Counting tamely ramified Galois extensions of $mathbb{Q}_p$ with a given Galois group.
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For a homework exercise, I'm to determine for each $p$ the number of non-isomorphic tamely ramified Galois extensions $K/mathbb{Q}_p$ such that $operatorname{Gal}(K/mathbb{Q}_p) cong mathcal{Q}_8$ (the quaternion group of order $8$).
I was out of class the day we covered this, and I don't see the topic discussed in Gouvêa's book, so all I have is a lone equation from a classmate's notes:
$$#{text{tame Galois extensions } K/mathbb{Q}_p , : , operatorname{Gal}(K/mathbb{Q}_p) cong G} = frac{#{(a,b) in G times G , : , aba^{-1}=b^{p},, langle a,b rangle = G}}{left|operatorname{Aut}(G)right|}$$
I know that $left|operatorname{Aut}(mathcal{Q}_8)right| = !left|mathcal{S}_4right| = 24$, so when $pequiv 1,2 !pmod{4}$, I get that there are $0$ such Galois extensions, and when $pequiv 3 !pmod{4}$, I get that there is only $1$. This seems odd to me.
Where does this equation come from? Is it even correct? I have tried to Google it, but have thus far been unsuccessful.
algebraic-number-theory p-adic-number-theory
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add a comment |
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For a homework exercise, I'm to determine for each $p$ the number of non-isomorphic tamely ramified Galois extensions $K/mathbb{Q}_p$ such that $operatorname{Gal}(K/mathbb{Q}_p) cong mathcal{Q}_8$ (the quaternion group of order $8$).
I was out of class the day we covered this, and I don't see the topic discussed in Gouvêa's book, so all I have is a lone equation from a classmate's notes:
$$#{text{tame Galois extensions } K/mathbb{Q}_p , : , operatorname{Gal}(K/mathbb{Q}_p) cong G} = frac{#{(a,b) in G times G , : , aba^{-1}=b^{p},, langle a,b rangle = G}}{left|operatorname{Aut}(G)right|}$$
I know that $left|operatorname{Aut}(mathcal{Q}_8)right| = !left|mathcal{S}_4right| = 24$, so when $pequiv 1,2 !pmod{4}$, I get that there are $0$ such Galois extensions, and when $pequiv 3 !pmod{4}$, I get that there is only $1$. This seems odd to me.
Where does this equation come from? Is it even correct? I have tried to Google it, but have thus far been unsuccessful.
algebraic-number-theory p-adic-number-theory
$endgroup$
1
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I am not familiar with that formula either, but I don‘t find your count surprising. $Q_8$ is not an easy group to find among your random collection of extensions. In fact, I’d like to see an explicit description of that quaternion-group extension of $mathbb Q_3$.
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– Lubin
Nov 30 '14 at 23:16
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@Lubin The construction of the unique $Q_8$-extension of $mathbb{Q}_p$ in the $pequiv, 3, (text{mod} 4)$ case goes back to Witt's 1936 paper on embeddings of biquadratic extensions into $Q_8$-extensions. The recipe is quoted in the paper "Quaternion Extensions" by Jensen & Yui - see th. I.1.1 and cor. II.3.6 (available here); note that th. I.1.1. corrects a typo in Witt's paper. It should be possible to unpack an explicit construction for the $p=3$ case from the results quoted in Jensen-Yui.
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– Giovanni Di Matteo
Jan 9 at 16:33
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Thanks, @GiovanniDiMatteo.
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– Lubin
Jan 9 at 21:18
add a comment |
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For a homework exercise, I'm to determine for each $p$ the number of non-isomorphic tamely ramified Galois extensions $K/mathbb{Q}_p$ such that $operatorname{Gal}(K/mathbb{Q}_p) cong mathcal{Q}_8$ (the quaternion group of order $8$).
