Mixing
cyclic extension of prime power of a local field
$begingroup$
Let $K$ be a non archimedian local field of characteristic $p>0$ with residue field $mathbb{F}_p$ and $lneq p$ be a prime.
It is known by local classfieldtheory that any abelian Galois extension $L|K$ lies in $Lsubset K^{nr}K_infty$, where $K^{nr}|K$ denotes the maximally unramified extension and $K_infty|K$ is a fixed Lubin-Tate extension, which is a tower of fields, which are totally ramified over $K$ with galois group $Gal(K_infty|K)congmathcal{O}_K^times$.
Can this fact be used to show that any cyclic Galois extension $L|K$ with degree $l^2$ is either unramified or totally ramified, or are there examples of $L|K$ with $Gal(L|K)=mathbb{Z}/l^2mathbb{Z}$, such that the inertia degree $f(L|K)=l$ and the ramification index $e(L|K)=l$?
I am searching explicitly for the case that $p=3$ and $l=2$.
galois-theory cyclic-groups class-field-theory local-field galois-extensions
asked Jan 26 at 13:04
EstusEstus
327110
$endgroup$
|
show 2 more comments
$begingroup$
Let $K$ be a non archimedian local field of characteristic $p>0$ with residue field $mathbb{F}_p$ and $lneq p$ be a prime.
It is known by local classfieldtheory that any abelian Galois extension $L|K$ lies in $Lsubset K^{nr}K_infty$, where $K^{nr}|K$ denotes the maximally unramified extension and $K_infty|K$ is a fixed Lubin-Tate extension, which is a tower of fields, which are totally ramified over $K$ with galois group $Gal(K_infty|K)congmathcal{O}_K^times$.
Can this fact be used to show that any cyclic Galois extension $L|K$ with degree $l^2$ is either unramified or totally ramified, or are there examples of $L|K$ with $Gal(L|K)=mathbb{Z}/l^2mathbb{Z}$, such that the inertia degree $f(L|K)=l$ and the ramification index $e(L|K)=l$?
I am searching explicitly for the case that $p=3$ and $l=2$.
galois-theory cyclic-groups class-field-theory local-field galois-extensions
asked Jan 26 at 13:04
EstusEstus
327110
$endgroup$
$begingroup$
Maybe like this in the situation $l^2nmid p-1$: Since $mathbb{F}_p$ is the residue field of $K$, we have $Gal(K^{nr}K_infty)=hat{mathbb{Z}}timesmathbb{Z}/(p-1)mathbb{Z}times (1+mathfrak{m}_K)$, where $mathfrak{m}_K$ is the maximal ideal of the integers of $K$. But $(1+mathfrak{m}_K)$ is an abelian pro-p-group, so the only continuous morphism to a finite group with order $l^2$ is the trivial one (Is this true?). So the only surjective continuous morphism to $mathbb{Z}/l^2mathbb{Z}$ can come from the projection $hat{mathbb{Z}}rightarrowmathbb{Z}/l^2mathbb{Z}$. Is this correct?
$endgroup$
– Estus
Jan 26 at 18:02
1
$begingroup$
Marius, won't you have non-trivial homomorphisms from $Bbb{Z}/(p-1)Bbb{Z}$ to the cyclic group of order $ell^2$ if $ellmid p-1$?
$endgroup$
– Jyrki Lahtonen
Jan 26 at 19:28
$begingroup$
Yes, exactly one, if $l=2$ and $p=3$ even. So there should be two extensions with cyclic galois group of order $4$. The unramified one given by $(pr,trivial):hat{mathbb{Z}}timesmathbb{Z}/2mathbb{Z}rightarrowmathbb{Z}/4mathbb{Z}$ and one given by $(pr,embedding):hat{mathbb{Z}}timesmathbb{Z}/2mathbb{Z}rightarrowmathbb{Z}/4mathbb{Z}$. Maybe that is your example. But I still have the hope that there is one given by an Eisenstein polynomial or something. :)
$endgroup$
– Estus
Jan 26 at 19:40
$begingroup$
I am approaching the limit of my understanding, but won't adjoining a zero of an Eisenstein polynomial yield a totally ramified extension?
$endgroup$
– Jyrki Lahtonen
Jan 26 at 19:44
$begingroup$
Yes, adjoining the zero of a eisenstein polynomial to a local field always yields a totally ramified extension.
$endgroup$
– Estus
Jan 26 at 19:47
|
show 2 more comments
$begingroup$
Let $K$ be a non archimedian local field of characteristic $p>0$ with residue field $mathbb{F}_p$ and $lneq p$ be a prime.
It is known by local classfieldtheory that any abelian Galois extension $L|K$ lies in $Lsubset K^{nr}K_infty$, where $K^{nr}|K$ denotes the maximally unramified extension and $K_infty|K$ is a fixed Lubin-Tate extension, which is a tower of fields, which are totally ramified over $K$ with galois group $Gal(K_infty|K)congmathcal{O}_K^times$.
Can this fact be used to show that any cyclic Galois extension $L|K$ with degree $l^2$ is either unramified or totally ramified, or are there examples of $L|K$ with $Gal(L|K)=mathbb{Z}/l^2mathbb{Z}$, such that the inertia degree $f(L|K)=l$ and the ramification index $e(L|K)=l$?
I am searching explicitly for the case that $p=3$ and $l=2$.
galois-theory cyclic-groups class-field-theory local-field galois-extensions
asked Jan 26 at 13:04
EstusEstus
327110
$endgroup$
Let $K$ be a non archimedian local field of characteristic $p>0$ with residue field $mathbb{F}_p$ and $lneq p$ be a prime.
