Extensions of indecomposable modules












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$begingroup$



Let $R$ be a unital ring. Suppose that $A$, $B$, and $C$ are unitary left $R$-modules such that there exists a non-split exact sequence
$$0to A overset{alpha}{longrightarrow}Boverset{beta}{longrightarrow}Cto 0.$$
If $A$ and $C$ are indecomposable $R$-modules, does it follow that $B$ is also indecomposable? (What if $R$ is non-unital or the modules are not necessarily unitary $R$-modules?)




Edit: As Jeremy Rickard shows, my work below is faulty due to a bad assumption (italicized in the text below). Because of this, the statement is not true even when $A$ and $C$ are modules of finite length.




Wrong Attempt



I know the answer if $A$ and $C$ have finite length. In that case, $B$ has finite length. Since $operatorname{im}alphacong A$ is an indecomposable submodule of $B$, there exists by the Krull-Schmidt theorem a direct sum decomposition $$B=B_1oplus B_2oplusldots oplus B_n$$ of $B$, where each $B_i$ is indecomposable, such that $operatorname{im}alphasubseteq B_1$. Then, $$Ccong (B_1/operatorname{im}alpha)oplus B_2oplus ldotsoplus B_n.$$ By indecomposability of $C$, either $n=2$ and $B_1=operatorname{im}alpha$ which gives $Ccong B_2$, or $n=1$ and $B$ is indecomposable. However, in the case $n=2$ and $B_1=operatorname{im}alpha$, it follows that the exact sequence splits since we have a retraction from $B$ to $A$ (given by $B=B_1oplus B_2overset{operatorname{proj}_1}{ -!!!-!!!twoheadrightarrow } B_1overset{alpha^{-1}}{longrightarrow} A$). I am struggling to see whether this result extends to the infinite length cases.






abstract-algebra ring-theory modules exact-sequence direct-sum







share|cite|improve this question











$endgroup$

















    0












    $begingroup$



    Let $R$ be a unital ring. Suppose that $A$, $B$, and $C$ are unitary left $R$-modules such that there exists a non-split exact sequence
    $$0to A overset{alpha}{longrightarrow}Boverset{beta}{longrightarrow}Cto 0.$$
    If $A$ and $C$ are indecomposable $R$-modules, does it follow that $B$ is also indecomposable? (What if $R$ is non-unital or the modules are not necessarily unitary $R$-modules?)




    Edit: As Jeremy Rickard shows, my work below is faulty due to a bad assumption (italicized in the text below). Because of this, the statement is not true even when $A$ and $C$ are modules of finite length.




    Wrong Attempt



    I know the answer if $A$ and $C$ have finite length. In that case, $B$ has finite length. Since $operatorname{im}alphacong A$ is an indecomposable submodule of $B$, there exists by the Krull-Schmidt theorem a direct sum decomposition $$B=B_1oplus B_2oplusldots oplus B_n$$ of $B$, where each $B_i$ is indecomposable, such that $operatorname{im}alphasubseteq B_1$. Then, $$Ccong (B_1/operatorname{im}alpha)oplus B_2oplus ldotsoplus B_n.$$ By indecomposability of $C$, either $n=2$ and $B_1=operatorname{im}alpha$ which gives $Ccong B_2$, or $n=1$ and $B$ is indecomposable. However, in the case $n=2$ and $B_1=operatorname{im}alpha$, it follows that the exact sequence splits since we have a retraction from $B$ to $A$ (given by $B=B_1oplus B_2overset{operatorname{proj}_1}{ -!!!-!!!twoheadrightarrow } B_1overset{alpha^{-1}}{longrightarrow} A$). I am struggling to see whether this result extends to the infinite length cases.






    abstract-algebra ring-theory modules exact-sequence direct-sum







    share|cite|improve this question











    $endgroup$















      0












      0








      0





      $begingroup$



      Let $R$ be a unital ring. Suppose that $A$, $B$, and $C$ are unitary left $R$-modules such that there exists a non-split exact sequence
      $$0to A overset{alpha}{longrightarrow}Boverset{beta}{longrightarrow}Cto 0.$$
      If $A$ and $C$ are indecomposable $R$-modules, does it follow that $B$ is also indecomposable? (What if $R$ is non-unital or the modules are not necessarily unitary $R$-modules?)