I was out of class the day we covered this, and I don't see the topic discussed in Gouvêa's book, so all I have is a lone equation from a classmate's notes:
$$#{text{tame Galois extensions } K/mathbb{Q}_p , : , operatorname{Gal}(K/mathbb{Q}_p) cong G} = frac{#{(a,b) in G times G , : , aba^{-1}=b^{p},, langle a,b rangle = G}}{left|operatorname{Aut}(G)right|}$$
I know that $left|operatorname{Aut}(mathcal{Q}_8)right| = !left|mathcal{S}_4right| = 24$, so when $pequiv 1,2 !pmod{4}$, I get that there are $0$ such Galois extensions, and when $pequiv 3 !pmod{4}$, I get that there is only $1$. This seems odd to me.
Where does this equation come from? Is it even correct? I have tried to Google it, but have thus far been unsuccessful.
algebraic-number-theory p-adic-number-theory
$endgroup$
For a homework exercise, I'm to determine for each $p$ the number of non-isomorphic tamely ramified Galois extensions $K/mathbb{Q}_p$ such that $operatorname{Gal}(K/mathbb{Q}_p) cong mathcal{Q}_8$ (the quaternion group of order $8$).
I was out of class the day we covered this, and I don't see the topic discussed in Gouvêa's book, so all I have is a lone equation from a classmate's notes:
$$#{text{tame Galois extensions } K/mathbb{Q}_p , : , operatorname{Gal}(K/mathbb{Q}_p) cong G} = frac{#{(a,b) in G times G , : , aba^{-1}=b^{p},, langle a,b rangle = G}}{left|operatorname{Aut}(G)right|}$$
I know that $left|operatorname{Aut}(mathcal{Q}_8)right| = !left|mathcal{S}_4right| = 24$, so when $pequiv 1,2 !pmod{4}$, I get that there are $0$ such Galois extensions, and when $pequiv 3 !pmod{4}$, I get that there is only $1$. This seems odd to me.
Where does this equation come from? Is it even correct? I have tried to Google it, but have thus far been unsuccessful.
algebraic-number-theory p-adic-number-theory
algebraic-number-theory p-adic-number-theory
asked Nov 29 '14 at 19:14
JoeDubJoeDub
618312
618312
1
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I am not familiar with that formula either, but I don‘t find your count surprising. $Q_8$ is not an easy group to find among your random collection of extensions. In fact, I’d like to see an explicit description of that quaternion-group extension of $mathbb Q_3$.
$endgroup$
– Lubin
Nov 30 '14 at 23:16
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@Lubin The construction of the unique $Q_8$-extension of $mathbb{Q}_p$ in the $pequiv, 3, (text{mod} 4)$ case goes back to Witt's 1936 paper on embeddings of biquadratic extensions into $Q_8$-extensions. The recipe is quoted in the paper "Quaternion Extensions" by Jensen & Yui - see th. I.1.1 and cor. II.3.6 (available here); note that th. I.1.1. corrects a typo in Witt's paper. It should be possible to unpack an explicit construction for the $p=3$ case from the results quoted in Jensen-Yui.
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– Giovanni Di Matteo
Jan 9 at 16:33
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Thanks, @GiovanniDiMatteo.
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– Lubin
Jan 9 at 21:18
add a comment |
1
$begingroup$
I am not familiar with that formula either, but I don‘t find your count surprising. $Q_8$ is not an easy group to find among your random collection of extensions. In fact, I’d like to see an explicit description of that quaternion-group extension of $mathbb Q_3$.
$endgroup$
– Lubin
Nov 30 '14 at 23:16
$begingroup$
@Lubin The construction of the unique $Q_8$-extension of $mathbb{Q}_p$ in the $pequiv, 3, (text{mod} 4)$ case goes back to Witt's 1936 paper on embeddings of biquadratic extensions into $Q_8$-extensions. The recipe is quoted in the paper "Quaternion Extensions" by Jensen & Yui - see th. I.1.1 and cor. II.3.6 (available here); note that th. I.1.1. corrects a typo in Witt's paper. It should be possible to unpack an explicit construction for the $p=3$ case from the results quoted in Jensen-Yui.