It is known by local classfieldtheory that any abelian Galois extension $L|K$ lies in $Lsubset K^{nr}K_infty$, where $K^{nr}|K$ denotes the maximally unramified extension and $K_infty|K$ is a fixed Lubin-Tate extension, which is a tower of fields, which are totally ramified over $K$ with galois group $Gal(K_infty|K)congmathcal{O}_K^times$.
Can this fact be used to show that any cyclic Galois extension $L|K$ with degree $l^2$ is either unramified or totally ramified, or are there examples of $L|K$ with $Gal(L|K)=mathbb{Z}/l^2mathbb{Z}$, such that the inertia degree $f(L|K)=l$ and the ramification index $e(L|K)=l$?
I am searching explicitly for the case that $p=3$ and $l=2$.
galois-theory cyclic-groups class-field-theory local-field galois-extensions
galois-theory cyclic-groups class-field-theory local-field galois-extensions
asked Jan 26 at 13:04
EstusEstus
327110
asked Jan 26 at 13:04
EstusEstus
327110
asked Jan 26 at 13:04
EstusEstus
327110
asked Jan 26 at 13:04
EstusEstus
327110
asked Jan 26 at 13:04
EstusEstus
327110
327110
$begingroup$
Maybe like this in the situation $l^2nmid p-1$: Since $mathbb{F}_p$ is the residue field of $K$, we have $Gal(K^{nr}K_infty)=hat{mathbb{Z}}timesmathbb{Z}/(p-1)mathbb{Z}times (1+mathfrak{m}_K)$, where $mathfrak{m}_K$ is the maximal ideal of the integers of $K$. But $(1+mathfrak{m}_K)$ is an abelian pro-p-group, so the only continuous morphism to a finite group with order $l^2$ is the trivial one (Is this true?). So the only surjective continuous morphism to $mathbb{Z}/l^2mathbb{Z}$ can come from the projection $hat{mathbb{Z}}rightarrowmathbb{Z}/l^2mathbb{Z}$. Is this correct?
$endgroup$
– Estus
Jan 26 at 18:02
1
$begingroup$
Marius, won't you have non-trivial homomorphisms from $Bbb{Z}/(p-1)Bbb{Z}$ to the cyclic group of order $ell^2$ if $ellmid p-1$?
$endgroup$
– Jyrki Lahtonen
Jan 26 at 19:28
$begingroup$
Yes, exactly one, if $l=2$ and $p=3$ even. So there should be two extensions with cyclic galois group of order $4$. The unramified one given by $(pr,trivial):hat{mathbb{Z}}timesmathbb{Z}/2mathbb{Z}rightarrowmathbb{Z}/4mathbb{Z}$ and one given by $(pr,embedding):hat{mathbb{Z}}timesmathbb{Z}/2mathbb{Z}rightarrowmathbb{Z}/4mathbb{Z}$. Maybe that is your example. But I still have the hope that there is one given by an Eisenstein polynomial or something. :)
$endgroup$
– Estus
Jan 26 at 19:40
$begingroup$
I am approaching the limit of my understanding, but won't adjoining a zero of an Eisenstein polynomial yield a totally ramified extension?
$endgroup$
– Jyrki Lahtonen
Jan 26 at 19:44
$begingroup$
Yes, adjoining the zero of a eisenstein polynomial to a local field always yields a totally ramified extension.
$endgroup$
– Estus
Jan 26 at 19:47
|
show 2 more comments
$begingroup$
Maybe like this in the situation $l^2nmid p-1$: Since $mathbb{F}_p$ is the residue field of $K$, we have $Gal(K^{nr}K_infty)=hat{mathbb{Z}}timesmathbb{Z}/(p-1)mathbb{Z}times (1+mathfrak{m}_K)$, where $mathfrak{m}_K$ is the maximal ideal of the integers of $K$. But $(1+mathfrak{m}_K)$ is an abelian pro-p-group, so the only continuous morphism to a finite group with order $l^2$ is the trivial one (Is this true?). So the only surjective continuous morphism to $mathbb{Z}/l^2mathbb{Z}$ can come from the projection $hat{mathbb{Z}}rightarrowmathbb{Z}/l^2mathbb{Z}$. Is this correct?
$endgroup$
– Estus
Jan 26 at 18:02
1
$begingroup$
Marius, won't you have non-trivial homomorphisms from $Bbb{Z}/(p-1)Bbb{Z}$ to the cyclic group of order $ell^2$ if $ellmid p-1$?
$endgroup$
– Jyrki Lahtonen
Jan 26 at 19:28
$begingroup$
Yes, exactly one, if $l=2$ and $p=3$ even. So there should be two extensions with cyclic galois group of order $4$. The unramified one given by $(pr,trivial):hat{mathbb{Z}}timesmathbb{Z}/2mathbb{Z}rightarrowmathbb{Z}/4mathbb{Z}$ and one given by $(pr,embedding):hat{mathbb{Z}}timesmathbb{Z}/2mathbb{Z}rightarrowmathbb{Z}/4mathbb{Z}$. Maybe that is your example. But I still have the hope that there is one given by an Eisenstein polynomial or something. :)
$endgroup$
– Estus
Jan 26 at 19:40
$begingroup$
I am approaching the limit of my understanding, but won't adjoining a zero of an Eisenstein polynomial yield a totally ramified extension?
$endgroup$
– Jyrki Lahtonen
Jan 26 at 19:44
$begingroup$
Yes, adjoining the zero of a eisenstein polynomial to a local field always yields a totally ramified extension.
