Ramification groups are normal subgroups of Galois group?

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Let $L/K$ be a finite Galois extension where $K= operatornameFrac(O)$ is complete and $O$ is a Dedekind domain. Let $Delta=Gal(L/K)$. Since $K$ is complete, there is a unique prime $Q$ associated to $O_L$ where $O_L$ is the integral closure of $O_K$ in $L$.
Then let $Delta_i=deltainDeltavertforall ain O_L,delta(a)-ain Q^i$.

$Delta_1$ is unique $p$-Sylow subgroup of $Delta_0$. It is clear that $Delta_0$ is the inertia group associated to prime $Q$ of $O_L$. If $Delta_1$ is unique $p$-Sylow of $Delta_0$, then $Delta_1$ is normal by all $p$-Sylow being conjugates.

$Delta_0$ is normal due to exact sequence $1toDelta_0toDeltato operatornameGal(k_L/k)to 1$ where $k$ is residue field of $K$ and $k_L$ is the residue field of $L$.

$textbfQ:$ The book says it is follows easily $Delta_1leqDelta$ is normal. Why $Delta_1$ is normal here? Normality is non-transitive. I knew $Delta_1leqDelta_0$ normal and $Delta_0leqDelta$ normal. $ef=[L:K]$. I knew $|Delta_0|=e$. However there is no reason to expect $pnmid f$.

Ref:Algebraic Number Theory by Taylor, Frohlich, Chpter 4 section 4.

abstract-algebra number-theory

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edited Jul 25 at 22:54

Bernard

110k635103

asked Jul 25 at 22:36

user45765

2,1942718

• 
  
  
  

  
  

  
  
  
  
  
  

  
  The definition of $Delta_i$ uses $forall ain O_L$. $Q$ is fixed by $Delta$ (as is $O_L$). Hence, one can replace $a$ by $ga$, and $Q$ by $gQ$ in the definition, for any fixed $gin Delta$. OK?
  – peter a g
  Jul 26 at 0:33
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @peterag Then you know $pnmid f$ by direct consequence. Why this is the case then? What do you mean by $gQ$ here? There is only 1 non trivial prime in $O_L$ by $K$ complete implying $L$ complete.($O_L$ is DVR.) $gQ=Q$ by $g$ is ring homomorphism.
  – user45765
  Jul 26 at 1:22
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @peterag I do not see that is the author's intention. It should be deduced from $p-$sylow part. I do not see any particular reason making this clear. $pnmid f$ is equivalent to $Delta_1$ normal.
  – user45765
  Jul 26 at 1:46
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  (before your new comment - I'm sorry I didn't read your question carefully! but I think the following still worthwhile. )As you say$gQ =Q$, because there is one prime. So if $delta g a - ga in gQ^i+1$, for all $a$, we have that $ (g^-1 delta g) a - a in Q^i+1$ for all $a$, and conversely. (I think you are missing a $+1$ in your definition, btw, if you are using the usual numbering.) Hence $Delta_i$ is normal. For your question about $pnot | f$: certainly there are unramified extensions of degree $p$. For instance $p=2$, the spl field of $x^2+x+1=0$ over $mathbb Q_2$.
  – peter a g
  Jul 26 at 1:50
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  Because $Delta_0 = gDelta_0g^-1$. OK?
  – peter a g
  Jul 26 at 2:08
  
  

  
  
  
  

 | 
show 3 more comments

up vote
2
down vote

favorite

1

Let $L/K$ be a finite Galois extension where $K= operatornameFrac(O)$ is complete and $O$ is a Dedekind domain. Let $Delta=Gal(L/K)$. Since $K$ is complete, there is a unique prime $Q$ associated to $O_L$ where $O_L$ is the integral closure of $O_K$ in $L$.
Then let $Delta_i=deltainDeltavertforall ain O_L,delta(a)-ain Q^i$.

$Delta_1$ is unique $p$-Sylow subgroup of $Delta_0$. It is clear that $Delta_0$ is the inertia group associated to prime $Q$ of $O_L$. If $Delta_1$ is unique $p$-Sylow of $Delta_0$, then $Delta_1$ is normal by all $p$-Sylow being conjugates.

$Delta_0$ is normal due to exact sequence $1toDelta_0toDeltato operatornameGal(k_L/k)to 1$ where $k$ is residue field of $K$ and $k_L$ is the residue field of $L$.

