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Non-Galois finite extensions of $F^operatornameur$
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Let $F$ be a $p$-adic field with algebraic closure $overlineF$, and let $F^operatornameur$ be the maximal unramified extension of $F$. Let $E$ be a finite extension of $F$. I'm a bit rusty on my local class field theory and was wondering: is the composite field $EF^operatornameur$ always a Galois extension of $F^operatornameur$?
If $[EF^operatornameur : F^textrmur]$ is not divisible by $p$, then I think the answer is yes. I recall that for each $n$ with $p nmid n$, there exists a unique extension $K$ of $F^textrmur$ of degree $n$, equal to $F^operatornameur(sqrt[n]varpi)$ for a uniformizer $varpi$ of $F$. Since all $n$th roots of elements of $mathcal O_F^ast$ lie in $F^textrmur$, we have that $K$ is defined independently of the choice of uniformizer, and is Galois over $F^operatornameur$.
So a counterexample would have something to do with wild ramification.
The reason I am asking is in another of my questions, I am considering the inertia group $operatornameGal(overlineF/EF^operatornameur)$ of $E$, and was hoping for it to be normalized by $operatornameGal(overlineF/F)$, or at least by the Weil group $W_F subseteq operatornameGal(overlineF/F)$.
galois-theory algebraic-number-theory p-adic-number-theory class-field-theory local-field
asked Jul 25 at 4:33
D_S
12.8k51550
add a comment |Â
up vote
2
down vote
favorite
Let $F$ be a $p$-adic field with algebraic closure $overlineF$, and let $F^operatornameur$ be the maximal unramified extension of $F$. Let $E$ be a finite extension of $F$. I'm a bit rusty on my local class field theory and was wondering: is the composite field $EF^operatornameur$ always a Galois extension of $F^operatornameur$?
If $[EF^operatornameur : F^textrmur]$ is not divisible by $p$, then I think the answer is yes. I recall that for each $n$ with $p nmid n$, there exists a unique extension $K$ of $F^textrmur$ of degree $n$, equal to $F^operatornameur(sqrt[n]varpi)$ for a uniformizer $varpi$ of $F$. Since all $n$th roots of elements of $mathcal O_F^ast$ lie in $F^textrmur$, we have that $K$ is defined independently of the choice of uniformizer, and is Galois over $F^operatornameur$.
So a counterexample would have something to do with wild ramification.
The reason I am asking is in another of my questions, I am considering the inertia group $operatornameGal(overlineF/EF^operatornameur)$ of $E$, and was hoping for it to be normalized by $operatornameGal(overlineF/F)$, or at least by the Weil group $W_F subseteq operatornameGal(overlineF/F)$.
galois-theory algebraic-number-theory p-adic-number-theory class-field-theory local-field
asked Jul 25 at 4:33
D_S
12.8k51550
Are you claiming the every extension of $F^ur$ of finite degree prime to $p$ is obtained by the above method? This seems to be incorrect, as you only get extension abelian over $F$ in this way, and there are many extensions of $F$ which are nonabelian of degree prime to $p$ (depending on what you mean by "$p$-adic field"). Maybe I will come with an explicit counterexample later.
– Gal Porat
Jul 25 at 7:47
add a comment |Â
up vote
2
down vote
favorite
up vote
2
down vote
favorite
Let $F$ be a $p$-adic field with algebraic closure $overlineF$, and let $F^operatornameur$ be the maximal unramified extension of $F$. Let $E$ be a finite extension of $F$. I'm a bit rusty on my local class field theory and was wondering: is the composite field $EF^operatornameur$ always a Galois extension of $F^operatornameur$?
If $[EF^operatornameur : F^textrmur]$ is not divisible by $p$, then I think the answer is yes. I recall that for each $n$ with $p nmid n$, there exists a unique extension $K$ of $F^textrmur$ of degree $n$, equal to $F^operatornameur(sqrt[n]varpi)$ for a uniformizer $varpi$ of $F$. Since all $n$th roots of elements of $mathcal O_F^ast$ lie in $F^textrmur$, we have that $K$ is defined independently of the choice of uniformizer, and is Galois over $F^operatornameur$.
So a counterexample would have something to do with wild ramification.
The reason I am asking is in another of my questions, I am considering the inertia group $operatornameGal(overlineF/EF^operatornameur)$ of $E$, and was hoping for it to be normalized by $operatornameGal(overlineF/F)$, or at least by the Weil group $W_F subseteq operatornameGal(overlineF/F)$.
galois-theory algebraic-number-theory p-adic-number-theory class-field-theory local-field
asked Jul 25 at 4:33
D_S
12.8k51550
Let $F$ be a $p$-adic field with algebraic closure $overlineF$, and let $F^operatornameur$ be the maximal unramified extension of $F$. Let $E$ be a finite extension of $F$. I'm a bit rusty on my local class field theory and was wondering: is the composite field $EF^operatornameur$ always a Galois extension of $F^operatornameur$?
If $[EF^operatornameur : F^textrmur]$ is not divisible by $p$, then I think the answer is yes. I recall that for each $n$ with $p nmid n$, there exists a unique extension $K$ of $F^textrmur$ of degree $n$, equal to $F^operatornameur(sqrt[n]varpi)$ for a uniformizer $varpi$ of $F$. Since all $n$th roots of elements of $mathcal O_F^ast$ lie in $F^textrmur$, we have that $K$ is defined independently of the choice of uniformizer, and is Galois over $F^operatornameur$.