$endgroup$
– Estus
Jan 26 at 19:47
$begingroup$
Maybe like this in the situation $l^2nmid p-1$: Since $mathbb{F}_p$ is the residue field of $K$, we have $Gal(K^{nr}K_infty)=hat{mathbb{Z}}timesmathbb{Z}/(p-1)mathbb{Z}times (1+mathfrak{m}_K)$, where $mathfrak{m}_K$ is the maximal ideal of the integers of $K$. But $(1+mathfrak{m}_K)$ is an abelian pro-p-group, so the only continuous morphism to a finite group with order $l^2$ is the trivial one (Is this true?). So the only surjective continuous morphism to $mathbb{Z}/l^2mathbb{Z}$ can come from the projection $hat{mathbb{Z}}rightarrowmathbb{Z}/l^2mathbb{Z}$. Is this correct?
$endgroup$
– Estus
Jan 26 at 18:02
$begingroup$
Maybe like this in the situation $l^2nmid p-1$: Since $mathbb{F}_p$ is the residue field of $K$, we have $Gal(K^{nr}K_infty)=hat{mathbb{Z}}timesmathbb{Z}/(p-1)mathbb{Z}times (1+mathfrak{m}_K)$, where $mathfrak{m}_K$ is the maximal ideal of the integers of $K$. But $(1+mathfrak{m}_K)$ is an abelian pro-p-group, so the only continuous morphism to a finite group with order $l^2$ is the trivial one (Is this true?). So the only surjective continuous morphism to $mathbb{Z}/l^2mathbb{Z}$ can come from the projection $hat{mathbb{Z}}rightarrowmathbb{Z}/l^2mathbb{Z}$. Is this correct?
$endgroup$
– Estus
Jan 26 at 18:02
1
1
$begingroup$
Marius, won't you have non-trivial homomorphisms from $Bbb{Z}/(p-1)Bbb{Z}$ to the cyclic group of order $ell^2$ if $ellmid p-1$?
$endgroup$
– Jyrki Lahtonen
Jan 26 at 19:28
$begingroup$
Marius, won't you have non-trivial homomorphisms from $Bbb{Z}/(p-1)Bbb{Z}$ to the cyclic group of order $ell^2$ if $ellmid p-1$?
$endgroup$
– Jyrki Lahtonen
Jan 26 at 19:28
$begingroup$
Yes, exactly one, if $l=2$ and $p=3$ even. So there should be two extensions with cyclic galois group of order $4$. The unramified one given by $(pr,trivial):hat{mathbb{Z}}timesmathbb{Z}/2mathbb{Z}rightarrowmathbb{Z}/4mathbb{Z}$ and one given by $(pr,embedding):hat{mathbb{Z}}timesmathbb{Z}/2mathbb{Z}rightarrowmathbb{Z}/4mathbb{Z}$. Maybe that is your example. But I still have the hope that there is one given by an Eisenstein polynomial or something. :)
$endgroup$
– Estus
Jan 26 at 19:40
$begingroup$
Yes, exactly one, if $l=2$ and $p=3$ even. So there should be two extensions with cyclic galois group of order $4$. The unramified one given by $(pr,trivial):hat{mathbb{Z}}timesmathbb{Z}/2mathbb{Z}rightarrowmathbb{Z}/4mathbb{Z}$ and one given by $(pr,embedding):hat{mathbb{Z}}timesmathbb{Z}/2mathbb{Z}rightarrowmathbb{Z}/4mathbb{Z}$. Maybe that is your example. But I still have the hope that there is one given by an Eisenstein polynomial or something. :)
$endgroup$
– Estus
Jan 26 at 19:40
$begingroup$
I am approaching the limit of my understanding, but won't adjoining a zero of an Eisenstein polynomial yield a totally ramified extension?
$endgroup$
– Jyrki Lahtonen
Jan 26 at 19:44
$begingroup$
I am approaching the limit of my understanding, but won't adjoining a zero of an Eisenstein polynomial yield a totally ramified extension?
$endgroup$
– Jyrki Lahtonen
Jan 26 at 19:44
$begingroup$
Yes, adjoining the zero of a eisenstein polynomial to a local field always yields a totally ramified extension.
$endgroup$
– Estus
Jan 26 at 19:47
$begingroup$
Yes, adjoining the zero of a eisenstein polynomial to a local field always yields a totally ramified extension.
$endgroup$
– Estus
Jan 26 at 19:47
|
show 2 more comments
2 Answers
2
active
oldest
votes
$begingroup$
I think the following is an example of a cyclic extension of degree $4$ over the field
$K=Bbb{F}_3((x))$ with $e=f=2$.
Let $i$ be a solution of $i^2+1=0$ in the extension field $Bbb{F}_9$. Let
$$
u=sqrt{(1+i)x},quad v=sqrt{(1-i)x}.
$$
It follows that $uv=pm ixnotin K$. Similarly $u^2,v^2notin K$, but $u^2+v^2=-xin K$. This implies that
$$
p(T)=(T^2-u^2)(T^2-v^2)=T^4+xT^2-x^2in K[T]
$$
is irreducible. None of the zeros $pm u,pm v$ are in $K$, ruling out linear factors, and no product of two zeros is in $K$ ruling out quadratic factors.
The coefficients $a=x$ and $b=-x^2$ satisfy the condition described here, part (b). Neither $b=-x^2$ nor $a^2-4b=-x^2$
is a square in $K$ but their product is. Therefore the Galois group of $p(T)$ over $K$
is cyclic of order four.
We easily see that $L=K(u)$ is the splitting field. The element $i=(u^2/x)-1in L$
as is $v=ix/u$. So $L$ has a subfield isomorphic to $Bbb{F}_9$. This forces the inertia degree to be divisible by two. On the other hand $u^2=(1+i)x$ forces the ramification degree $e$ to be at least two also. As $[L:K]=ef=4$, the conclusion is that we must have $e=f=2$.