      Edit: As Jeremy Rickard shows, my work below is faulty due to a bad assumption (italicized in the text below). Because of this, the statement is not true even when $A$ and $C$ are modules of finite length.




      Wrong Attempt



      I know the answer if $A$ and $C$ have finite length. In that case, $B$ has finite length. Since $operatorname{im}alphacong A$ is an indecomposable submodule of $B$, there exists by the Krull-Schmidt theorem a direct sum decomposition $$B=B_1oplus B_2oplusldots oplus B_n$$ of $B$, where each $B_i$ is indecomposable, such that $operatorname{im}alphasubseteq B_1$. Then, $$Ccong (B_1/operatorname{im}alpha)oplus B_2oplus ldotsoplus B_n.$$ By indecomposability of $C$, either $n=2$ and $B_1=operatorname{im}alpha$ which gives $Ccong B_2$, or $n=1$ and $B$ is indecomposable. However, in the case $n=2$ and $B_1=operatorname{im}alpha$, it follows that the exact sequence splits since we have a retraction from $B$ to $A$ (given by $B=B_1oplus B_2overset{operatorname{proj}_1}{ -!!!-!!!twoheadrightarrow } B_1overset{alpha^{-1}}{longrightarrow} A$). I am struggling to see whether this result extends to the infinite length cases.






      abstract-algebra ring-theory modules exact-sequence direct-sum







      share|cite|improve this question











      $endgroup$





      Let $R$ be a unital ring. Suppose that $A$, $B$, and $C$ are unitary left $R$-modules such that there exists a non-split exact sequence
      $$0to A overset{alpha}{longrightarrow}Boverset{beta}{longrightarrow}Cto 0.$$
      If $A$ and $C$ are indecomposable $R$-modules, does it follow that $B$ is also indecomposable? (What if $R$ is non-unital or the modules are not necessarily unitary $R$-modules?)




      Edit: As Jeremy Rickard shows, my work below is faulty due to a bad assumption (italicized in the text below). Because of this, the statement is not true even when $A$ and $C$ are modules of finite length.




      Wrong Attempt



      I know the answer if $A$ and $C$ have finite length. In that case, $B$ has finite length. Since $operatorname{im}alphacong A$ is an indecomposable submodule of $B$, there exists by the Krull-Schmidt theorem a direct sum decomposition $$B=B_1oplus B_2oplusldots oplus B_n$$ of $B$, where each $B_i$ is indecomposable, such that $operatorname{im}alphasubseteq B_1$. Then, $$Ccong (B_1/operatorname{im}alpha)oplus B_2oplus ldotsoplus B_n.$$ By indecomposability of $C$, either $n=2$ and $B_1=operatorname{im}alpha$ which gives $Ccong B_2$, or $n=1$ and $B$ is indecomposable. However, in the case $n=2$ and $B_1=operatorname{im}alpha$, it follows that the exact sequence splits since we have a retraction from $B$ to $A$ (given by $B=B_1oplus B_2overset{operatorname{proj}_1}{ -!!!-!!!twoheadrightarrow } B_1overset{alpha^{-1}}{longrightarrow} A$). I am struggling to see whether this result extends to the infinite length cases.






      abstract-algebra ring-theory modules exact-sequence direct-sum




      abstract-algebra ring-theory modules exact-sequence direct-sum






      share|cite|improve this question















      share|cite|improve this question













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      share|cite|improve this question








      edited Dec 3 '18 at 18:59

























      asked Dec 3 '18 at 18:02







      user593746





























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          $begingroup$

          Even for finite length modules, it's not true.