So a counterexample would have something to do with wild ramification.
The reason I am asking is in another of my questions, I am considering the inertia group $operatornameGal(overlineF/EF^operatornameur)$ of $E$, and was hoping for it to be normalized by $operatornameGal(overlineF/F)$, or at least by the Weil group $W_F subseteq operatornameGal(overlineF/F)$.
galois-theory algebraic-number-theory p-adic-number-theory class-field-theory local-field
asked Jul 25 at 4:33
D_S
12.8k51550
asked Jul 25 at 4:33
D_S
12.8k51550
asked Jul 25 at 4:33
D_S
12.8k51550
asked Jul 25 at 4:33
D_S
12.8k51550
12.8k51550
Are you claiming the every extension of $F^ur$ of finite degree prime to $p$ is obtained by the above method? This seems to be incorrect, as you only get extension abelian over $F$ in this way, and there are many extensions of $F$ which are nonabelian of degree prime to $p$ (depending on what you mean by "$p$-adic field"). Maybe I will come with an explicit counterexample later.
– Gal Porat
Jul 25 at 7:47
add a comment |Â
Are you claiming the every extension of $F^ur$ of finite degree prime to $p$ is obtained by the above method? This seems to be incorrect, as you only get extension abelian over $F$ in this way, and there are many extensions of $F$ which are nonabelian of degree prime to $p$ (depending on what you mean by "$p$-adic field"). Maybe I will come with an explicit counterexample later.
– Gal Porat
Jul 25 at 7:47
Are you claiming the every extension of $F^ur$ of finite degree prime to $p$ is obtained by the above method? This seems to be incorrect, as you only get extension abelian over $F$ in this way, and there are many extensions of $F$ which are nonabelian of degree prime to $p$ (depending on what you mean by "$p$-adic field"). Maybe I will come with an explicit counterexample later.
– Gal Porat
Jul 25 at 7:47
Are you claiming the every extension of $F^ur$ of finite degree prime to $p$ is obtained by the above method? This seems to be incorrect, as you only get extension abelian over $F$ in this way, and there are many extensions of $F$ which are nonabelian of degree prime to $p$ (depending on what you mean by "$p$-adic field"). Maybe I will come with an explicit counterexample later.
– Gal Porat
Jul 25 at 7:47
add a comment |Â
1 Answer
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up vote
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Behold! $F=Bbb Q_p$, $E=Bbb Q_p(sqrt[p]p)$, then $EF^ur=F^ur(sqrt[p]p)/F^ur$ is not normal, since it doesn't contain the $p$-th roots of unity and hence doesn't contain all roots of $x^p-p$.
answered Jul 27 at 19:07
MatheinBoulomenos
7,7211934
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
3
down vote
Behold! $F=Bbb Q_p$, $E=Bbb Q_p(sqrt[p]p)$, then $EF^ur=F^ur(sqrt[p]p)/F^ur$ is not normal, since it doesn't contain the $p$-th roots of unity and hence doesn't contain all roots of $x^p-p$.
answered Jul 27 at 19:07
MatheinBoulomenos
7,7211934
add a comment |Â
up vote
3
down vote
Behold! $F=Bbb Q_p$, $E=Bbb Q_p(sqrt[p]p)$, then $EF^ur=F^ur(sqrt[p]p)/F^ur$ is not normal, since it doesn't contain the $p$-th roots of unity and hence doesn't contain all roots of $x^p-p$.
answered Jul 27 at 19:07
MatheinBoulomenos
7,7211934
add a comment |Â
up vote
3
down vote
up vote
3
down vote
Behold! $F=Bbb Q_p$, $E=Bbb Q_p(sqrt[p]p)$, then $EF^ur=F^ur(sqrt[p]p)/F^ur$ is not normal, since it doesn't contain the $p$-th roots of unity and hence doesn't contain all roots of $x^p-p$.
answered Jul 27 at 19:07
MatheinBoulomenos
7,7211934
Behold! $F=Bbb Q_p$, $E=Bbb Q_p(sqrt[p]p)$, then $EF^ur=F^ur(sqrt[p]p)/F^ur$ is not normal, since it doesn't contain the $p$-th roots of unity and hence doesn't contain all roots of $x^p-p$.
answered Jul 27 at 19:07
MatheinBoulomenos
7,7211934
answered Jul 27 at 19:07
MatheinBoulomenos
7,7211934
answered Jul 27 at 19:07
MatheinBoulomenos
7,7211934
answered Jul 27 at 19:07
MatheinBoulomenos
7,7211934
7,7211934
add a comment |Â
add a comment |Â
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Are you claiming the every extension of $F^ur$ of finite degree prime to $p$ is obtained by the above method? This seems to be incorrect, as you only get extension abelian over $F$ in this way, and there are many extensions of $F$ which are nonabelian of degree prime to $p$ (depending on what you mean by "$p$-adic field"). Maybe I will come with an explicit counterexample later.
– Gal Porat
Jul 25 at 7:47