It is a bit unnerving that the leading coefficient of $u$ appears to be $sqrt{1+i}notinBbb{F}_9$. The explanation is surely that $sqrt xnotin L$! Mentioning this because something similar may allow us to generalize this construction to other $(p,ell)$ pairs? A key ingredient seems to be that $L=Bbb{F}_9((u))$, but $u$ is not really a fractional power of the original local parameter $x$. We have this extra twist coming from the coefficient $1+i$ under the square root. Similar twisting may give us extensions with $e=f=ell$ whenever $ellmid p-1$, but that needs a bit more work.
answered Jan 26 at 20:15
Jyrki LahtonenJyrki Lahtonen
110k13171386
$endgroup$
add a comment |
$begingroup$
Expanding the idea from the last paragraph of my first attempt to an example of a cyclic extension $L/K$ of characteristic $p$ local fields such that $e=f=ell$. As Marius foresaw in the comments, we need $ell$ to be a factor of $p-1$.
Let $K=Bbb{F}_p((x))$. Let $E=Bbb{F}_{p^{ell^2}}$ be an extended constant field, and let $M=E((x^{1/ell}))$. As we assume $ellmid p-1$ there exists a root of unity $zeta$ of order $ell$ in the prime field $Bbb{F}_p$. The extensions $E((x))/K$ and
$K(x^{1/ell})/K$ are known to be Galois with cyclic Galois groups of respective orders $ell^2$ and $ell$. One is unramified and the other is totally ramified, so they are linearly disjoint. Hence their compositum $M$ is also Galois with Galois group $G=Bbb{Z}/langle ell^2rangletimes Bbb{Z}/langle ellrangle$. More precisely, the $K$-automorphism $sigma_{a,b}$ of $M$ associated to $(a,b)in G$
is the automorphism fully described by
$$
sigma_{a,b}(z)=z^{p^a} text{for all $zin E$},quad sigma_{a,b}(x^{1/ell})=zeta^b x^{1/ell}.
$$
Let us consider the subgroup $Hle G$ generated by $tau:=sigma_{ell,1}$. Clearly
$tau$ is of order $ell$, and $G/Hsimeq Bbb{Z}/langle ell^2rangle$.
Next I want to identify the fixed field $L=M^H$. Clearly $Bbb{F}_{p^ell}subset E$
is fixed by all the automorphisms in $H$. To get all of $L$ I need the following observation.
Fact. There exists an element $etain E$ such that $eta^{p^ell-1}=1/zeta$.
Proof. Consider the homomorphism of cyclic groups $f:E^*to E^*$ given by
$f(z)=z^{p^ell-1}$. Because $pequiv1pmodell$
$$
N:=frac{p^{ell^2}-1}{p^ell-1}=1+p^ell+p^{2ell}+cdots+p^{(ell-1)ell}equiv1+1+cdots+1equiv0pmod{ell}.
$$
The image of $f$ is the unique cyclic subgroup of $E^*$ of order $N$. We just saw that $ellmid N$ implying that all $ell$th roots of unity are in $operatorname{Im}(f)$. QED
The rest is straightforward. With $eta$ an element promised by the above observation let $u=eta x^{1/ell}$. Then
$$
sigma_{ell,1}(u)=eta^{p^ell}zeta x^{1/ell}=eta(eta^{p^ell-1}zeta)x^{1/ell}=u.
$$
So the field $Bbb{F}_{p^ell}((u))subseteq M^H$. Clearly this is a degree $ell^2$ extension of $K$, so $L=Bbb{F}_{p^ell}((u))$.
Here $Gal(L/K)simeq G/H$ is cyclic of order $ell^2$ as prescribed.
Equally clearly $e(L/K)=ell=f(L/K)$.
This is surely a more illuminating and generalizable construction. The methods in the other answer are ad hoc.
answered Jan 27 at 5:51
Jyrki LahtonenJyrki Lahtonen
110k13171386
$endgroup$
add a comment |
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2 Answers
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2 Answers
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$begingroup$
I think the following is an example of a cyclic extension of degree $4$ over the field
$K=Bbb{F}_3((x))$ with $e=f=2$.
Let $i$ be a solution of $i^2+1=0$ in the extension field $Bbb{F}_9$. Let
$$
u=sqrt{(1+i)x},quad v=sqrt{(1-i)x}.
$$
It follows that $uv=pm ixnotin K$. Similarly $u^2,v^2notin K$, but $u^2+v^2=-xin K$. This implies that
$$
p(T)=(T^2-u^2)(T^2-v^2)=T^4+xT^2-x^2in K[T]
$$
is irreducible. None of the zeros $pm u,pm v$ are in $K$, ruling out linear factors, and no product of two zeros is in $K$ ruling out quadratic factors.
The coefficients $a=x$ and $b=-x^2$ satisfy the condition described here, part (b). Neither $b=-x^2$ nor $a^2-4b=-x^2$
is a square in $K$ but their product is. Therefore the Galois group of $p(T)$ over $K$
is cyclic of order four.
We easily see that $L=K(u)$ is the splitting field. The element $i=(u^2/x)-1in L$
as is $v=ix/u$. So $L$ has a subfield isomorphic to $Bbb{F}_9$. This forces the inertia degree to be divisible by two. On the other hand $u^2=(1+i)x$ forces the ramification degree $e$ to be at least two also. As $[L:K]=ef=4$, the conclusion is that we must have $e=f=2$.
