $R/(b_1) oplus cdots oplus R/(b_m)$ part has to be the same as the $R/(a_1) oplus cdots oplus R/(a_n)$ part.






    abstract-algebra modules finitely-generated







    share|cite|improve this question



























      1












      1








      1







      In my book it says that if $M$ is finitely generated over $R$, a P.I.D., then



      $$M cong R^r oplus R/(a_1) oplus cdots oplus R/(a_n)$$



      and that



      $$operatorname{Tor}(M) cong R/(a_1) oplus cdots oplus R/(a_n)$$



      I don't understand how the second part follows from the first. First, I don't see why $operatorname{Tor}(M)$ has to be finitely generated; I know that a submodule of a finitely generated module is not necessarily finitely generated.



      Secondly, if we assume that $operatorname{Tor}(M)$ is finitely generated, then we can apply the theorem and get



      $$operatorname{Tor}(M) cong R^k oplus R/(b_1) oplus cdots oplus R/(b_m)$$



      I understand that since it's a torsion module, $k$ should equal zero. But I don't see why the $R/(b_1) oplus cdots oplus R/(b_m)$ part has to be the same as the $R/(a_1) oplus cdots oplus R/(a_n)$ part.






      abstract-algebra modules finitely-generated







      share|cite|improve this question















      In my book it says that if $M$ is finitely generated over $R$, a P.I.D., then



      $$M cong R^r oplus R/(a_1) oplus cdots oplus R/(a_n)$$



      and that



      $$operatorname{Tor}(M) cong R/(a_1) oplus cdots oplus R/(a_n)$$



      I don't understand how the second part follows from the first. First, I don't see why $operatorname{Tor}(M)$ has to be finitely generated; I know that a submodule of a finitely generated module is not necessarily finitely generated.



      Secondly, if we assume that $operatorname{Tor}(M)$ is finitely generated, then we can apply the theorem and get



      $$operatorname{Tor}(M) cong R^k oplus R/(b_1) oplus cdots oplus R/(b_m)$$



      I understand that since it's a torsion module, $k$ should equal zero. But I don't see why the $R/(b_1) oplus cdots oplus R/(b_m)$ part has to be the same as the $R/(a_1) oplus cdots oplus R/(a_n)$ part.






      abstract-algebra modules finitely-generated




      abstract-algebra modules finitely-generated






      share|cite|improve this question















      share|cite|improve this question













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      edited Nov 18 at 15:14









      Bernard

      118k639112




      118k639112










      asked Nov 18 at 14:32









      Ovi

      12.4k1038111




      12.4k1038111






















          2 Answers
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          Instead of thinking of Tor$(M)$, just think of Tor$(R^k oplus R/(b_1) oplus cdots oplus R/(b_m))$. It is clear that this is just $R/(b_1) oplus cdots oplus R/(b_m)$, and therefore so is the torsion submodule of $M$.






          share|cite|improve this answer





























            1















            1. A P.I.D. is a noetherian ring, and a finitely generated module over a noetherian ring is noetherian, i.e. all its submodules are finitely generated.

            2. There is a uniqueness part in the Structure theorem for finitely generated modules over a P.I.D.






            share|cite|improve this answer























            • Ok got the first part; but what I said in my question is part of the existence theorem, which is given before the uniqueness theorem. Even so, let's assume uniqueness. To get that $operatorname{Tor}(M) cong R/(a_1) oplus cdots oplus R/(a_2)$, wouldn't we have to show that $M cong R^r oplus operatorname{Tor}(M)$ also?
              – Ovi
              Nov 18 at 14:46










            • But this is the case: a finitely generated module over a P.I.D. (or more generally, a Dedekind ring) is the direct sum of its torsion submodule $T$ and a projective module (projective modules are free in the case of P.I.D.s) simply because we hace a surjective homomorphism $Mto R^k$ with kernel $T$.
              – Bernard
              Nov 18 at 15:26













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            2 Answers
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            active

            oldest

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            2 Answers
            2






            active

            oldest

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            active

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            active

            oldest

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            1














            Instead of thinking of Tor$(M)$, just think of Tor$(R^k oplus R/(b_1) oplus cdots oplus R/(b_m))$. It is clear that this is just $R/(b_1) oplus cdots oplus R/(b_m)$, and therefore so is the torsion submodule of $M$.






            share|cite|improve this answer


























              1














              Instead of thinking of Tor$(M)$, just think of Tor$(R^k oplus R/(b_1) oplus cdots oplus R/(b_m))$. It is clear that this is just $R/(b_1) oplus cdots oplus R/(b_m)$, and therefore so is the torsion submodule of $M$.






              share|cite|improve this answer
























                1












                1








                1






                Instead of thinking of Tor$(M)$, just think of Tor$(R^k oplus R/(b_1) oplus cdots oplus R/(b_m))$. It is clear that this is just $R/(b_1) oplus cdots oplus R/(b_m)$, and therefore so is the torsion submodule of $M$.






                share|cite|improve this answer












                Instead of thinking of Tor$(M)$, just think of Tor$(R^k oplus R/(b_1) oplus cdots oplus R/(b_m))$. It is clear that this is just $R/(b_1) oplus cdots oplus R/(b_m)$, and therefore so is the torsion submodule of $M$.







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Nov 25 at 4:20









                Ovi

                12.4k1038111




                12.4k1038111























                    1















                    1. A P.I.D. is a noetherian ring, and a finitely generated module over a noetherian ring is noetherian, i.e. all its submodules are finitely generated.