          For example, ($R=mathbb{Z}$) there is a non-split exact sequence
          $$0tomathbb{Z}/4mathbb{Z}stackrel{alpha}{to} mathbb{Z}/8mathbb{Z}oplusmathbb{Z}/2mathbb{Z}tomathbb{Z}/4mathbb{Z}to 0,$$
          where $alpha(n)=(2n,n)$.
          Your claim that $text{im}(alpha)$ must be contained in an indecomposable summand of $B$ isn't necessarily true.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Thank you very much. Does my claim for the finite-length case at least hold if $R$ is an algebra over a field?
            $endgroup$
            – user593746
            Dec 3 '18 at 18:40






          • 1




            $begingroup$
            I just realized I could create a similar construction if $R$ is an algebra over $mathbb{R}$: $$0 to mathbb{R}[x]/big((x^2+1)^2big) overset{alpha}{longrightarrow} mathbb{R}[x]/big((x^2+1)^3big)oplus mathbb{R}[x]/big(x^2+1big) overset{beta}{longrightarrow} mathbb{R}[x]/big((x^2+1)^2big) to 0$$ via $alpha(f)=big((x^2+1)f,fbig)$ and $beta(f,g)=f-(x^2+1)g$.
            $endgroup$
            – user593746
            Dec 3 '18 at 18:57













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          $begingroup$

          Even for finite length modules, it's not true.



          For example, ($R=mathbb{Z}$) there is a non-split exact sequence
          $$0tomathbb{Z}/4mathbb{Z}stackrel{alpha}{to} mathbb{Z}/8mathbb{Z}oplusmathbb{Z}/2mathbb{Z}tomathbb{Z}/4mathbb{Z}to 0,$$
          where $alpha(n)=(2n,n)$.
          Your claim that $text{im}(alpha)$ must be contained in an indecomposable summand of $B$ isn't necessarily true.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Thank you very much. Does my claim for the finite-length case at least hold if $R$ is an algebra over a field?
            $endgroup$
            – user593746
            Dec 3 '18 at 18:40






          • 1




            $begingroup$
            I just realized I could create a similar construction if $R$ is an algebra over $mathbb{R}$: $$0 to mathbb{R}[x]/big((x^2+1)^2big) overset{alpha}{longrightarrow} mathbb{R}[x]/big((x^2+1)^3big)oplus mathbb{R}[x]/big(x^2+1big) overset{beta}{longrightarrow} mathbb{R}[x]/big((x^2+1)^2big) to 0$$ via $alpha(f)=big((x^2+1)f,fbig)$ and $beta(f,g)=f-(x^2+1)g$.
            $endgroup$
            – user593746
            Dec 3 '18 at 18:57


















          1












          $begingroup$

          Even for finite length modules, it's not true.



          For example, ($R=mathbb{Z}$) there is a non-split exact sequence
          $$0tomathbb{Z}/4mathbb{Z}stackrel{alpha}{to} mathbb{Z}/8mathbb{Z}oplusmathbb{Z}/2mathbb{Z}tomathbb{Z}/4mathbb{Z}to 0,$$
          where $alpha(n)=(2n,n)$.
          Your claim that $text{im}(alpha)$ must be contained in an indecomposable summand of $B$ isn't necessarily true.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Thank you very much. Does my claim for the finite-length case at least hold if $R$ is an algebra over a field?
            $endgroup$
            – user593746
            Dec 3 '18 at 18:40






          • 1




            $begingroup$
            I just realized I could create a similar construction if $R$ is an algebra over $mathbb{R}$: $$0 to mathbb{R}[x]/big((x^2+1)^2big) overset{alpha}{longrightarrow} mathbb{R}[x]/big((x^2+1)^3big)oplus mathbb{R}[x]/big(x^2+1big) overset{beta}{longrightarrow} mathbb{R}[x]/big((x^2+1)^2big) to 0$$ via $alpha(f)=big((x^2+1)f,fbig)$ and $beta(f,g)=f-(x^2+1)g$.
            $endgroup$
            – user593746
            Dec 3 '18 at 18:57
















          1












          1








          1





          $begingroup$

          Even for finite length modules, it's not true.



          For example, ($R=mathbb{Z}$) there is a non-split exact sequence
          $$0tomathbb{Z}/4mathbb{Z}stackrel{alpha}{to} mathbb{Z}/8mathbb{Z}oplusmathbb{Z}/2mathbb{Z}tomathbb{Z}/4mathbb{Z}to 0,$$
          where $alpha(n)=(2n,n)$.
          Your claim that $text{im}(alpha)$ must be contained in an indecomposable summand of $B$ isn't necessarily true.






          share|cite|improve this answer









          $endgroup$



          Even for finite length modules, it's not true.