It is a bit unnerving that the leading coefficient of $u$ appears to be $sqrt{1+i}notinBbb{F}_9$. The explanation is surely that $sqrt xnotin L$! Mentioning this because something similar may allow us to generalize this construction to other $(p,ell)$ pairs? A key ingredient seems to be that $L=Bbb{F}_9((u))$, but $u$ is not really a fractional power of the original local parameter $x$. We have this extra twist coming from the coefficient $1+i$ under the square root. Similar twisting may give us extensions with $e=f=ell$ whenever $ellmid p-1$, but that needs a bit more work.
answered Jan 26 at 20:15
Jyrki LahtonenJyrki Lahtonen
110k13171386
$endgroup$
add a comment |
$begingroup$
I think the following is an example of a cyclic extension of degree $4$ over the field
$K=Bbb{F}_3((x))$ with $e=f=2$.
Let $i$ be a solution of $i^2+1=0$ in the extension field $Bbb{F}_9$. Let
$$
u=sqrt{(1+i)x},quad v=sqrt{(1-i)x}.
$$
It follows that $uv=pm ixnotin K$. Similarly $u^2,v^2notin K$, but $u^2+v^2=-xin K$. This implies that
$$
p(T)=(T^2-u^2)(T^2-v^2)=T^4+xT^2-x^2in K[T]
$$
is irreducible. None of the zeros $pm u,pm v$ are in $K$, ruling out linear factors, and no product of two zeros is in $K$ ruling out quadratic factors.
The coefficients $a=x$ and $b=-x^2$ satisfy the condition described here, part (b). Neither $b=-x^2$ nor $a^2-4b=-x^2$
is a square in $K$ but their product is. Therefore the Galois group of $p(T)$ over $K$
is cyclic of order four.
We easily see that $L=K(u)$ is the splitting field. The element $i=(u^2/x)-1in L$
as is $v=ix/u$. So $L$ has a subfield isomorphic to $Bbb{F}_9$. This forces the inertia degree to be divisible by two. On the other hand $u^2=(1+i)x$ forces the ramification degree $e$ to be at least two also. As $[L:K]=ef=4$, the conclusion is that we must have $e=f=2$.
It is a bit unnerving that the leading coefficient of $u$ appears to be $sqrt{1+i}notinBbb{F}_9$. The explanation is surely that $sqrt xnotin L$! Mentioning this because something similar may allow us to generalize this construction to other $(p,ell)$ pairs? A key ingredient seems to be that $L=Bbb{F}_9((u))$, but $u$ is not really a fractional power of the original local parameter $x$. We have this extra twist coming from the coefficient $1+i$ under the square root. Similar twisting may give us extensions with $e=f=ell$ whenever $ellmid p-1$, but that needs a bit more work.
answered Jan 26 at 20:15
Jyrki LahtonenJyrki Lahtonen
110k13171386
$endgroup$
add a comment |
$begingroup$
I think the following is an example of a cyclic extension of degree $4$ over the field
$K=Bbb{F}_3((x))$ with $e=f=2$.
Let $i$ be a solution of $i^2+1=0$ in the extension field $Bbb{F}_9$. Let
$$
u=sqrt{(1+i)x},quad v=sqrt{(1-i)x}.
$$
It follows that $uv=pm ixnotin K$. Similarly $u^2,v^2notin K$, but $u^2+v^2=-xin K$. This implies that
$$
p(T)=(T^2-u^2)(T^2-v^2)=T^4+xT^2-x^2in K[T]
$$
is irreducible. None of the zeros $pm u,pm v$ are in $K$, ruling out linear factors, and no product of two zeros is in $K$ ruling out quadratic factors.
The coefficients $a=x$ and $b=-x^2$ satisfy the condition described here, part (b). Neither $b=-x^2$ nor $a^2-4b=-x^2$
is a square in $K$ but their product is. Therefore the Galois group of $p(T)$ over $K$
is cyclic of order four.
We easily see that $L=K(u)$ is the splitting field. The element $i=(u^2/x)-1in L$
as is $v=ix/u$. So $L$ has a subfield isomorphic to $Bbb{F}_9$. This forces the inertia degree to be divisible by two. On the other hand $u^2=(1+i)x$ forces the ramification degree $e$ to be at least two also. As $[L:K]=ef=4$, the conclusion is that we must have $e=f=2$.
It is a bit unnerving that the leading coefficient of $u$ appears to be $sqrt{1+i}notinBbb{F}_9$. The explanation is surely that $sqrt xnotin L$! Mentioning this because something similar may allow us to generalize this construction to other $(p,ell)$ pairs? A key ingredient seems to be that $L=Bbb{F}_9((u))$, but $u$ is not really a fractional power of the original local parameter $x$. We have this extra twist coming from the coefficient $1+i$ under the square root. Similar twisting may give us extensions with $e=f=ell$ whenever $ellmid p-1$, but that needs a bit more work.
answered Jan 26 at 20:15
Jyrki LahtonenJyrki Lahtonen
110k13171386
$endgroup$
I think the following is an example of a cyclic extension of degree $4$ over the field
$K=Bbb{F}_3((x))$ with $e=f=2$.
Let $i$ be a solution of $i^2+1=0$ in the extension field $Bbb{F}_9$. Let
$$
u=sqrt{(1+i)x},quad v=sqrt{(1-i)x}.
$$
It follows that $uv=pm ixnotin K$. Similarly $u^2,v^2notin K$, but $u^2+v^2=-xin K$. This implies that
$$
p(T)=(T^2-u^2)(T^2-v^2)=T^4+xT^2-x^2in K[T]
$$
is irreducible. None of the zeros $pm u,pm v$ are in $K$, ruling out linear factors, and no product of two zeros is in $K$ ruling out quadratic factors.
