Decomposition a Module into a Pure Submodule and another Submodule












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I am having trouble doing the following exercise.



Let $F$ be a field and let $M$ be a finitely generated module over the polynomial ring $F[x]$. Let $N$ be a pure submodule of $M$. Then, there exists a submodule $L$ of $M$ such that $N+L = M$ and $N cap L =0$.



By pure, I mean that if an element $y$ in $N$ and and element $a$ in $F$ are such that there exists an element $x$ in $M$ where $ax=y$, then there exists an element $z$ in $N$ such that $az = y$.










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    5












    $begingroup$


    I am having trouble doing the following exercise.



    Let $F$ be a field and let $M$ be a finitely generated module over the polynomial ring $F[x]$. Let $N$ be a pure submodule of $M$. Then, there exists a submodule $L$ of $M$ such that $N+L = M$ and $N cap L =0$.



    By pure, I mean that if an element $y$ in $N$ and and element $a$ in $F$ are such that there exists an element $x$ in $M$ where $ax=y$, then there exists an element $z$ in $N$ such that $az = y$.










    share|cite|improve this question











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      5












      5








      5


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      $begingroup$


      I am having trouble doing the following exercise.



      Let $F$ be a field and let $M$ be a finitely generated module over the polynomial ring $F[x]$. Let $N$ be a pure submodule of $M$. Then, there exists a submodule $L$ of $M$ such that $N+L = M$ and $N cap L =0$.



      By pure, I mean that if an element $y$ in $N$ and and element $a$ in $F$ are such that there exists an element $x$ in $M$ where $ax=y$, then there exists an element $z$ in $N$ such that $az = y$.










      share|cite|improve this question











      $endgroup$




      I am having trouble doing the following exercise.



      Let $F$ be a field and let $M$ be a finitely generated module over the polynomial ring $F[x]$. Let $N$ be a pure submodule of $M$. Then, there exists a submodule $L$ of $M$ such that $N+L = M$ and $N cap L =0$.



      By pure, I mean that if an element $y$ in $N$ and and element $a$ in $F$ are such that there exists an element $x$ in $M$ where $ax=y$, then there exists an element $z$ in $N$ such that $az = y$.







      linear-algebra abstract-algebra






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      edited Dec 12 '18 at 15:42







      LinearGuy

















      asked Dec 4 '18 at 20:19









      LinearGuyLinearGuy

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      13211






















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          $begingroup$

          I will write down some hints. Firstly, since $F$ is a field we have that $F[x]$ is PID.



          Consider the following short exact sequence: $0 longrightarrow N overset{i}longrightarrow M overset{pi}{longrightarrow} M / N longrightarrow 0$, where $i$ is the inclusion and $pi$ is the quotient map If this sequence splits then $M=N oplus M/N$.



          In order to find such morphism $ucolon M/N to M$ such that it satisfies the $2$nd equivalent statement of Splitting Lemma consider $M/N$ as a finitely generated module over the PID $F[x]$ and aplpy the Structure theorem for finitely generated modules over a principal ideal domain and the use purity of $N$.






          share|cite|improve this answer











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          • $begingroup$
            If this answer does not help me let me know so i can enlighten you.
            $endgroup$
            – Cornelius
            Dec 11 '18 at 16:28











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          active

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          1












          $begingroup$

          I will write down some hints. Firstly, since $F$ is a field we have that $F[x]$ is PID.



          Consider the following short exact sequence: $0 longrightarrow N overset{i}longrightarrow M overset{pi}{longrightarrow} M / N longrightarrow 0$, where $i$ is the inclusion and $pi$ is the quotient map If this sequence splits then $M=N oplus M/N$.



          In order to find such morphism $ucolon M/N to M$ such that it satisfies the $2$nd equivalent statement of Splitting Lemma consider $M/N$ as a finitely generated module over the PID $F[x]$ and aplpy the Structure theorem for finitely generated modules over a principal ideal domain and the use purity of $N$.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            If this answer does not help me let me know so i can enlighten you.
            $endgroup$
            – Cornelius
            Dec 11 '18 at 16:28
















          1












          $begingroup$

          I will write down some hints. Firstly, since $F$ is a field we have that $F[x]$ is PID.



          Consider the following short exact sequence: $0 longrightarrow N overset{i}longrightarrow M overset{pi}{longrightarrow} M / N longrightarrow 0$, where $i$ is the inclusion and $pi$ is the quotient map If this sequence splits then $M=N oplus M/N$.



          In order to find such morphism $ucolon M/N to M$ such that it satisfies the $2$nd equivalent statement of Splitting Lemma consider $M/N$ as a finitely generated module over the PID $F[x]$ and aplpy the Structure theorem for finitely generated modules over a principal ideal domain and the use purity of $N$.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            If this answer does not help me let me know so i can enlighten you.
            $endgroup$
            – Cornelius
            Dec 11 '18 at 16:28














          1












          1








          1





          $begingroup$

          I will write down some hints. Firstly, since $F$ is a field we have that $F[x]$ is PID.



          Consider the following short exact sequence: $0 longrightarrow N overset{i}longrightarrow M overset{pi}{longrightarrow} M / N longrightarrow 0$, where $i$ is the inclusion and $pi$ is the quotient map If this sequence splits then $M=N oplus M/N$.



          In order to find such morphism $ucolon M/N to M$ such that it satisfies the $2$nd equivalent statement of Splitting Lemma consider $M/N$ as a finitely generated module over the PID $F[x]$ and aplpy the Structure theorem for finitely generated modules over a principal ideal domain and the use purity of $N$.






          share|cite|improve this answer











          $endgroup$



          I will write down some hints. Firstly, since $F$ is a field we have that $F[x]$ is PID.



          Consider the following short exact sequence: $0 longrightarrow N overset{i}longrightarrow M overset{pi}{longrightarrow} M / N longrightarrow 0$, where $i$ is the inclusion and $pi$ is the quotient map If this sequence splits then $M=N oplus M/N$.



          In order to find such morphism $ucolon M/N to M$ such that it satisfies the $2$nd equivalent statement of Splitting Lemma consider $M/N$ as a finitely generated module over the PID $F[x]$ and aplpy the Structure theorem for finitely generated modules over a principal ideal domain and the use purity of $N$.







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Dec 12 '18 at 8:02

























          answered Dec 11 '18 at 16:28









          CorneliusCornelius

          1957




          1957












          • $begingroup$
            If this answer does not help me let me know so i can enlighten you.
            $endgroup$
            – Cornelius
            Dec 11 '18 at 16:28


















          • $begingroup$
            If this answer does not help me let me know so i can enlighten you.
            $endgroup$
            – Cornelius
            Dec 11 '18 at 16:28
















          $begingroup$
          If this answer does not help me let me know so i can enlighten you.
          $endgroup$
          – Cornelius
          Dec 11 '18 at 16:28




          $begingroup$
          If this answer does not help me let me know so i can enlighten you.
          $endgroup$
          – Cornelius
          Dec 11 '18 at 16:28


















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