The Kernel is inside the radical when we have an essential epimorphism
$begingroup$
This is a proposition in Auslander's book (Representation Theory of Artin algebras). I want proof that:
If $f$ is an essential epimorphism then Ker$f subset rad A$, where
$f: Arightarrow B$, and $A$, $B$ are finitely generated modules over a left Artin ring.
I try to prove this proposition unsuccessfully. In the book says that is an easy consequence of the following fact: $rad A= radLambda*A$, when $A$ is a finitely generated module over a left Artin ring $Lambda$. I really appreciate any help for proving this, I use this result every time in my new studies.
abstract-algebra modules representation-theory artinian
asked Jan 14 at 21:16
Júlio César M. MarquesJúlio César M. Marques
616
$endgroup$
add a comment |
$begingroup$
This is a proposition in Auslander's book (Representation Theory of Artin algebras). I want proof that:
If $f$ is an essential epimorphism then Ker$f subset rad A$, where
$f: Arightarrow B$, and $A$, $B$ are finitely generated modules over a left Artin ring.
I try to prove this proposition unsuccessfully. In the book says that is an easy consequence of the following fact: $rad A= radLambda*A$, when $A$ is a finitely generated module over a left Artin ring $Lambda$. I really appreciate any help for proving this, I use this result every time in my new studies.
abstract-algebra modules representation-theory artinian
asked Jan 14 at 21:16
Júlio César M. MarquesJúlio César M. Marques
616
$endgroup$
add a comment |
$begingroup$
This is a proposition in Auslander's book (Representation Theory of Artin algebras). I want proof that:
If $f$ is an essential epimorphism then Ker$f subset rad A$, where
$f: Arightarrow B$, and $A$, $B$ are finitely generated modules over a left Artin ring.
I try to prove this proposition unsuccessfully. In the book says that is an easy consequence of the following fact: $rad A= radLambda*A$, when $A$ is a finitely generated module over a left Artin ring $Lambda$. I really appreciate any help for proving this, I use this result every time in my new studies.
abstract-algebra modules representation-theory artinian
asked Jan 14 at 21:16
Júlio César M. MarquesJúlio César M. Marques
616
$endgroup$
This is a proposition in Auslander's book (Representation Theory of Artin algebras). I want proof that:
If $f$ is an essential epimorphism then Ker$f subset rad A$, where
$f: Arightarrow B$, and $A$, $B$ are finitely generated modules over a left Artin ring.
I try to prove this proposition unsuccessfully. In the book says that is an easy consequence of the following fact: $rad A= radLambda*A$, when $A$ is a finitely generated module over a left Artin ring $Lambda$. I really appreciate any help for proving this, I use this result every time in my new studies.
abstract-algebra modules representation-theory artinian
abstract-algebra modules representation-theory artinian
asked Jan 14 at 21:16
Júlio César M. MarquesJúlio César M. Marques
616
asked Jan 14 at 21:16
Júlio César M. MarquesJúlio César M. Marques
616
edited Jan 17 at 12:40
Júlio César M. Marques
asked Jan 14 at 21:16
Júlio César M. MarquesJúlio César M. Marques
616
asked Jan 14 at 21:16
Júlio César M. MarquesJúlio César M. Marques
616
asked Jan 14 at 21:16
Júlio César M. MarquesJúlio César M. Marques
616
616
add a comment |
add a comment |
1 Answer
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$begingroup$
Take $X$ a maximal submodule of $A$. Suppose there is $ain$ Ker$f$ and $anotin X$. So we have $X+$ Ker$f=A$ and consider the following:
$g:Xrightarrow A$ an imbedding and $f: Arightarrow B$.
Since $f$ is an epimorphism, given $bin B$ there is $ain A$ that $b=f(a)=f(x+d)$, where $xin X$ and $din$Ker$f$. Then $b=f(a)=f(x+d)=f(x)=f(g(x))$, and $fg$ is epimorphism, so on $g$ is epimorphism since $f$ is essential. But this is a contradiction with definition of $g$.
Then Ker$fsubset$ rad$A$.
answered 2 days ago
Júlio César M. MarquesJúlio César M. Marques
616
$endgroup$
add a comment |
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1 Answer
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$begingroup$
Take $X$ a maximal submodule of $A$. Suppose there is $ain$ Ker$f$ and $anotin X$. So we have $X+$ Ker$f=A$ and consider the following:
$g:Xrightarrow A$ an imbedding and $f: Arightarrow B$.
Since $f$ is an epimorphism, given $bin B$ there is $ain A$ that $b=f(a)=f(x+d)$, where $xin X$ and $din$Ker$f$. Then $b=f(a)=f(x+d)=f(x)=f(g(x))$, and $fg$ is epimorphism, so on $g$ is epimorphism since $f$ is essential. But this is a contradiction with definition of $g$.
Then Ker$fsubset$ rad$A$.
answered 2 days ago
Júlio César M. MarquesJúlio César M. Marques
616
$endgroup$
add a comment |
$begingroup$
Take $X$ a maximal submodule of $A$. Suppose there is $ain$ Ker$f$ and $anotin X$. So we have $X+$ Ker$f=A$ and consider the following:
$g:Xrightarrow A$ an imbedding and $f: Arightarrow B$.
Since $f$ is an epimorphism, given $bin B$ there is $ain A$ that $b=f(a)=f(x+d)$, where $xin X$ and $din$Ker$f$. Then $b=f(a)=f(x+d)=f(x)=f(g(x))$, and $fg$ is epimorphism, so on $g$ is epimorphism since $f$ is essential. But this is a contradiction with definition of $g$.
Then Ker$fsubset$ rad$A$.
answered 2 days ago
Júlio César M. MarquesJúlio César M. Marques
616
$endgroup$
add a comment |
$begingroup$
Take $X$ a maximal submodule of $A$. Suppose there is $ain$ Ker$f$ and $anotin X$. So we have $X+$ Ker$f=A$ and consider the following:
$g:Xrightarrow A$ an imbedding and $f: Arightarrow B$.
Since $f$ is an epimorphism, given $bin B$ there is $ain A$ that $b=f(a)=f(x+d)$, where $xin X$ and $din$Ker$f$. Then $b=f(a)=f(x+d)=f(x)=f(g(x))$, and $fg$ is epimorphism, so on $g$ is epimorphism since $f$ is essential. But this is a contradiction with definition of $g$.
Then Ker$fsubset$ rad$A$.
answered 2 days ago
Júlio César M. MarquesJúlio César M. Marques
616
$endgroup$
Take $X$ a maximal submodule of $A$. Suppose there is $ain$ Ker$f$ and $anotin X$. So we have $X+$ Ker$f=A$ and consider the following:
$g:Xrightarrow A$ an imbedding and $f: Arightarrow B$.
Since $f$ is an epimorphism, given $bin B$ there is $ain A$ that $b=f(a)=f(x+d)$, where $xin X$ and $din$Ker$f$. Then $b=f(a)=f(x+d)=f(x)=f(g(x))$, and $fg$ is epimorphism, so on $g$ is epimorphism since $f$ is essential. But this is a contradiction with definition of $g$.
Then Ker$fsubset$ rad$A$.
answered 2 days ago
Júlio César M. MarquesJúlio César M. Marques
616
answered 2 days ago
Júlio César M. MarquesJúlio César M. Marques
616
answered 2 days ago
Júlio César M. MarquesJúlio César M. Marques
616
answered 2 days ago
Júlio César M. MarquesJúlio César M. Marques
616
616
add a comment |
add a comment |
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