The coefficients $a=x$ and $b=-x^2$ satisfy the condition described here, part (b). Neither $b=-x^2$ nor $a^2-4b=-x^2$
is a square in $K$ but their product is. Therefore the Galois group of $p(T)$ over $K$
is cyclic of order four.
We easily see that $L=K(u)$ is the splitting field. The element $i=(u^2/x)-1in L$
as is $v=ix/u$. So $L$ has a subfield isomorphic to $Bbb{F}_9$. This forces the inertia degree to be divisible by two. On the other hand $u^2=(1+i)x$ forces the ramification degree $e$ to be at least two also. As $[L:K]=ef=4$, the conclusion is that we must have $e=f=2$.
It is a bit unnerving that the leading coefficient of $u$ appears to be $sqrt{1+i}notinBbb{F}_9$. The explanation is surely that $sqrt xnotin L$! Mentioning this because something similar may allow us to generalize this construction to other $(p,ell)$ pairs? A key ingredient seems to be that $L=Bbb{F}_9((u))$, but $u$ is not really a fractional power of the original local parameter $x$. We have this extra twist coming from the coefficient $1+i$ under the square root. Similar twisting may give us extensions with $e=f=ell$ whenever $ellmid p-1$, but that needs a bit more work.
answered Jan 26 at 20:15
Jyrki LahtonenJyrki Lahtonen
110k13171386
edited Jan 27 at 4:43
answered Jan 26 at 20:15
Jyrki LahtonenJyrki Lahtonen
110k13171386
answered Jan 26 at 20:15
Jyrki LahtonenJyrki Lahtonen
110k13171386
answered Jan 26 at 20:15
Jyrki LahtonenJyrki Lahtonen
110k13171386
110k13171386
add a comment |
add a comment |
$begingroup$
Expanding the idea from the last paragraph of my first attempt to an example of a cyclic extension $L/K$ of characteristic $p$ local fields such that $e=f=ell$. As Marius foresaw in the comments, we need $ell$ to be a factor of $p-1$.
Let $K=Bbb{F}_p((x))$. Let $E=Bbb{F}_{p^{ell^2}}$ be an extended constant field, and let $M=E((x^{1/ell}))$. As we assume $ellmid p-1$ there exists a root of unity $zeta$ of order $ell$ in the prime field $Bbb{F}_p$. The extensions $E((x))/K$ and
$K(x^{1/ell})/K$ are known to be Galois with cyclic Galois groups of respective orders $ell^2$ and $ell$. One is unramified and the other is totally ramified, so they are linearly disjoint. Hence their compositum $M$ is also Galois with Galois group $G=Bbb{Z}/langle ell^2rangletimes Bbb{Z}/langle ellrangle$. More precisely, the $K$-automorphism $sigma_{a,b}$ of $M$ associated to $(a,b)in G$
is the automorphism fully described by
$$
sigma_{a,b}(z)=z^{p^a} text{for all $zin E$},quad sigma_{a,b}(x^{1/ell})=zeta^b x^{1/ell}.
$$
Let us consider the subgroup $Hle G$ generated by $tau:=sigma_{ell,1}$. Clearly
$tau$ is of order $ell$, and $G/Hsimeq Bbb{Z}/langle ell^2rangle$.
Next I want to identify the fixed field $L=M^H$. Clearly $Bbb{F}_{p^ell}subset E$
is fixed by all the automorphisms in $H$. To get all of $L$ I need the following observation.
Fact. There exists an element $etain E$ such that $eta^{p^ell-1}=1/zeta$.
Proof. Consider the homomorphism of cyclic groups $f:E^*to E^*$ given by
$f(z)=z^{p^ell-1}$. Because $pequiv1pmodell$
$$
N:=frac{p^{ell^2}-1}{p^ell-1}=1+p^ell+p^{2ell}+cdots+p^{(ell-1)ell}equiv1+1+cdots+1equiv0pmod{ell}.
$$
The image of $f$ is the unique cyclic subgroup of $E^*$ of order $N$. We just saw that $ellmid N$ implying that all $ell$th roots of unity are in $operatorname{Im}(f)$. QED
The rest is straightforward. With $eta$ an element promised by the above observation let $u=eta x^{1/ell}$. Then
$$
sigma_{ell,1}(u)=eta^{p^ell}zeta x^{1/ell}=eta(eta^{p^ell-1}zeta)x^{1/ell}=u.
$$
So the field $Bbb{F}_{p^ell}((u))subseteq M^H$. Clearly this is a degree $ell^2$ extension of $K$, so $L=Bbb{F}_{p^ell}((u))$.
Here $Gal(L/K)simeq G/H$ is cyclic of order $ell^2$ as prescribed.
Equally clearly $e(L/K)=ell=f(L/K)$.
This is surely a more illuminating and generalizable construction. The methods in the other answer are ad hoc.
answered Jan 27 at 5:51
Jyrki LahtonenJyrki Lahtonen
110k13171386
$endgroup$
add a comment |
$begingroup$
Expanding the idea from the last paragraph of my first attempt to an example of a cyclic extension $L/K$ of characteristic $p$ local fields such that $e=f=ell$. As Marius foresaw in the comments, we need $ell$ to be a factor of $p-1$.