          For example, ($R=mathbb{Z}$) there is a non-split exact sequence
          $$0tomathbb{Z}/4mathbb{Z}stackrel{alpha}{to} mathbb{Z}/8mathbb{Z}oplusmathbb{Z}/2mathbb{Z}tomathbb{Z}/4mathbb{Z}to 0,$$
          where $alpha(n)=(2n,n)$.
          Your claim that $text{im}(alpha)$ must be contained in an indecomposable summand of $B$ isn't necessarily true.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Dec 3 '18 at 18:20









          Jeremy RickardJeremy Rickard

          16.1k11643




          16.1k11643












          • $begingroup$
            Thank you very much. Does my claim for the finite-length case at least hold if $R$ is an algebra over a field?
            $endgroup$
            – user593746
            Dec 3 '18 at 18:40






          • 1




            $begingroup$
            I just realized I could create a similar construction if $R$ is an algebra over $mathbb{R}$: $$0 to mathbb{R}[x]/big((x^2+1)^2big) overset{alpha}{longrightarrow} mathbb{R}[x]/big((x^2+1)^3big)oplus mathbb{R}[x]/big(x^2+1big) overset{beta}{longrightarrow} mathbb{R}[x]/big((x^2+1)^2big) to 0$$ via $alpha(f)=big((x^2+1)f,fbig)$ and $beta(f,g)=f-(x^2+1)g$.
            $endgroup$
            – user593746
            Dec 3 '18 at 18:57




















          • $begingroup$
            Thank you very much. Does my claim for the finite-length case at least hold if $R$ is an algebra over a field?
            $endgroup$
            – user593746
            Dec 3 '18 at 18:40






          • 1




            $begingroup$
            I just realized I could create a similar construction if $R$ is an algebra over $mathbb{R}$: $$0 to mathbb{R}[x]/big((x^2+1)^2big) overset{alpha}{longrightarrow} mathbb{R}[x]/big((x^2+1)^3big)oplus mathbb{R}[x]/big(x^2+1big) overset{beta}{longrightarrow} mathbb{R}[x]/big((x^2+1)^2big) to 0$$ via $alpha(f)=big((x^2+1)f,fbig)$ and $beta(f,g)=f-(x^2+1)g$.
            $endgroup$
            – user593746
            Dec 3 '18 at 18:57


















          $begingroup$
          Thank you very much. Does my claim for the finite-length case at least hold if $R$ is an algebra over a field?
          $endgroup$
          – user593746
          Dec 3 '18 at 18:40




          $begingroup$
          Thank you very much. Does my claim for the finite-length case at least hold if $R$ is an algebra over a field?
          $endgroup$
          – user593746
          Dec 3 '18 at 18:40




          1




          1




          $begingroup$
          I just realized I could create a similar construction if $R$ is an algebra over $mathbb{R}$: $$0 to mathbb{R}[x]/big((x^2+1)^2big) overset{alpha}{longrightarrow} mathbb{R}[x]/big((x^2+1)^3big)oplus mathbb{R}[x]/big(x^2+1big) overset{beta}{longrightarrow} mathbb{R}[x]/big((x^2+1)^2big) to 0$$ via $alpha(f)=big((x^2+1)f,fbig)$ and $beta(f,g)=f-(x^2+1)g$.
          $endgroup$
          – user593746
          Dec 3 '18 at 18:57






          $begingroup$
          I just realized I could create a similar construction if $R$ is an algebra over $mathbb{R}$: $$0 to mathbb{R}[x]/big((x^2+1)^2big) overset{alpha}{longrightarrow} mathbb{R}[x]/big((x^2+1)^3big)oplus mathbb{R}[x]/big(x^2+1big) overset{beta}{longrightarrow} mathbb{R}[x]/big((x^2+1)^2big) to 0$$ via $alpha(f)=big((x^2+1)f,fbig)$ and $beta(f,g)=f-(x^2+1)g$.
          $endgroup$
          – user593746
          Dec 3 '18 at 18:57




















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