                    2. There is a uniqueness part in the Structure theorem for finitely generated modules over a P.I.D.






                    share|cite|improve this answer























                    • Ok got the first part; but what I said in my question is part of the existence theorem, which is given before the uniqueness theorem. Even so, let's assume uniqueness. To get that $operatorname{Tor}(M) cong R/(a_1) oplus cdots oplus R/(a_2)$, wouldn't we have to show that $M cong R^r oplus operatorname{Tor}(M)$ also?
                      – Ovi
                      Nov 18 at 14:46










                    • But this is the case: a finitely generated module over a P.I.D. (or more generally, a Dedekind ring) is the direct sum of its torsion submodule $T$ and a projective module (projective modules are free in the case of P.I.D.s) simply because we hace a surjective homomorphism $Mto R^k$ with kernel $T$.
                      – Bernard
                      Nov 18 at 15:26


















                    1















                    1. A P.I.D. is a noetherian ring, and a finitely generated module over a noetherian ring is noetherian, i.e. all its submodules are finitely generated.

                    2. There is a uniqueness part in the Structure theorem for finitely generated modules over a P.I.D.






                    share|cite|improve this answer























                    • Ok got the first part; but what I said in my question is part of the existence theorem, which is given before the uniqueness theorem. Even so, let's assume uniqueness. To get that $operatorname{Tor}(M) cong R/(a_1) oplus cdots oplus R/(a_2)$, wouldn't we have to show that $M cong R^r oplus operatorname{Tor}(M)$ also?
                      – Ovi
                      Nov 18 at 14:46










                    • But this is the case: a finitely generated module over a P.I.D. (or more generally, a Dedekind ring) is the direct sum of its torsion submodule $T$ and a projective module (projective modules are free in the case of P.I.D.s) simply because we hace a surjective homomorphism $Mto R^k$ with kernel $T$.
                      – Bernard
                      Nov 18 at 15:26
















                    1












                    1








                    1







                    1. A P.I.D. is a noetherian ring, and a finitely generated module over a noetherian ring is noetherian, i.e. all its submodules are finitely generated.

                    2. There is a uniqueness part in the Structure theorem for finitely generated modules over a P.I.D.






                    share|cite|improve this answer















                    1. A P.I.D. is a noetherian ring, and a finitely generated module over a noetherian ring is noetherian, i.e. all its submodules are finitely generated.

                    2. There is a uniqueness part in the Structure theorem for finitely generated modules over a P.I.D.







                    share|cite|improve this answer














                    share|cite|improve this answer



                    share|cite|improve this answer








                    edited Nov 25 at 11:07

























                    answered Nov 18 at 14:40









                    Bernard

                    118k639112




                    118k639112












                    • Ok got the first part; but what I said in my question is part of the existence theorem, which is given before the uniqueness theorem. Even so, let's assume uniqueness. To get that $operatorname{Tor}(M) cong R/(a_1) oplus cdots oplus R/(a_2)$, wouldn't we have to show that $M cong R^r oplus operatorname{Tor}(M)$ also?
                      – Ovi
                      Nov 18 at 14:46










                    • But this is the case: a finitely generated module over a P.I.D. (or more generally, a Dedekind ring) is the direct sum of its torsion submodule $T$ and a projective module (projective modules are free in the case of P.I.D.s) simply because we hace a surjective homomorphism $Mto R^k$ with kernel $T$.
                      – Bernard
                      Nov 18 at 15:26




















                    • Ok got the first part; but what I said in my question is part of the existence theorem, which is given before the uniqueness theorem. Even so, let's assume uniqueness. To get that $operatorname{Tor}(M) cong R/(a_1) oplus cdots oplus R/(a_2)$, wouldn't we have to show that $M cong R^r oplus operatorname{Tor}(M)$ also?
                      – Ovi
                      Nov 18 at 14:46










                    • But this is the case: a finitely generated module over a P.I.D. (or more generally, a Dedekind ring) is the direct sum of its torsion submodule $T$ and a projective module (projective modules are free in the case of P.I.D.s) simply because we hace a surjective homomorphism $Mto R^k$ with kernel $T$.
                      – Bernard
                      Nov 18 at 15:26


















                    Ok got the first part; but what I said in my question is part of the existence theorem, which is given before the uniqueness theorem. Even so, let's assume uniqueness. To get that $operatorname{Tor}(M) cong R/(a_1) oplus cdots oplus R/(a_2)$, wouldn't we have to show that $M cong R^r oplus operatorname{Tor}(M)$ also?
                    – Ovi
                    Nov 18 at 14:46




                    Ok got the first part; but what I said in my question is part of the existence theorem, which is given before the uniqueness theorem. Even so, let's assume uniqueness. To get that $operatorname{Tor}(M) cong R/(a_1) oplus cdots oplus R/(a_2)$, wouldn't we have to show that $M cong R^r oplus operatorname{Tor}(M)$ also?
                    – Ovi
                    Nov 18 at 14:46












                    But this is the case: a finitely generated module over a P.I.D. (or more generally, a Dedekind ring) is the direct sum of its torsion submodule $T$ and a projective module (projective modules are free in the case of P.I.D.s) simply because we hace a surjective homomorphism $Mto R^k$ with kernel $T$.
                    – Bernard
                    Nov 18 at 15:26






                    But this is the case: a finitely generated module over a P.I.D. (or more generally, a Dedekind ring) is the direct sum of its torsion submodule $T$ and a projective module (projective modules are free in the case of P.I.D.s) simply because we hace a surjective homomorphism $Mto R^k$ with kernel $T$.
                    – Bernard
                    Nov 18 at 15:26




















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