Let $K=Bbb{F}_p((x))$. Let $E=Bbb{F}_{p^{ell^2}}$ be an extended constant field, and let $M=E((x^{1/ell}))$. As we assume $ellmid p-1$ there exists a root of unity $zeta$ of order $ell$ in the prime field $Bbb{F}_p$. The extensions $E((x))/K$ and
$K(x^{1/ell})/K$ are known to be Galois with cyclic Galois groups of respective orders $ell^2$ and $ell$. One is unramified and the other is totally ramified, so they are linearly disjoint. Hence their compositum $M$ is also Galois with Galois group $G=Bbb{Z}/langle ell^2rangletimes Bbb{Z}/langle ellrangle$. More precisely, the $K$-automorphism $sigma_{a,b}$ of $M$ associated to $(a,b)in G$
is the automorphism fully described by
$$
sigma_{a,b}(z)=z^{p^a} text{for all $zin E$},quad sigma_{a,b}(x^{1/ell})=zeta^b x^{1/ell}.
$$
Let us consider the subgroup $Hle G$ generated by $tau:=sigma_{ell,1}$. Clearly
$tau$ is of order $ell$, and $G/Hsimeq Bbb{Z}/langle ell^2rangle$.
Next I want to identify the fixed field $L=M^H$. Clearly $Bbb{F}_{p^ell}subset E$
is fixed by all the automorphisms in $H$. To get all of $L$ I need the following observation.
Fact. There exists an element $etain E$ such that $eta^{p^ell-1}=1/zeta$.
Proof. Consider the homomorphism of cyclic groups $f:E^*to E^*$ given by
$f(z)=z^{p^ell-1}$. Because $pequiv1pmodell$
$$
N:=frac{p^{ell^2}-1}{p^ell-1}=1+p^ell+p^{2ell}+cdots+p^{(ell-1)ell}equiv1+1+cdots+1equiv0pmod{ell}.
$$
The image of $f$ is the unique cyclic subgroup of $E^*$ of order $N$. We just saw that $ellmid N$ implying that all $ell$th roots of unity are in $operatorname{Im}(f)$. QED
The rest is straightforward. With $eta$ an element promised by the above observation let $u=eta x^{1/ell}$. Then
$$
sigma_{ell,1}(u)=eta^{p^ell}zeta x^{1/ell}=eta(eta^{p^ell-1}zeta)x^{1/ell}=u.
$$
So the field $Bbb{F}_{p^ell}((u))subseteq M^H$. Clearly this is a degree $ell^2$ extension of $K$, so $L=Bbb{F}_{p^ell}((u))$.
Here $Gal(L/K)simeq G/H$ is cyclic of order $ell^2$ as prescribed.
Equally clearly $e(L/K)=ell=f(L/K)$.
This is surely a more illuminating and generalizable construction. The methods in the other answer are ad hoc.
answered Jan 27 at 5:51
Jyrki LahtonenJyrki Lahtonen
110k13171386
$endgroup$
add a comment |
$begingroup$
Expanding the idea from the last paragraph of my first attempt to an example of a cyclic extension $L/K$ of characteristic $p$ local fields such that $e=f=ell$. As Marius foresaw in the comments, we need $ell$ to be a factor of $p-1$.
Let $K=Bbb{F}_p((x))$. Let $E=Bbb{F}_{p^{ell^2}}$ be an extended constant field, and let $M=E((x^{1/ell}))$. As we assume $ellmid p-1$ there exists a root of unity $zeta$ of order $ell$ in the prime field $Bbb{F}_p$. The extensions $E((x))/K$ and
$K(x^{1/ell})/K$ are known to be Galois with cyclic Galois groups of respective orders $ell^2$ and $ell$. One is unramified and the other is totally ramified, so they are linearly disjoint. Hence their compositum $M$ is also Galois with Galois group $G=Bbb{Z}/langle ell^2rangletimes Bbb{Z}/langle ellrangle$. More precisely, the $K$-automorphism $sigma_{a,b}$ of $M$ associated to $(a,b)in G$
is the automorphism fully described by
$$
sigma_{a,b}(z)=z^{p^a} text{for all $zin E$},quad sigma_{a,b}(x^{1/ell})=zeta^b x^{1/ell}.
$$
Let us consider the subgroup $Hle G$ generated by $tau:=sigma_{ell,1}$. Clearly
$tau$ is of order $ell$, and $G/Hsimeq Bbb{Z}/langle ell^2rangle$.
Next I want to identify the fixed field $L=M^H$. Clearly $Bbb{F}_{p^ell}subset E$
is fixed by all the automorphisms in $H$. To get all of $L$ I need the following observation.
Fact. There exists an element $etain E$ such that $eta^{p^ell-1}=1/zeta$.
Proof. Consider the homomorphism of cyclic groups $f:E^*to E^*$ given by
$f(z)=z^{p^ell-1}$. Because $pequiv1pmodell$
$$
N:=frac{p^{ell^2}-1}{p^ell-1}=1+p^ell+p^{2ell}+cdots+p^{(ell-1)ell}equiv1+1+cdots+1equiv0pmod{ell}.
$$
The image of $f$ is the unique cyclic subgroup of $E^*$ of order $N$. We just saw that $ellmid N$ implying that all $ell$th roots of unity are in $operatorname{Im}(f)$. QED
The rest is straightforward. With $eta$ an element promised by the above observation let $u=eta x^{1/ell}$. Then
$$
sigma_{ell,1}(u)=eta^{p^ell}zeta x^{1/ell}=eta(eta^{p^ell-1}zeta)x^{1/ell}=u.
$$
So the field $Bbb{F}_{p^ell}((u))subseteq M^H$. Clearly this is a degree $ell^2$ extension of $K$, so $L=Bbb{F}_{p^ell}((u))$.
Here $Gal(L/K)simeq G/H$ is cyclic of order $ell^2$ as prescribed.
Equally clearly $e(L/K)=ell=f(L/K)$.
This is surely a more illuminating and generalizable construction. The methods in the other answer are ad hoc.
answered Jan 27 at 5:51
Jyrki LahtonenJyrki Lahtonen
110k13171386
$endgroup$
Expanding the idea from the last paragraph of my first attempt to an example of a cyclic extension $L/K$ of characteristic $p$ local fields such that $e=f=ell$. As Marius foresaw in the comments, we need $ell$ to be a factor of $p-1$.
