Projective cover of modules
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
asked Nov 28 '18 at 0:40
Davis We
705
add a comment |
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
asked Nov 28 '18 at 0:40
Davis We
705
What have you tried so far?
– Y. Forman
Nov 28 '18 at 0:42
add a comment |
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
asked Nov 28 '18 at 0:40
Davis We
705
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
asked Nov 28 '18 at 0:40
Davis We
705
asked Nov 28 '18 at 0:40
Davis We
705
asked Nov 28 '18 at 0:40
Davis We
705
asked Nov 28 '18 at 0:40
Davis We
705
asked Nov 28 '18 at 0:40
Davis We
705
705
What have you tried so far?
– Y. Forman
Nov 28 '18 at 0:42
add a comment |
What have you tried so far?
– Y. Forman
Nov 28 '18 at 0:42
What have you tried so far?
– Y. Forman
Nov 28 '18 at 0:42
What have you tried so far?
– Y. Forman
Nov 28 '18 at 0:42
add a comment |
1 Answer
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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.
answered Nov 28 '18 at 1:32
BWW
9,21622237
What do you mean " $2 mathbb{Z}$ is not superfluous in $mathbb{Z}$ "?
– Davis We
Nov 29 '18 at 3:50
By the way, thank's for ask :)
– Davis We
Nov 29 '18 at 3:50
@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 $L=M$.
– BWW
Nov 29 '18 at 4:09
add a comment |
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1 Answer
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1 Answer
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active
oldest
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oldest
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active
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votes
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.
answered Nov 28 '18 at 1:32
BWW
9,21622237
What do you mean " $2 mathbb{Z}$ is not superfluous in $mathbb{Z}$ "?
– Davis We
Nov 29 '18 at 3:50
By the way, thank's for ask :)
– Davis We
Nov 29 '18 at 3:50
@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 $L=M$.
– BWW
Nov 29 '18 at 4:09
add a comment |
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.
answered Nov 28 '18 at 1:32
BWW
9,21622237
What do you mean " $2 mathbb{Z}$ is not superfluous in $mathbb{Z}$ "?
– Davis We
Nov 29 '18 at 3:50
By the way, thank's for ask :)
– Davis We
Nov 29 '18 at 3:50
@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 $L=M$.
– BWW
Nov 29 '18 at 4:09
add a comment |
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.
answered Nov 28 '18 at 1:32
BWW
9,21622237
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.
answered Nov 28 '18 at 1:32
BWW
9,21622237
answered Nov 28 '18 at 1:32
BWW
9,21622237
answered Nov 28 '18 at 1:32
BWW
9,21622237
answered Nov 28 '18 at 1:32
BWW
9,21622237
9,21622237
What do you mean " $2 mathbb{Z}$ is not superfluous in $mathbb{Z}$ "?
– Davis We
Nov 29 '18 at 3:50
By the way, thank's for ask :)
– Davis We
Nov 29 '18 at 3:50
@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 $L=M$.
– BWW
Nov 29 '18 at 4:09
add a comment |
What do you mean " $2 mathbb{Z}$ is not superfluous in $mathbb{Z}$ "?
– Davis We
Nov 29 '18 at 3:50
By the way, thank's for ask :)
– Davis We
Nov 29 '18 at 3:50
@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 $L=M$.
– BWW
Nov 29 '18 at 4:09
What do you mean " $2 mathbb{Z}$ is not superfluous in $mathbb{Z}$ "?
– Davis We
Nov 29 '18 at 3:50
What do you mean " $2 mathbb{Z}$ is not superfluous in $mathbb{Z}$ "?
– Davis We
Nov 29 '18 at 3:50
By the way, thank's for ask :)
– Davis We
Nov 29 '18 at 3:50
By the way, thank's for ask :)
– Davis We
Nov 29 '18 at 3:50
@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 $L=M$.
– BWW
Nov 29 '18 at 4:09
@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 $L=M$.
– BWW
Nov 29 '18 at 4:09
add a comment |
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What have you tried so far?
– Y. Forman
Nov 28 '18 at 0:42