$endgroup$
– Giovanni Di Matteo
Jan 9 at 16:33
$begingroup$
Thanks, @GiovanniDiMatteo.
$endgroup$
– Lubin
Jan 9 at 21:18
1
1
$begingroup$
I am not familiar with that formula either, but I don‘t find your count surprising. $Q_8$ is not an easy group to find among your random collection of extensions. In fact, I’d like to see an explicit description of that quaternion-group extension of $mathbb Q_3$.
$endgroup$
– Lubin
Nov 30 '14 at 23:16
$begingroup$
I am not familiar with that formula either, but I don‘t find your count surprising. $Q_8$ is not an easy group to find among your random collection of extensions. In fact, I’d like to see an explicit description of that quaternion-group extension of $mathbb Q_3$.
$endgroup$
– Lubin
Nov 30 '14 at 23:16
$begingroup$
@Lubin The construction of the unique $Q_8$-extension of $mathbb{Q}_p$ in the $pequiv, 3, (text{mod} 4)$ case goes back to Witt's 1936 paper on embeddings of biquadratic extensions into $Q_8$-extensions. The recipe is quoted in the paper "Quaternion Extensions" by Jensen & Yui - see th. I.1.1 and cor. II.3.6 (available here); note that th. I.1.1. corrects a typo in Witt's paper. It should be possible to unpack an explicit construction for the $p=3$ case from the results quoted in Jensen-Yui.
$endgroup$
– Giovanni Di Matteo
Jan 9 at 16:33
$begingroup$
@Lubin The construction of the unique $Q_8$-extension of $mathbb{Q}_p$ in the $pequiv, 3, (text{mod} 4)$ case goes back to Witt's 1936 paper on embeddings of biquadratic extensions into $Q_8$-extensions. The recipe is quoted in the paper "Quaternion Extensions" by Jensen & Yui - see th. I.1.1 and cor. II.3.6 (available here); note that th. I.1.1. corrects a typo in Witt's paper. It should be possible to unpack an explicit construction for the $p=3$ case from the results quoted in Jensen-Yui.
$endgroup$
– Giovanni Di Matteo
Jan 9 at 16:33
$begingroup$
Thanks, @GiovanniDiMatteo.
$endgroup$
– Lubin
Jan 9 at 21:18
$begingroup$
Thanks, @GiovanniDiMatteo.
$endgroup$
– Lubin
Jan 9 at 21:18
add a comment |
1 Answer
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I think the formula you cite is true and is a consequence of the following observations :
Let $F/mathbf{Q}_p$ be a finite sub-extension of $overline{mathbf{Q}}_p/mathbf{Q}_p$, let $G_F = text{Gal}(overline{mathbf{Q}}_p/F)$, and let $k_F$ denote the residual field of $F$. Let $G$ denote a fixed finite group. Let's say that a $G$-extension of $F$ is a finite Galois sub-extension $K/F$ of $overline{mathbf{Q}}_p/F$ with $text{Gal}(K/F)simeq G$.
Assertion 1 : (see section 1.1 of [1]) The set of $G$-extensions of $F$ is in bijection with set of surjective homomorphisms $f:G_Fto G$, modulo $text{Aut}(G)$.
Assertion 2 : Under this correspondence, the tamely ramified $G$-extensions correspond to homomorphisms $f$ which factor through the quotient $G_Ftwoheadrightarrowtext{Gal}(F^{text{tame}}/F)$, where $F^{text{tame}}$ is the maximal tamely ramified extension of $F$.
Assertion 3 : We have $ text{Gal}(F^{text{tame}}/F) = langle x, y, |, yxy^{-1} = x^qrangle$ as a profinite group (i.e. $x$ and $y$ are topological generators), where $q = |k_F|$. This is theorem 2.(i) of Iwasawa 's paper [2].
