Counting tamely ramified Galois extensions of $mathbb{Q}_p$ with a given Galois group.












7












$begingroup$


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.










share|cite|improve this question









$endgroup$








  • 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
















7












$begingroup$


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.










share|cite|improve this question









$endgroup$








  • 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














7












7








7


5



$begingroup$


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.










share|cite|improve this question









$endgroup$




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






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Nov 29 '14 at 19:14









JoeDubJoeDub

618312




618312








  • 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




    $begingroup$
    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










1 Answer
1






active

oldest

votes


















6












$begingroup$

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 :




  1. 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.

  2. 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).


  3. 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








share|cite|improve this answer











$endgroup$













    Your Answer





    StackExchange.ifUsing("editor", function () {
    return StackExchange.using("mathjaxEditing", function () {
    StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
    StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
    });
    });
    }, "mathjax-editing");

    StackExchange.ready(function() {
    var channelOptions = {
    tags: "".split(" "),
    id: "69"
    };
    initTagRenderer("".split(" "), "".split(" "), channelOptions);

    StackExchange.using("externalEditor", function() {
    // Have to fire editor after snippets, if snippets enabled
    if (StackExchange.settings.snippets.snippetsEnabled) {
    StackExchange.using("snippets", function() {
    createEditor();
    });
    }
    else {
    createEditor();
    }
    });

    function createEditor() {
    StackExchange.prepareEditor({
    heartbeatType: 'answer',
    autoActivateHeartbeat: false,
    convertImagesToLinks: true,
    noModals: true,
    showLowRepImageUploadWarning: true,
    reputationToPostImages: 10,
    bindNavPrevention: true,
    postfix: "",
    imageUploader: {
    brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
    contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
    allowUrls: true
    },
    noCode: true, onDemand: true,
    discardSelector: ".discard-answer"
    ,immediatelyShowMarkdownHelp:true
    });


    }
    });














    draft saved

    draft discarded


















    StackExchange.ready(
    function () {
    StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f1044011%2fcounting-tamely-ramified-galois-extensions-of-mathbbq-p-with-a-given-galois%23new-answer', 'question_page');
    }
    );

    Post as a guest















    Required, but never shown

























    1 Answer
    1






    active

    oldest

    votes








    1 Answer
    1






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    6












    $begingroup$

    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 :




    1. 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.

    2. 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).


    3. 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








    share|cite|improve this answer











    $endgroup$


















      6












      $begingroup$

      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 :




      1. 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.

      2. 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).


      3. 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








      share|cite|improve this answer











      $endgroup$
















        6












        6








        6





        $begingroup$

        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 :




        1. 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.

        2. 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).


        3. 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








        share|cite|improve this answer











        $endgroup$



        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 :




        1. 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.

        2. 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).


        3. 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









        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited Dec 30 '18 at 18:34

























        answered Dec 29 '18 at 19:41









        Giovanni Di MatteoGiovanni Di Matteo

        7816




        7816






























            draft saved

            draft discarded




















































            Thanks for contributing an answer to Mathematics Stack Exchange!


            • Please be sure to answer the question. Provide details and share your research!

            But avoid



            • Asking for help, clarification, or responding to other answers.

            • Making statements based on opinion; back them up with references or personal experience.


            Use MathJax to format equations. MathJax reference.


            To learn more, see our tips on writing great answers.




            draft saved


            draft discarded














            StackExchange.ready(
            function () {
            StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f1044011%2fcounting-tamely-ramified-galois-extensions-of-mathbbq-p-with-a-given-galois%23new-answer', 'question_page');
            }
            );

            Post as a guest















            Required, but never shown





















































            Required, but never shown














            Required, but never shown












            Required, but never shown







            Required, but never shown

































            Required, but never shown














            Required, but never shown












            Required, but never shown







            Required, but never shown







            Popular posts from this blog

            Tonle Sap (See)

            I get strange results when I access the Sqlitedatabase with Unity C# via XAMPP

            Guatemaltekische Davis-Cup-Mannschaft