Projective cover of modules












0












$begingroup$


Give me a hand please, how can i prove this sentence.



"Show $mathbb{Z}_{2}$ doesn't have projective cover as $mathbb{Z}$-module."






abstract-algebra modules projective-module







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  • $begingroup$
    What have you tried so far?
    $endgroup$
    – Y. Forman
    Nov 28 '18 at 0:42
















0












$begingroup$


Give me a hand please, how can i prove this sentence.



"Show $mathbb{Z}_{2}$ doesn't have projective cover as $mathbb{Z}$-module."






abstract-algebra modules projective-module







share|cite|improve this question









$endgroup$












  • $begingroup$
    What have you tried so far?
    $endgroup$
    – Y. Forman
    Nov 28 '18 at 0:42














0












0








0





$begingroup$


Give me a hand please, how can i prove this sentence.



"Show $mathbb{Z}_{2}$ doesn't have projective cover as $mathbb{Z}$-module."






abstract-algebra modules projective-module







share|cite|improve this question









$endgroup$




Give me a hand please, how can i prove this sentence.



"Show $mathbb{Z}_{2}$ doesn't have projective cover as $mathbb{Z}$-module."






abstract-algebra modules projective-module




abstract-algebra modules projective-module






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Nov 28 '18 at 0:40









Davis WeDavis We

705




705












  • $begingroup$
    What have you tried so far?
    $endgroup$
    – Y. Forman
    Nov 28 '18 at 0:42


















  • $begingroup$
    What have you tried so far?
    $endgroup$
    – Y. Forman
    Nov 28 '18 at 0:42
















$begingroup$
What have you tried so far?
$endgroup$
– Y. Forman
Nov 28 '18 at 0:42




$begingroup$
What have you tried so far?
$endgroup$
– Y. Forman
Nov 28 '18 at 0:42










1 Answer
1






active

oldest

votes


















1












$begingroup$

I think it's helpful to use the fundamental lemma for projective covers.



Suppose $pcolon Ptomathbb{Z}_2$ is a projective cover. There is also the canonical surjection $pi:mathbb{Z}tomathbb{Z}_2$ given by reduction mod $2$. Since $mathbb{Z}$ is projective, the Fundamental Lemma of Projective Covers gives that there exists a decomposition $mathbb{Z}=P'oplus P''$, where $Psimeq P'$, $pi(P'')=0$, and the restriction $pi|_{P'}colon P'tomathbb{Z}_2$ is a projective cover.




But $mathbb{Z}$ is indecomposable, so in any case $mathbb{Z}=P'$. Hence one would get $picolonmathbb{Z}tomathbb{Z}_2$ is a projective cover, but this is a contradiction since $kerpi=2mathbb{Z}$ is not superfluous in $mathbb{Z}$, since $2mathbb{Z}+3mathbb{Z}=mathbb{Z}$, for instance.







share|cite|improve this answer









$endgroup$













  • $begingroup$
    What do you mean " $2 mathbb{Z}$ is not superfluous in $mathbb{Z}$ "?
    $endgroup$
    – Davis We
    Nov 29 '18 at 3:50










  • $begingroup$
    By the way, thank's for ask :)
    $endgroup$
    – Davis We
    Nov 29 '18 at 3:50










  • $begingroup$
    @Davis In a category of modules, $(P,p)$ is a projective cover when $p$ is a superfluous epimorphism, meaning that $ker p$ is a superfluous submodule of $P$. Generally, $N$ is a superfluous submodule of $M$ if for any submodule $L$ of $M$ such that $N+L=M$, necessarily $L=M$.
    $endgroup$
    – BWW
    Nov 29 '18 at 4:09













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1 Answer
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1 Answer
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active

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active

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active

oldest

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1












$begingroup$

I think it's helpful to use the fundamental lemma for projective covers.



Suppose $pcolon Ptomathbb{Z}_2$ is a projective cover. There is also the canonical surjection $pi:mathbb{Z}tomathbb{Z}_2$ given by reduction mod $2$. Since $mathbb{Z}$ is projective, the Fundamental Lemma of Projective Covers gives that there exists a decomposition $mathbb{Z}=P'oplus P''$, where $Psimeq P'$, $pi(P'')=0$, and the restriction $pi|_{P'}colon P'tomathbb{Z}_2$ is a projective cover.