Putting these things together, we get the following formula : if $F/mathbb{Q}_p$ is a finite extension and $q = |k_F|$, then the number of tamely ramified $G$-extensions of $F$ is equal to
$$frac{#{ (a,b) in Gtimes G | aba^{-1} = b^q text{ and }langle a,b rangle = G}}{|text{Aut}(G)|}$$
Note that this number depends only on the group $G$ and the integer $q$. The interpretation of the exponent $q$ coming from Iwasawa's theorem allows us to deduce a few interesting consequences (see remarks 2 and 3 below).
Specializing to the case when $F=mathbf{Q}_p$, we have $q=p$ and I think we recover the formula you stated, which is valid for any finite group $G$.
Remarks :
- If $|G|$ is prime to the residue characteristic of $F$, then all $G$-extensions of $F$ are tamely ramified (see section 3.2 of [1]). Therefore, if $G$ is a finite group with $|G|$ prime to $p$, then the above formula for $F/mathbf{Q}_p$ finite counts the total number of all $G$-extensions of $F$. In particular, the unique tamely ramified $Q_8$-extension you find in the $F=mathbf{Q}_p$ and $pequiv 3 ,(text{mod} 4)$ case is the only $Q_8$-extension.
- If $F/mathbf{Q}_p$ is totally ramified, then the residual field of $F$ is equal to $mathbf{F}_p$; in particular, in light of Iwasawa's result and OP's calculation, we get the following for free : if $pequiv 3 (text{mod} 4)$ and if $F/mathbf{Q}_p$ is totally ramified, then there is a unique octic tamely ramified Galois extension $K/F$ with $text{Gal}(K/F)simeq Q_8$, and this extension is necessarily the unique $Q_8$-extension of $F$ by remark (1).
More generally, if $F/mathbf{Q}_p$ is finite and if $F'/F$ is totally ramified (so that $k_{F'} = k_{F}$), then the number of tamely ramified $G$-extensions of $F$ is equal to the number of tamely ramified $G$-extensions of $F'$. And again, when $|G|$ is prime to $p$, all of the $G$-extensions of $F$ (resp. $F'$) are tamely ramified.
[1]: M. Yamagishi, On the number of Galois p-extensions of a local field, Proceedings of the AMS (1995). http://www.ams.org/journals/proc/1995-123-08/S0002-9939-1995-1264832-0/S0002-9939-1995-1264832-0.pdf
[2]: K. Iwasawa, On Galois groups of local fields, Trans. Amer. Math. Soc. 80 ( 1955), 448-469.
http://www.ams.org/journals/tran/1955-080-02/S0002-9947-1955-0075239-5/S0002-9947-1955-0075239-5.pdf
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I think the formula you cite is true and is a consequence of the following observations :
Let $F/mathbf{Q}_p$ be a finite sub-extension of $overline{mathbf{Q}}_p/mathbf{Q}_p$, let $G_F = text{Gal}(overline{mathbf{Q}}_p/F)$, and let $k_F$ denote the residual field of $F$. Let $G$ denote a fixed finite group. Let's say that a $G$-extension of $F$ is a finite Galois sub-extension $K/F$ of $overline{mathbf{Q}}_p/F$ with $text{Gal}(K/F)simeq G$.
Assertion 1 : (see section 1.1 of [1]) The set of $G$-extensions of $F$ is in bijection with set of surjective homomorphisms $f:G_Fto G$, modulo $text{Aut}(G)$.
Assertion 2 : Under this correspondence, the tamely ramified $G$-extensions correspond to homomorphisms $f$ which factor through the quotient $G_Ftwoheadrightarrowtext{Gal}(F^{text{tame}}/F)$, where $F^{text{tame}}$ is the maximal tamely ramified extension of $F$.
Assertion 3 : We have $ text{Gal}(F^{text{tame}}/F) = langle x, y, |, yxy^{-1} = x^qrangle$ as a profinite group (i.e. $x$ and $y$ are topological generators), where $q = |k_F|$. This is theorem 2.(i) of Iwasawa 's paper [2].
