$rad(P)$ non-projective induces all submodules of P non-projective?

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Let $A$ be a finite dimensional algebra. $P$ is an idecomposable projective $A$-module such that its radical $rad(P)$ is non-projective.



Is it right that every non-zero proper submodule of $P$ is not projective?







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    Let $A$ be a finite dimensional algebra. $P$ is an idecomposable projective $A$-module such that its radical $rad(P)$ is non-projective.



    Is it right that every non-zero proper submodule of $P$ is not projective?







    share|cite|improve this question






















      up vote
      0
      down vote

      favorite









      up vote
      0
      down vote

      favorite











      Let $A$ be a finite dimensional algebra. $P$ is an idecomposable projective $A$-module such that its radical $rad(P)$ is non-projective.



      Is it right that every non-zero proper submodule of $P$ is not projective?







      share|cite|improve this question












      Let $A$ be a finite dimensional algebra. $P$ is an idecomposable projective $A$-module such that its radical $rad(P)$ is non-projective.



      Is it right that every non-zero proper submodule of $P$ is not projective?









      share|cite|improve this question











      share|cite|improve this question




      share|cite|improve this question










      asked Aug 11 at 3:05









      Xiaosong Peng

      639414




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          No. Consider $A$ the quotient of the path algebra of the square (i. e. 4 vertices and arrows $a: 1to 2, b: 2to 4, c: 1to 3, d:3to 4$) modulo the ideal spanned by $ba-dc$. Then $operatornamerad (P_1)$ is not projective while $P_2, P_3, P_4$ are all submodules of $P_1$.






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            1 Answer
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            1 Answer
            1






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            active

            oldest

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            up vote
            2
            down vote



            accepted










            No. Consider $A$ the quotient of the path algebra of the square (i. e. 4 vertices and arrows $a: 1to 2, b: 2to 4, c: 1to 3, d:3to 4$) modulo the ideal spanned by $ba-dc$. Then $operatornamerad (P_1)$ is not projective while $P_2, P_3, P_4$ are all submodules of $P_1$.






            share|cite|improve this answer
























              up vote
              2
              down vote



              accepted










              No. Consider $A$ the quotient of the path algebra of the square (i. e. 4 vertices and arrows $a: 1to 2, b: 2to 4, c: 1to 3, d:3to 4$) modulo the ideal spanned by $ba-dc$. Then $operatornamerad (P_1)$ is not projective while $P_2, P_3, P_4$ are all submodules of $P_1$.






              share|cite|improve this answer






















                up vote
                2
                down vote



                accepted







                up vote
                2
                down vote



                accepted






                No. Consider $A$ the quotient of the path algebra of the square (i. e. 4 vertices and arrows $a: 1to 2, b: 2to 4, c: 1to 3, d:3to 4$) modulo the ideal spanned by $ba-dc$. Then $operatornamerad (P_1)$ is not projective while $P_2, P_3, P_4$ are all submodules of $P_1$.






                share|cite|improve this answer












                No. Consider $A$ the quotient of the path algebra of the square (i. e. 4 vertices and arrows $a: 1to 2, b: 2to 4, c: 1to 3, d:3to 4$) modulo the ideal spanned by $ba-dc$. Then $operatornamerad (P_1)$ is not projective while $P_2, P_3, P_4$ are all submodules of $P_1$.







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                share|cite|improve this answer



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                answered Aug 11 at 4:53









                Julian Kuelshammer

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