$textbfQ:$ The book says it is follows easily $Delta_1leqDelta$ is normal. Why $Delta_1$ is normal here? Normality is non-transitive. I knew $Delta_1leqDelta_0$ normal and $Delta_0leqDelta$ normal. $ef=[L:K]$. I knew $|Delta_0|=e$. However there is no reason to expect $pnmid f$.

Ref:Algebraic Number Theory by Taylor, Frohlich, Chpter 4 section 4.

abstract-algebra number-theory

share|cite|improve this question

edited Jul 25 at 22:54

Bernard

110k635103

asked Jul 25 at 22:36

user45765

2,1942718

• 
  
  
  

  
  

  
  
  
  
  
  

  
  The definition of $Delta_i$ uses $forall ain O_L$. $Q$ is fixed by $Delta$ (as is $O_L$). Hence, one can replace $a$ by $ga$, and $Q$ by $gQ$ in the definition, for any fixed $gin Delta$. OK?
  – peter a g
  Jul 26 at 0:33
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @peterag Then you know $pnmid f$ by direct consequence. Why this is the case then? What do you mean by $gQ$ here? There is only 1 non trivial prime in $O_L$ by $K$ complete implying $L$ complete.($O_L$ is DVR.) $gQ=Q$ by $g$ is ring homomorphism.
  – user45765
  Jul 26 at 1:22
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @peterag I do not see that is the author's intention. It should be deduced from $p-$sylow part. I do not see any particular reason making this clear. $pnmid f$ is equivalent to $Delta_1$ normal.
  – user45765
  Jul 26 at 1:46
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  (before your new comment - I'm sorry I didn't read your question carefully! but I think the following still worthwhile. )As you say$gQ =Q$, because there is one prime. So if $delta g a - ga in gQ^i+1$, for all $a$, we have that $ (g^-1 delta g) a - a in Q^i+1$ for all $a$, and conversely. (I think you are missing a $+1$ in your definition, btw, if you are using the usual numbering.) Hence $Delta_i$ is normal. For your question about $pnot | f$: certainly there are unramified extensions of degree $p$. For instance $p=2$, the spl field of $x^2+x+1=0$ over $mathbb Q_2$.
  – peter a g
  Jul 26 at 1:50
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  Because $Delta_0 = gDelta_0g^-1$. OK?
  – peter a g
  Jul 26 at 2:08
  
  

  
  
  
  

 | 
show 3 more comments

up vote
2
down vote

favorite

1

up vote
2
down vote

favorite

1

1

Let $L/K$ be a finite Galois extension where $K= operatornameFrac(O)$ is complete and $O$ is a Dedekind domain. Let $Delta=Gal(L/K)$. Since $K$ is complete, there is a unique prime $Q$ associated to $O_L$ where $O_L$ is the integral closure of $O_K$ in $L$.
Then let $Delta_i=deltainDeltavertforall ain O_L,delta(a)-ain Q^i$.

$Delta_1$ is unique $p$-Sylow subgroup of $Delta_0$. It is clear that $Delta_0$ is the inertia group associated to prime $Q$ of $O_L$. If $Delta_1$ is unique $p$-Sylow of $Delta_0$, then $Delta_1$ is normal by all $p$-Sylow being conjugates.

$Delta_0$ is normal due to exact sequence $1toDelta_0toDeltato operatornameGal(k_L/k)to 1$ where $k$ is residue field of $K$ and $k_L$ is the residue field of $L$.

$textbfQ:$ The book says it is follows easily $Delta_1leqDelta$ is normal. Why $Delta_1$ is normal here? Normality is non-transitive. I knew $Delta_1leqDelta_0$ normal and $Delta_0leqDelta$ normal. $ef=[L:K]$. I knew $|Delta_0|=e$. However there is no reason to expect $pnmid f$.

Ref:Algebraic Number Theory by Taylor, Frohlich, Chpter 4 section 4.

abstract-algebra number-theory

share|cite|improve this question

edited Jul 25 at 22:54

Bernard

110k635103

asked Jul 25 at 22:36

user45765

2,1942718

Let $L/K$ be a finite Galois extension where $K= operatornameFrac(O)$ is complete and $O$ is a Dedekind domain. Let $Delta=Gal(L/K)$. Since $K$ is complete, there is a unique prime $Q$ associated to $O_L$ where $O_L$ is the integral closure of $O_K$ in $L$.
Then let $Delta_i=deltainDeltavertforall ain O_L,delta(a)-ain Q^i$.