Putting these things together, we get the following formula : if $F/mathbb{Q}_p$ is a finite extension and $q = |k_F|$, then the number of tamely ramified $G$-extensions of $F$ is equal to
$$frac{#{ (a,b) in Gtimes G | aba^{-1} = b^q text{ and }langle a,b rangle = G}}{|text{Aut}(G)|}$$
Note that this number depends only on the group $G$ and the integer $q$. The interpretation of the exponent $q$ coming from Iwasawa's theorem allows us to deduce a few interesting consequences (see remarks 2 and 3 below).
Specializing to the case when $F=mathbf{Q}_p$, we have $q=p$ and I think we recover the formula you stated, which is valid for any finite group $G$.
Remarks :
- If $|G|$ is prime to the residue characteristic of $F$, then all $G$-extensions of $F$ are tamely ramified (see section 3.2 of [1]). Therefore, if $G$ is a finite group with $|G|$ prime to $p$, then the above formula for $F/mathbf{Q}_p$ finite counts the total number of all $G$-extensions of $F$. In particular, the unique tamely ramified $Q_8$-extension you find in the $F=mathbf{Q}_p$ and $pequiv 3 ,(text{mod} 4)$ case is the only $Q_8$-extension.
- If $F/mathbf{Q}_p$ is totally ramified, then the residual field of $F$ is equal to $mathbf{F}_p$; in particular, in light of Iwasawa's result and OP's calculation, we get the following for free : if $pequiv 3 (text{mod} 4)$ and if $F/mathbf{Q}_p$ is totally ramified, then there is a unique octic tamely ramified Galois extension $K/F$ with $text{Gal}(K/F)simeq Q_8$, and this extension is necessarily the unique $Q_8$-extension of $F$ by remark (1).
More generally, if $F/mathbf{Q}_p$ is finite and if $F'/F$ is totally ramified (so that $k_{F'} = k_{F}$), then the number of tamely ramified $G$-extensions of $F$ is equal to the number of tamely ramified $G$-extensions of $F'$. And again, when $|G|$ is prime to $p$, all of the $G$-extensions of $F$ (resp. $F'$) are tamely ramified.
[1]: M. Yamagishi, On the number of Galois p-extensions of a local field, Proceedings of the AMS (1995). http://www.ams.org/journals/proc/1995-123-08/S0002-9939-1995-1264832-0/S0002-9939-1995-1264832-0.pdf
[2]: K. Iwasawa, On Galois groups of local fields, Trans. Amer. Math. Soc. 80 ( 1955), 448-469.
http://www.ams.org/journals/tran/1955-080-02/S0002-9947-1955-0075239-5/S0002-9947-1955-0075239-5.pdf
$endgroup$
add a comment |
$begingroup$
I think the formula you cite is true and is a consequence of the following observations :
Let $F/mathbf{Q}_p$ be a finite sub-extension of $overline{mathbf{Q}}_p/mathbf{Q}_p$, let $G_F = text{Gal}(overline{mathbf{Q}}_p/F)$, and let $k_F$ denote the residual field of $F$. Let $G$ denote a fixed finite group. Let's say that a $G$-extension of $F$ is a finite Galois sub-extension $K/F$ of $overline{mathbf{Q}}_p/F$ with $text{Gal}(K/F)simeq G$.
Assertion 1 : (see section 1.1 of [1]) The set of $G$-extensions of $F$ is in bijection with set of surjective homomorphisms $f:G_Fto G$, modulo $text{Aut}(G)$.
Assertion 2 : Under this correspondence, the tamely ramified $G$-extensions correspond to homomorphisms $f$ which factor through the quotient $G_Ftwoheadrightarrowtext{Gal}(F^{text{tame}}/F)$, where $F^{text{tame}}$ is the maximal tamely ramified extension of $F$.
Assertion 3 : We have $ text{Gal}(F^{text{tame}}/F) = langle x, y, |, yxy^{-1} = x^qrangle$ as a profinite group (i.e. $x$ and $y$ are topological generators), where $q = |k_F|$. This is theorem 2.(i) of Iwasawa 's paper [2].