Let $K=Bbb{F}_p((x))$. Let $E=Bbb{F}_{p^{ell^2}}$ be an extended constant field, and let $M=E((x^{1/ell}))$. As we assume $ellmid p-1$ there exists a root of unity $zeta$ of order $ell$ in the prime field $Bbb{F}_p$. The extensions $E((x))/K$ and
$K(x^{1/ell})/K$ are known to be Galois with cyclic Galois groups of respective orders $ell^2$ and $ell$. One is unramified and the other is totally ramified, so they are linearly disjoint. Hence their compositum $M$ is also Galois with Galois group $G=Bbb{Z}/langle ell^2rangletimes Bbb{Z}/langle ellrangle$. More precisely, the $K$-automorphism $sigma_{a,b}$ of $M$ associated to $(a,b)in G$
is the automorphism fully described by
$$
sigma_{a,b}(z)=z^{p^a} text{for all $zin E$},quad sigma_{a,b}(x^{1/ell})=zeta^b x^{1/ell}.
$$
Let us consider the subgroup $Hle G$ generated by $tau:=sigma_{ell,1}$. Clearly
$tau$ is of order $ell$, and $G/Hsimeq Bbb{Z}/langle ell^2rangle$.
Next I want to identify the fixed field $L=M^H$. Clearly $Bbb{F}_{p^ell}subset E$
is fixed by all the automorphisms in $H$. To get all of $L$ I need the following observation.
Fact. There exists an element $etain E$ such that $eta^{p^ell-1}=1/zeta$.
Proof. Consider the homomorphism of cyclic groups $f:E^*to E^*$ given by
$f(z)=z^{p^ell-1}$. Because $pequiv1pmodell$
$$
N:=frac{p^{ell^2}-1}{p^ell-1}=1+p^ell+p^{2ell}+cdots+p^{(ell-1)ell}equiv1+1+cdots+1equiv0pmod{ell}.
$$
The image of $f$ is the unique cyclic subgroup of $E^*$ of order $N$. We just saw that $ellmid N$ implying that all $ell$th roots of unity are in $operatorname{Im}(f)$. QED
The rest is straightforward. With $eta$ an element promised by the above observation let $u=eta x^{1/ell}$. Then
$$
sigma_{ell,1}(u)=eta^{p^ell}zeta x^{1/ell}=eta(eta^{p^ell-1}zeta)x^{1/ell}=u.
$$
So the field $Bbb{F}_{p^ell}((u))subseteq M^H$. Clearly this is a degree $ell^2$ extension of $K$, so $L=Bbb{F}_{p^ell}((u))$.
Here $Gal(L/K)simeq G/H$ is cyclic of order $ell^2$ as prescribed.
Equally clearly $e(L/K)=ell=f(L/K)$.
This is surely a more illuminating and generalizable construction. The methods in the other answer are ad hoc.
answered Jan 27 at 5:51
Jyrki LahtonenJyrki Lahtonen
110k13171386
answered Jan 27 at 5:51
Jyrki LahtonenJyrki Lahtonen
110k13171386
answered Jan 27 at 5:51
Jyrki LahtonenJyrki Lahtonen
110k13171386
answered Jan 27 at 5:51
Jyrki LahtonenJyrki Lahtonen
110k13171386
110k13171386
add a comment |
add a comment |
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$begingroup$
Maybe like this in the situation $l^2nmid p-1$: Since $mathbb{F}_p$ is the residue field of $K$, we have $Gal(K^{nr}K_infty)=hat{mathbb{Z}}timesmathbb{Z}/(p-1)mathbb{Z}times (1+mathfrak{m}_K)$, where $mathfrak{m}_K$ is the maximal ideal of the integers of $K$. But $(1+mathfrak{m}_K)$ is an abelian pro-p-group, so the only continuous morphism to a finite group with order $l^2$ is the trivial one (Is this true?). So the only surjective continuous morphism to $mathbb{Z}/l^2mathbb{Z}$ can come from the projection $hat{mathbb{Z}}rightarrowmathbb{Z}/l^2mathbb{Z}$. Is this correct?
$endgroup$
– Estus
Jan 26 at 18:02
1
$begingroup$
Marius, won't you have non-trivial homomorphisms from $Bbb{Z}/(p-1)Bbb{Z}$ to the cyclic group of order $ell^2$ if $ellmid p-1$?
$endgroup$
– Jyrki Lahtonen
Jan 26 at 19:28
$begingroup$
Yes, exactly one, if $l=2$ and $p=3$ even. So there should be two extensions with cyclic galois group of order $4$. The unramified one given by $(pr,trivial):hat{mathbb{Z}}timesmathbb{Z}/2mathbb{Z}rightarrowmathbb{Z}/4mathbb{Z}$ and one given by $(pr,embedding):hat{mathbb{Z}}timesmathbb{Z}/2mathbb{Z}rightarrowmathbb{Z}/4mathbb{Z}$. Maybe that is your example. But I still have the hope that there is one given by an Eisenstein polynomial or something. :)
$endgroup$
– Estus
Jan 26 at 19:40
$begingroup$
I am approaching the limit of my understanding, but won't adjoining a zero of an Eisenstein polynomial yield a totally ramified extension?
$endgroup$
– Jyrki Lahtonen
Jan 26 at 19:44
$begingroup$
Yes, adjoining the zero of a eisenstein polynomial to a local field always yields a totally ramified extension.
$endgroup$
– Estus
Jan 26 at 19:47