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Cyclic Galois extension in Weibel example6.3.8
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Let $Ksubset L$ be a finite cyclic Galois extension with Galois group $G$. suppose $G$ is generated by $sigma$ and the order of $G$ is $m$. Denote $operatornametr(x)=x+sigma x+cdots+sigma^m-1x$ where $xin L$. Then there is an exact sequence $Lxrightarrow sigma-1Lxrightarrow operatornametrK rightarrow 0$
There is an isomorphism $Lcong mathbbZGotimes _mathbbZK$ as $G$ module. for free resolution $mathbbZGxrightarrow sigma -1mathbbZGrightarrow mathbbZrightarrow 0$ by tensor L and use the isomorphism,we have exact sequence $Lxrightarrow sigma-1Lxrightarrow f Krightarrow 0$.
I can't prove $f$ can be replaced by $operatornametr$. What I know is $ker(operatornametr)=operatornameIm(sigma -1)$ by Hilbert's Theorem 90. So how to show $operatornametr$ is epic?
Thank you in advance.
abstract-algebra homological-algebra
edited Jul 31 at 3:55
Michael Hardy
204k23185460
asked Jul 31 at 3:10
Sky
939210
add a comment |Â
up vote
0
down vote
favorite
Let $Ksubset L$ be a finite cyclic Galois extension with Galois group $G$. suppose $G$ is generated by $sigma$ and the order of $G$ is $m$. Denote $operatornametr(x)=x+sigma x+cdots+sigma^m-1x$ where $xin L$. Then there is an exact sequence $Lxrightarrow sigma-1Lxrightarrow operatornametrK rightarrow 0$
There is an isomorphism $Lcong mathbbZGotimes _mathbbZK$ as $G$ module. for free resolution $mathbbZGxrightarrow sigma -1mathbbZGrightarrow mathbbZrightarrow 0$ by tensor L and use the isomorphism,we have exact sequence $Lxrightarrow sigma-1Lxrightarrow f Krightarrow 0$.
I can't prove $f$ can be replaced by $operatornametr$. What I know is $ker(operatornametr)=operatornameIm(sigma -1)$ by Hilbert's Theorem 90. So how to show $operatornametr$ is epic?
Thank you in advance.
abstract-algebra homological-algebra
edited Jul 31 at 3:55
Michael Hardy
204k23185460
asked Jul 31 at 3:10
Sky
939210
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Let $Ksubset L$ be a finite cyclic Galois extension with Galois group $G$. suppose $G$ is generated by $sigma$ and the order of $G$ is $m$. Denote $operatornametr(x)=x+sigma x+cdots+sigma^m-1x$ where $xin L$. Then there is an exact sequence $Lxrightarrow sigma-1Lxrightarrow operatornametrK rightarrow 0$
There is an isomorphism $Lcong mathbbZGotimes _mathbbZK$ as $G$ module. for free resolution $mathbbZGxrightarrow sigma -1mathbbZGrightarrow mathbbZrightarrow 0$ by tensor L and use the isomorphism,we have exact sequence $Lxrightarrow sigma-1Lxrightarrow f Krightarrow 0$.
I can't prove $f$ can be replaced by $operatornametr$. What I know is $ker(operatornametr)=operatornameIm(sigma -1)$ by Hilbert's Theorem 90. So how to show $operatornametr$ is epic?
Thank you in advance.
abstract-algebra homological-algebra
edited Jul 31 at 3:55
Michael Hardy
204k23185460
asked Jul 31 at 3:10
Sky
939210
Let $Ksubset L$ be a finite cyclic Galois extension with Galois group $G$. suppose $G$ is generated by $sigma$ and the order of $G$ is $m$. Denote $operatornametr(x)=x+sigma x+cdots+sigma^m-1x$ where $xin L$. Then there is an exact sequence $Lxrightarrow sigma-1Lxrightarrow operatornametrK rightarrow 0$
There is an isomorphism $Lcong mathbbZGotimes _mathbbZK$ as $G$ module. for free resolution $mathbbZGxrightarrow sigma -1mathbbZGrightarrow mathbbZrightarrow 0$ by tensor L and use the isomorphism,we have exact sequence $Lxrightarrow sigma-1Lxrightarrow f Krightarrow 0$.
I can't prove $f$ can be replaced by $operatornametr$. What I know is $ker(operatornametr)=operatornameIm(sigma -1)$ by Hilbert's Theorem 90. So how to show $operatornametr$ is epic?
Thank you in advance.
abstract-algebra homological-algebra
edited Jul 31 at 3:55
Michael Hardy
204k23185460
asked Jul 31 at 3:10
Sky
939210
edited Jul 31 at 3:55
Michael Hardy
204k23185460
edited Jul 31 at 3:55
Michael Hardy
204k23185460
edited Jul 31 at 3:55
Michael Hardy
204k23185460
204k23185460
asked Jul 31 at 3:10
Sky
939210
asked Jul 31 at 3:10
Sky
939210
asked Jul 31 at 3:10
Sky
939210
939210
add a comment |Â
add a comment |Â
1 Answer
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The trace map is non-zero (and so onto) due to Dedekind's
lemma on linear independence of automorphisms.
The automorphisms $sigma^j$ for $0le jle m-1$ are linearly
independent over $L$ and in particular
$xmapstosum_j=0^m-1sigma^j(x)$ cannot be identically zero.
answered Jul 31 at 3:17
Lord Shark the Unknown
84.4k950111
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
The trace map is non-zero (and so onto) due to Dedekind's
lemma on linear independence of automorphisms.
The automorphisms $sigma^j$ for $0le jle m-1$ are linearly
independent over $L$ and in particular
$xmapstosum_j=0^m-1sigma^j(x)$ cannot be identically zero.
answered Jul 31 at 3:17
Lord Shark the Unknown
84.4k950111
add a comment |Â
up vote
2
down vote
The trace map is non-zero (and so onto) due to Dedekind's
lemma on linear independence of automorphisms.
The automorphisms $sigma^j$ for $0le jle m-1$ are linearly
independent over $L$ and in particular
$xmapstosum_j=0^m-1sigma^j(x)$ cannot be identically zero.
answered Jul 31 at 3:17
Lord Shark the Unknown
84.4k950111
add a comment |Â
up vote
2
down vote
up vote
2
down vote
The trace map is non-zero (and so onto) due to Dedekind's
lemma on linear independence of automorphisms.
The automorphisms $sigma^j$ for $0le jle m-1$ are linearly
independent over $L$ and in particular
$xmapstosum_j=0^m-1sigma^j(x)$ cannot be identically zero.
answered Jul 31 at 3:17
Lord Shark the Unknown
84.4k950111
The trace map is non-zero (and so onto) due to Dedekind's
lemma on linear independence of automorphisms.
The automorphisms $sigma^j$ for $0le jle m-1$ are linearly
independent over $L$ and in particular
$xmapstosum_j=0^m-1sigma^j(x)$ cannot be identically zero.
answered Jul 31 at 3:17
Lord Shark the Unknown
84.4k950111
answered Jul 31 at 3:17
Lord Shark the Unknown
84.4k950111
answered Jul 31 at 3:17
Lord Shark the Unknown
84.4k950111
answered Jul 31 at 3:17
Lord Shark the Unknown
84.4k950111
84.4k950111
add a comment |Â
add a comment |Â
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