Putting these things together, we get the following formula : if $F/mathbb{Q}_p$ is a finite extension and $q = |k_F|$, then the number of tamely ramified $G$-extensions of $F$ is equal to
$$frac{#{ (a,b) in Gtimes G | aba^{-1} = b^q text{ and }langle a,b rangle = G}}{|text{Aut}(G)|}$$
Note that this number depends only on the group $G$ and the integer $q$. The interpretation of the exponent $q$ coming from Iwasawa's theorem allows us to deduce a few interesting consequences (see remarks 2 and 3 below).
Specializing to the case when $F=mathbf{Q}_p$, we have $q=p$ and I think we recover the formula you stated, which is valid for any finite group $G$.
Remarks :
- If $|G|$ is prime to the residue characteristic of $F$, then all $G$-extensions of $F$ are tamely ramified (see section 3.2 of [1]). Therefore, if $G$ is a finite group with $|G|$ prime to $p$, then the above formula for $F/mathbf{Q}_p$ finite counts the total number of all $G$-extensions of $F$. In particular, the unique tamely ramified $Q_8$-extension you find in the $F=mathbf{Q}_p$ and $pequiv 3 ,(text{mod} 4)$ case is the only $Q_8$-extension.
- If $F/mathbf{Q}_p$ is totally ramified, then the residual field of $F$ is equal to $mathbf{F}_p$; in particular, in light of Iwasawa's result and OP's calculation, we get the following for free : if $pequiv 3 (text{mod} 4)$ and if $F/mathbf{Q}_p$ is totally ramified, then there is a unique octic tamely ramified Galois extension $K/F$ with $text{Gal}(K/F)simeq Q_8$, and this extension is necessarily the unique $Q_8$-extension of $F$ by remark (1).
More generally, if $F/mathbf{Q}_p$ is finite and if $F'/F$ is totally ramified (so that $k_{F'} = k_{F}$), then the number of tamely ramified $G$-extensions of $F$ is equal to the number of tamely ramified $G$-extensions of $F'$. And again, when $|G|$ is prime to $p$, all of the $G$-extensions of $F$ (resp. $F'$) are tamely ramified.
[1]: M. Yamagishi, On the number of Galois p-extensions of a local field, Proceedings of the AMS (1995). http://www.ams.org/journals/proc/1995-123-08/S0002-9939-1995-1264832-0/S0002-9939-1995-1264832-0.pdf
[2]: K. Iwasawa, On Galois groups of local fields, Trans. Amer. Math. Soc. 80 ( 1955), 448-469.
http://www.ams.org/journals/tran/1955-080-02/S0002-9947-1955-0075239-5/S0002-9947-1955-0075239-5.pdf
$endgroup$
add a comment |
$begingroup$
I think the formula you cite is true and is a consequence of the following observations :
Let $F/mathbf{Q}_p$ be a finite sub-extension of $overline{mathbf{Q}}_p/mathbf{Q}_p$, let $G_F = text{Gal}(overline{mathbf{Q}}_p/F)$, and let $k_F$ denote the residual field of $F$. Let $G$ denote a fixed finite group. Let's say that a $G$-extension of $F$ is a finite Galois sub-extension $K/F$ of $overline{mathbf{Q}}_p/F$ with $text{Gal}(K/F)simeq G$.
Assertion 1 : (see section 1.1 of [1]) The set of $G$-extensions of $F$ is in bijection with set of surjective homomorphisms $f:G_Fto G$, modulo $text{Aut}(G)$.
Assertion 2 : Under this correspondence, the tamely ramified $G$-extensions correspond to homomorphisms $f$ which factor through the quotient $G_Ftwoheadrightarrowtext{Gal}(F^{text{tame}}/F)$, where $F^{text{tame}}$ is the maximal tamely ramified extension of $F$.
Assertion 3 : We have $ text{Gal}(F^{text{tame}}/F) = langle x, y, |, yxy^{-1} = x^qrangle$ as a profinite group (i.e. $x$ and $y$ are topological generators), where $q = |k_F|$. This is theorem 2.(i) of Iwasawa 's paper [2].