But $mathbb{Z}$ is indecomposable, so in any case $mathbb{Z}=P'$. Hence one would get $picolonmathbb{Z}tomathbb{Z}_2$ is a projective cover, but this is a contradiction since $kerpi=2mathbb{Z}$ is not superfluous in $mathbb{Z}$, since $2mathbb{Z}+3mathbb{Z}=mathbb{Z}$, for instance.







share|cite|improve this answer









$endgroup$













  • $begingroup$
    What do you mean " $2 mathbb{Z}$ is not superfluous in $mathbb{Z}$ "?
    $endgroup$
    – Davis We
    Nov 29 '18 at 3:50










  • $begingroup$
    By the way, thank's for ask :)
    $endgroup$
    – Davis We
    Nov 29 '18 at 3:50










  • $begingroup$
    @Davis In a category of modules, $(P,p)$ is a projective cover when $p$ is a superfluous epimorphism, meaning that $ker p$ is a superfluous submodule of $P$. Generally, $N$ is a superfluous submodule of $M$ if for any submodule $L$ of $M$ such that $N+L=M$, necessarily $L=M$.
    $endgroup$
    – BWW
    Nov 29 '18 at 4:09


















1












$begingroup$

I think it's helpful to use the fundamental lemma for projective covers.



Suppose $pcolon Ptomathbb{Z}_2$ is a projective cover. There is also the canonical surjection $pi:mathbb{Z}tomathbb{Z}_2$ given by reduction mod $2$. Since $mathbb{Z}$ is projective, the Fundamental Lemma of Projective Covers gives that there exists a decomposition $mathbb{Z}=P'oplus P''$, where $Psimeq P'$, $pi(P'')=0$, and the restriction $pi|_{P'}colon P'tomathbb{Z}_2$ is a projective cover.




But $mathbb{Z}$ is indecomposable, so in any case $mathbb{Z}=P'$. Hence one would get $picolonmathbb{Z}tomathbb{Z}_2$ is a projective cover, but this is a contradiction since $kerpi=2mathbb{Z}$ is not superfluous in $mathbb{Z}$, since $2mathbb{Z}+3mathbb{Z}=mathbb{Z}$, for instance.







share|cite|improve this answer









$endgroup$













  • $begingroup$
    What do you mean " $2 mathbb{Z}$ is not superfluous in $mathbb{Z}$ "?
    $endgroup$
    – Davis We
    Nov 29 '18 at 3:50










  • $begingroup$
    By the way, thank's for ask :)
    $endgroup$
    – Davis We
    Nov 29 '18 at 3:50










  • $begingroup$
    @Davis In a category of modules, $(P,p)$ is a projective cover when $p$ is a superfluous epimorphism, meaning that $ker p$ is a superfluous submodule of $P$. Generally, $N$ is a superfluous submodule of $M$ if for any submodule $L$ of $M$ such that $N+L=M$, necessarily $L=M$.
    $endgroup$
    – BWW
    Nov 29 '18 at 4:09
















1












1








1





$begingroup$

I think it's helpful to use the fundamental lemma for projective covers.



Suppose $pcolon Ptomathbb{Z}_2$ is a projective cover. There is also the canonical surjection $pi:mathbb{Z}tomathbb{Z}_2$ given by reduction mod $2$. Since $mathbb{Z}$ is projective, the Fundamental Lemma of Projective Covers gives that there exists a decomposition $mathbb{Z}=P'oplus P''$, where $Psimeq P'$, $pi(P'')=0$, and the restriction $pi|_{P'}colon P'tomathbb{Z}_2$ is a projective cover.




But $mathbb{Z}$ is indecomposable, so in any case $mathbb{Z}=P'$. Hence one would get $picolonmathbb{Z}tomathbb{Z}_2$ is a projective cover, but this is a contradiction since $kerpi=2mathbb{Z}$ is not superfluous in $mathbb{Z}$, since $2mathbb{Z}+3mathbb{Z}=mathbb{Z}$, for instance.







share|cite|improve this answer









$endgroup$



I think it's helpful to use the fundamental lemma for projective covers.