$Delta_1$ is unique $p$-Sylow subgroup of $Delta_0$. It is clear that $Delta_0$ is the inertia group associated to prime $Q$ of $O_L$. If $Delta_1$ is unique $p$-Sylow of $Delta_0$, then $Delta_1$ is normal by all $p$-Sylow being conjugates.

$Delta_0$ is normal due to exact sequence $1toDelta_0toDeltato operatornameGal(k_L/k)to 1$ where $k$ is residue field of $K$ and $k_L$ is the residue field of $L$.

$textbfQ:$ The book says it is follows easily $Delta_1leqDelta$ is normal. Why $Delta_1$ is normal here? Normality is non-transitive. I knew $Delta_1leqDelta_0$ normal and $Delta_0leqDelta$ normal. $ef=[L:K]$. I knew $|Delta_0|=e$. However there is no reason to expect $pnmid f$.

Ref:Algebraic Number Theory by Taylor, Frohlich, Chpter 4 section 4.

abstract-algebra number-theory

share|cite|improve this question

edited Jul 25 at 22:54

Bernard

110k635103

asked Jul 25 at 22:36

user45765

2,1942718

share|cite|improve this question

share|cite|improve this question

edited Jul 25 at 22:54

Bernard

110k635103

edited Jul 25 at 22:54

Bernard

110k635103

edited Jul 25 at 22:54

Bernard

110k635103

110k635103

asked Jul 25 at 22:36

user45765

2,1942718

asked Jul 25 at 22:36

user45765

2,1942718

asked Jul 25 at 22:36

user45765

2,1942718

2,1942718

• 
  
  
  

  
  

  
  
  
  
  
  

  
  The definition of $Delta_i$ uses $forall ain O_L$. $Q$ is fixed by $Delta$ (as is $O_L$). Hence, one can replace $a$ by $ga$, and $Q$ by $gQ$ in the definition, for any fixed $gin Delta$. OK?
  – peter a g
  Jul 26 at 0:33
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @peterag Then you know $pnmid f$ by direct consequence. Why this is the case then? What do you mean by $gQ$ here? There is only 1 non trivial prime in $O_L$ by $K$ complete implying $L$ complete.($O_L$ is DVR.) $gQ=Q$ by $g$ is ring homomorphism.
  – user45765
  Jul 26 at 1:22
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @peterag I do not see that is the author's intention. It should be deduced from $p-$sylow part. I do not see any particular reason making this clear. $pnmid f$ is equivalent to $Delta_1$ normal.
  – user45765
  Jul 26 at 1:46
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  (before your new comment - I'm sorry I didn't read your question carefully! but I think the following still worthwhile. )As you say$gQ =Q$, because there is one prime. So if $delta g a - ga in gQ^i+1$, for all $a$, we have that $ (g^-1 delta g) a - a in Q^i+1$ for all $a$, and conversely. (I think you are missing a $+1$ in your definition, btw, if you are using the usual numbering.) Hence $Delta_i$ is normal. For your question about $pnot | f$: certainly there are unramified extensions of degree $p$. For instance $p=2$, the spl field of $x^2+x+1=0$ over $mathbb Q_2$.
  – peter a g
  Jul 26 at 1:50
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  Because $Delta_0 = gDelta_0g^-1$. OK?
  – peter a g
  Jul 26 at 2:08
  
  

  
  
  
  

 | 
show 3 more comments

• 
  
  
  

  
  

  
  
  
  
  
  