Putting these things together, we get the following formula : if $F/mathbb{Q}_p$ is a finite extension and $q = |k_F|$, then the number of tamely ramified $G$-extensions of $F$ is equal to
$$frac{#{ (a,b) in Gtimes G | aba^{-1} = b^q text{ and }langle a,b rangle = G}}{|text{Aut}(G)|}$$
Note that this number depends only on the group $G$ and the integer $q$. The interpretation of the exponent $q$ coming from Iwasawa's theorem allows us to deduce a few interesting consequences (see remarks 2 and 3 below).
Specializing to the case when $F=mathbf{Q}_p$, we have $q=p$ and I think we recover the formula you stated, which is valid for any finite group $G$.
Remarks :
- If $|G|$ is prime to the residue characteristic of $F$, then all $G$-extensions of $F$ are tamely ramified (see section 3.2 of [1]). Therefore, if $G$ is a finite group with $|G|$ prime to $p$, then the above formula for $F/mathbf{Q}_p$ finite counts the total number of all $G$-extensions of $F$. In particular, the unique tamely ramified $Q_8$-extension you find in the $F=mathbf{Q}_p$ and $pequiv 3 ,(text{mod} 4)$ case is the only $Q_8$-extension.
- If $F/mathbf{Q}_p$ is totally ramified, then the residual field of $F$ is equal to $mathbf{F}_p$; in particular, in light of Iwasawa's result and OP's calculation, we get the following for free : if $pequiv 3 (text{mod} 4)$ and if $F/mathbf{Q}_p$ is totally ramified, then there is a unique octic tamely ramified Galois extension $K/F$ with $text{Gal}(K/F)simeq Q_8$, and this extension is necessarily the unique $Q_8$-extension of $F$ by remark (1).
More generally, if $F/mathbf{Q}_p$ is finite and if $F'/F$ is totally ramified (so that $k_{F'} = k_{F}$), then the number of tamely ramified $G$-extensions of $F$ is equal to the number of tamely ramified $G$-extensions of $F'$. And again, when $|G|$ is prime to $p$, all of the $G$-extensions of $F$ (resp. $F'$) are tamely ramified.
[1]: M. Yamagishi, On the number of Galois p-extensions of a local field, Proceedings of the AMS (1995). http://www.ams.org/journals/proc/1995-123-08/S0002-9939-1995-1264832-0/S0002-9939-1995-1264832-0.pdf
[2]: K. Iwasawa, On Galois groups of local fields, Trans. Amer. Math. Soc. 80 ( 1955), 448-469.
http://www.ams.org/journals/tran/1955-080-02/S0002-9947-1955-0075239-5/S0002-9947-1955-0075239-5.pdf
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I think the formula you cite is true and is a consequence of the following observations :
Let $F/mathbf{Q}_p$ be a finite sub-extension of $overline{mathbf{Q}}_p/mathbf{Q}_p$, let $G_F = text{Gal}(overline{mathbf{Q}}_p/F)$, and let $k_F$ denote the residual field of $F$. Let $G$ denote a fixed finite group. Let's say that a $G$-extension of $F$ is a finite Galois sub-extension $K/F$ of $overline{mathbf{Q}}_p/F$ with $text{Gal}(K/F)simeq G$.
Assertion 1 : (see section 1.1 of [1]) The set of $G$-extensions of $F$ is in bijection with set of surjective homomorphisms $f:G_Fto G$, modulo $text{Aut}(G)$.
Assertion 2 : Under this correspondence, the tamely ramified $G$-extensions correspond to homomorphisms $f$ which factor through the quotient $G_Ftwoheadrightarrowtext{Gal}(F^{text{tame}}/F)$, where $F^{text{tame}}$ is the maximal tamely ramified extension of $F$.
Assertion 3 : We have $ text{Gal}(F^{text{tame}}/F) = langle x, y, |, yxy^{-1} = x^qrangle$ as a profinite group (i.e. $x$ and $y$ are topological generators), where $q = |k_F|$. This is theorem 2.(i) of Iwasawa 's paper [2].