Suppose $pcolon Ptomathbb{Z}_2$ is a projective cover. There is also the canonical surjection $pi:mathbb{Z}tomathbb{Z}_2$ given by reduction mod $2$. Since $mathbb{Z}$ is projective, the Fundamental Lemma of Projective Covers gives that there exists a decomposition $mathbb{Z}=P'oplus P''$, where $Psimeq P'$, $pi(P'')=0$, and the restriction $pi|_{P'}colon P'tomathbb{Z}_2$ is a projective cover.




But $mathbb{Z}$ is indecomposable, so in any case $mathbb{Z}=P'$. Hence one would get $picolonmathbb{Z}tomathbb{Z}_2$ is a projective cover, but this is a contradiction since $kerpi=2mathbb{Z}$ is not superfluous in $mathbb{Z}$, since $2mathbb{Z}+3mathbb{Z}=mathbb{Z}$, for instance.








share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Nov 28 '18 at 1:32









BWWBWW

9,23622238




9,23622238












  • $begingroup$
    What do you mean " $2 mathbb{Z}$ is not superfluous in $mathbb{Z}$ "?
    $endgroup$
    – Davis We
    Nov 29 '18 at 3:50










  • $begingroup$
    By the way, thank's for ask :)
    $endgroup$
    – Davis We
    Nov 29 '18 at 3:50










  • $begingroup$
    @Davis In a category of modules, $(P,p)$ is a projective cover when $p$ is a superfluous epimorphism, meaning that $ker p$ is a superfluous submodule of $P$. Generally, $N$ is a superfluous submodule of $M$ if for any submodule $L$ of $M$ such that $N+L=M$, necessarily $L=M$.
    $endgroup$
    – BWW
    Nov 29 '18 at 4:09




















  • $begingroup$
    What do you mean " $2 mathbb{Z}$ is not superfluous in $mathbb{Z}$ "?
    $endgroup$
    – Davis We
    Nov 29 '18 at 3:50










  • $begingroup$
    By the way, thank's for ask :)
    $endgroup$
    – Davis We
    Nov 29 '18 at 3:50










  • $begingroup$
    @Davis In a category of modules, $(P,p)$ is a projective cover when $p$ is a superfluous epimorphism, meaning that $ker p$ is a superfluous submodule of $P$. Generally, $N$ is a superfluous submodule of $M$ if for any submodule $L$ of $M$ such that $N+L=M$, necessarily $L=M$.
    $endgroup$
    – BWW
    Nov 29 '18 at 4:09


















$begingroup$
What do you mean " $2 mathbb{Z}$ is not superfluous in $mathbb{Z}$ "?
$endgroup$
– Davis We
Nov 29 '18 at 3:50




$begingroup$
What do you mean " $2 mathbb{Z}$ is not superfluous in $mathbb{Z}$ "?
$endgroup$
– Davis We
Nov 29 '18 at 3:50












$begingroup$
By the way, thank's for ask :)
$endgroup$
– Davis We
Nov 29 '18 at 3:50




$begingroup$
By the way, thank's for ask :)
$endgroup$
– Davis We
Nov 29 '18 at 3:50












$begingroup$
@Davis In a category of modules, $(P,p)$ is a projective cover when $p$ is a superfluous epimorphism, meaning that $ker p$ is a superfluous submodule of $P$. Generally, $N$ is a superfluous submodule of $M$ if for any submodule $L$ of $M$ such that $N+L=M$, necessarily $L=M$.
$endgroup$
– BWW
Nov 29 '18 at 4:09






$begingroup$
@Davis In a category of modules, $(P,p)$ is a projective cover when $p$ is a superfluous epimorphism, meaning that $ker p$ is a superfluous submodule of $P$. Generally, $N$ is a superfluous submodule of $M$ if for any submodule $L$ of $M$ such that $N+L=M$, necessarily $L=M$.
$endgroup$
– BWW
Nov 29 '18 at 4:09




















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