  
  The definition of $Delta_i$ uses $forall ain O_L$. $Q$ is fixed by $Delta$ (as is $O_L$). Hence, one can replace $a$ by $ga$, and $Q$ by $gQ$ in the definition, for any fixed $gin Delta$. OK?
  – peter a g
  Jul 26 at 0:33
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @peterag Then you know $pnmid f$ by direct consequence. Why this is the case then? What do you mean by $gQ$ here? There is only 1 non trivial prime in $O_L$ by $K$ complete implying $L$ complete.($O_L$ is DVR.) $gQ=Q$ by $g$ is ring homomorphism.
  – user45765
  Jul 26 at 1:22
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @peterag I do not see that is the author's intention. It should be deduced from $p-$sylow part. I do not see any particular reason making this clear. $pnmid f$ is equivalent to $Delta_1$ normal.
  – user45765
  Jul 26 at 1:46
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  (before your new comment - I'm sorry I didn't read your question carefully! but I think the following still worthwhile. )As you say$gQ =Q$, because there is one prime. So if $delta g a - ga in gQ^i+1$, for all $a$, we have that $ (g^-1 delta g) a - a in Q^i+1$ for all $a$, and conversely. (I think you are missing a $+1$ in your definition, btw, if you are using the usual numbering.) Hence $Delta_i$ is normal. For your question about $pnot | f$: certainly there are unramified extensions of degree $p$. For instance $p=2$, the spl field of $x^2+x+1=0$ over $mathbb Q_2$.
  – peter a g
  Jul 26 at 1:50
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  Because $Delta_0 = gDelta_0g^-1$. OK?
  – peter a g
  Jul 26 at 2:08
  
  

  
  
  
  

The definition of $Delta_i$ uses $forall ain O_L$. $Q$ is fixed by $Delta$ (as is $O_L$). Hence, one can replace $a$ by $ga$, and $Q$ by $gQ$ in the definition, for any fixed $gin Delta$. OK?
– peter a g
Jul 26 at 0:33

The definition of $Delta_i$ uses $forall ain O_L$. $Q$ is fixed by $Delta$ (as is $O_L$). Hence, one can replace $a$ by $ga$, and $Q$ by $gQ$ in the definition, for any fixed $gin Delta$. OK?
– peter a g
Jul 26 at 0:33

@peterag Then you know $pnmid f$ by direct consequence. Why this is the case then? What do you mean by $gQ$ here? There is only 1 non trivial prime in $O_L$ by $K$ complete implying $L$ complete.($O_L$ is DVR.) $gQ=Q$ by $g$ is ring homomorphism.
– user45765
Jul 26 at 1:22

@peterag Then you know $pnmid f$ by direct consequence. Why this is the case then? What do you mean by $gQ$ here? There is only 1 non trivial prime in $O_L$ by $K$ complete implying $L$ complete.($O_L$ is DVR.) $gQ=Q$ by $g$ is ring homomorphism.
– user45765
Jul 26 at 1:22

@peterag I do not see that is the author's intention. It should be deduced from $p-$sylow part. I do not see any particular reason making this clear. $pnmid f$ is equivalent to $Delta_1$ normal.
– user45765
Jul 26 at 1:46

@peterag I do not see that is the author's intention. It should be deduced from $p-$sylow part. I do not see any particular reason making this clear. $pnmid f$ is equivalent to $Delta_1$ normal.
– user45765
Jul 26 at 1:46

1

1

(before your new comment - I'm sorry I didn't read your question carefully! but I think the following still worthwhile. )As you say$gQ =Q$, because there is one prime. So if $delta g a - ga in gQ^i+1$, for all $a$, we have that $ (g^-1 delta g) a - a in Q^i+1$ for all $a$, and conversely. (I think you are missing a $+1$ in your definition, btw, if you are using the usual numbering.) Hence $Delta_i$ is normal. For your question about $pnot | f$: certainly there are unramified extensions of degree $p$. For instance $p=2$, the spl field of $x^2+x+1=0$ over $mathbb Q_2$.
– peter a g
Jul 26 at 1:50

(before your new comment - I'm sorry I didn't read your question carefully! but I think the following still worthwhile. )As you say$gQ =Q$, because there is one prime. So if $delta g a - ga in gQ^i+1$, for all $a$, we have that $ (g^-1 delta g) a - a in Q^i+1$ for all $a$, and conversely. (I think you are missing a $+1$ in your definition, btw, if you are using the usual numbering.) Hence $Delta_i$ is normal. For your question about $pnot | f$: certainly there are unramified extensions of degree $p$. For instance $p=2$, the spl field of $x^2+x+1=0$ over $mathbb Q_2$.
– peter a g
Jul 26 at 1:50

1

1

Because $Delta_0 = gDelta_0g^-1$. OK?
– peter a g
Jul 26 at 2:08

Because $Delta_0 = gDelta_0g^-1$. OK?
– peter a g
Jul 26 at 2:08

 | 
show 3 more comments

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