Putting these things together, we get the following formula : if $F/mathbb{Q}_p$ is a finite extension and $q = |k_F|$, then the number of tamely ramified $G$-extensions of $F$ is equal to
$$frac{#{ (a,b) in Gtimes G | aba^{-1} = b^q text{ and }langle a,b rangle = G}}{|text{Aut}(G)|}$$
Note that this number depends only on the group $G$ and the integer $q$. The interpretation of the exponent $q$ coming from Iwasawa's theorem allows us to deduce a few interesting consequences (see remarks 2 and 3 below).
Specializing to the case when $F=mathbf{Q}_p$, we have $q=p$ and I think we recover the formula you stated, which is valid for any finite group $G$.
Remarks :
- If $|G|$ is prime to the residue characteristic of $F$, then all $G$-extensions of $F$ are tamely ramified (see section 3.2 of [1]). Therefore, if $G$ is a finite group with $|G|$ prime to $p$, then the above formula for $F/mathbf{Q}_p$ finite counts the total number of all $G$-extensions of $F$. In particular, the unique tamely ramified $Q_8$-extension you find in the $F=mathbf{Q}_p$ and $pequiv 3 ,(text{mod} 4)$ case is the only $Q_8$-extension.
- If $F/mathbf{Q}_p$ is totally ramified, then the residual field of $F$ is equal to $mathbf{F}_p$; in particular, in light of Iwasawa's result and OP's calculation, we get the following for free : if $pequiv 3 (text{mod} 4)$ and if $F/mathbf{Q}_p$ is totally ramified, then there is a unique octic tamely ramified Galois extension $K/F$ with $text{Gal}(K/F)simeq Q_8$, and this extension is necessarily the unique $Q_8$-extension of $F$ by remark (1).
More generally, if $F/mathbf{Q}_p$ is finite and if $F'/F$ is totally ramified (so that $k_{F'} = k_{F}$), then the number of tamely ramified $G$-extensions of $F$ is equal to the number of tamely ramified $G$-extensions of $F'$. And again, when $|G|$ is prime to $p$, all of the $G$-extensions of $F$ (resp. $F'$) are tamely ramified.
[1]: M. Yamagishi, On the number of Galois p-extensions of a local field, Proceedings of the AMS (1995). http://www.ams.org/journals/proc/1995-123-08/S0002-9939-1995-1264832-0/S0002-9939-1995-1264832-0.pdf
[2]: K. Iwasawa, On Galois groups of local fields, Trans. Amer. Math. Soc. 80 ( 1955), 448-469.
http://www.ams.org/journals/tran/1955-080-02/S0002-9947-1955-0075239-5/S0002-9947-1955-0075239-5.pdf
edited Dec 30 '18 at 18:34
answered Dec 29 '18 at 19:41
Giovanni Di MatteoGiovanni Di Matteo
7816
7816
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I am not familiar with that formula either, but I don‘t find your count surprising. $Q_8$ is not an easy group to find among your random collection of extensions. In fact, I’d like to see an explicit description of that quaternion-group extension of $mathbb Q_3$.
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– Lubin
Nov 30 '14 at 23:16
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@Lubin The construction of the unique $Q_8$-extension of $mathbb{Q}_p$ in the $pequiv, 3, (text{mod} 4)$ case goes back to Witt's 1936 paper on embeddings of biquadratic extensions into $Q_8$-extensions. The recipe is quoted in the paper "Quaternion Extensions" by Jensen & Yui - see th. I.1.1 and cor. II.3.6 (available here); note that th. I.1.1. corrects a typo in Witt's paper. It should be possible to unpack an explicit construction for the $p=3$ case from the results quoted in Jensen-Yui.
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– Giovanni Di Matteo
Jan 9 at 16:33
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Thanks, @GiovanniDiMatteo.
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– Lubin
Jan 9 at 21:18