Computing degrees and ramification indices of some extensions of $mathbb{Q}_2$












2














Let $K=mathbb{Q}_2$ and $F = K(zeta_3,alpha)$ where $zeta$ is a primitive third root of unity and $alpha$ is a cubic root of $2$, i.e. $alpha^3 = 2$.Let $K_1 = K(zeta_3)$, $K_2 = K(alpha)$ and $L=K_1(beta)$ where $beta$ is defined by $beta^3 = zeta_3 alpha$.



Now I would like to compute the degrees and ramification indices of $L/K_1$.



Discoveries and attempts:




  • I noticed that $L$ contains $F$ since $beta^3/zeta_3 = alpha$.

  • I know that $F/K_1$ is a totally ramified extension of degree $3$. This can be seen by taking a look at the extensions $F/K_1/K$ and $F/K_2/K$ and noticing that $K_1/K$ is unramified of degree $2$ and $K_2/K$ is totally ramified of degree $3$.


Could you please explain what can I do now to get information about $L/K_1$ resp. $L/F$? Thank you!






abstract-algebra algebraic-number-theory extension-field local-field ramification







share|cite|improve this question






















  • $mathbb{Z}_2[zeta_3]/mathbb{Z}_2$ is unramified of degree $2$ (that is $(2)$ is the unique prime ideal of $mathbb{Z}_2[zeta_3]$ so $[mathbb{Z}_2[zeta_3]/(2) : mathbb{Z}_2/(2)]=[mathbb{Z}_2[zeta_3]:mathbb{Z}_2]$). $ mathbb{Z}_2[zeta_3,2^{1/3}]/mathbb{Z}_2[zeta_3]$ is totally ramified of degree $3$ ($pi=2^{1/3}$ then $(pi)$ is the unique prime ideal of $mathbb{Z}_2[zeta_3,2^{1/3}]$ and $pi^3 / 2 in mathbb{Z}_2[zeta_3,2^{1/3}]^times$). $mathbb{Z}_2[zeta_32^{1/3}]$ is isomorphic to $mathbb{Z}_2[2^{1/3}]$ since $2^{1/3},zeta_32^{1/3}$ have the same minimal polynomial
    – reuns
    Dec 23 at 1:43












  • @reuns: Thank you for your response! Could you please explain how your observation can be used for this problem? I am not able to see any relation.
    – Diglett
    Dec 23 at 11:52
















2














Let $K=mathbb{Q}_2$ and $F = K(zeta_3,alpha)$ where $zeta$ is a primitive third root of unity and $alpha$ is a cubic root of $2$, i.e. $alpha^3 = 2$.Let $K_1 = K(zeta_3)$, $K_2 = K(alpha)$ and $L=K_1(beta)$ where $beta$ is defined by $beta^3 = zeta_3 alpha$.



Now I would like to compute the degrees and ramification indices of $L/K_1$.



Discoveries and attempts:




  • I noticed that $L$ contains $F$ since $beta^3/zeta_3 = alpha$.

  • I know that $F/K_1$ is a totally ramified extension of degree $3$. This can be seen by taking a look at the extensions $F/K_1/K$ and $F/K_2/K$ and noticing that $K_1/K$ is unramified of degree $2$ and $K_2/K$ is totally ramified of degree $3$.


Could you please explain what can I do now to get information about $L/K_1$ resp. $L/F$? Thank you!






abstract-algebra algebraic-number-theory extension-field local-field ramification







share|cite|improve this question






















  • $mathbb{Z}_2[zeta_3]/mathbb{Z}_2$ is unramified of degree $2$ (that is $(2)$ is the unique prime ideal of $mathbb{Z}_2[zeta_3]$ so $[mathbb{Z}_2[zeta_3]/(2) : mathbb{Z}_2/(2)]=[mathbb{Z}_2[zeta_3]:mathbb{Z}_2]$). $ mathbb{Z}_2[zeta_3,2^{1/3}]/mathbb{Z}_2[zeta_3]$ is totally ramified of degree $3$ ($pi=2^{1/3}$ then $(pi)$ is the unique prime ideal of $mathbb{Z}_2[zeta_3,2^{1/3}]$ and $pi^3 / 2 in mathbb{Z}_2[zeta_3,2^{1/3}]^times$). $mathbb{Z}_2[zeta_32^{1/3}]$ is isomorphic to $mathbb{Z}_2[2^{1/3}]$ since $2^{1/3},zeta_32^{1/3}$ have the same minimal polynomial
    – reuns
    Dec 23 at 1:43












  • @reuns: Thank you for your response! Could you please explain how your observation can be used for this problem? I am not able to see any relation.
    – Diglett
    Dec 23 at 11:52














2












2








2







Let $K=mathbb{Q}_2$ and $F = K(zeta_3,alpha)$ where $zeta$ is a primitive third root of unity and $alpha$ is a cubic root of $2$, i.e. $alpha^3 = 2$.Let $K_1 = K(zeta_3)$, $K_2 = K(alpha)$ and $L=K_1(beta)$ where $beta$ is defined by $beta^3 = zeta_3 alpha$.



Now I would like to compute the degrees and ramification indices of $L/K_1$.



Discoveries and attempts:




  • I noticed that $L$ contains $F$ since $beta^3/zeta_3 = alpha$.

  • I know that $F/K_1$ is a totally ramified extension of degree $3$. This can be seen by taking a look at the extensions $F/K_1/K$ and $F/K_2/K$ and noticing that $K_1/K$ is unramified of degree $2$ and $K_2/K$ is totally ramified of degree $3$.


Could you please explain what can I do now to get information about $L/K_1$ resp. $L/F$? Thank you!






abstract-algebra algebraic-number-theory extension-field local-field ramification







share|cite|improve this question













Let $K=mathbb{Q}_2$ and $F = K(zeta_3,alpha)$ where $zeta$ is a primitive third root of unity and $alpha$ is a cubic root of $2$, i.e. $alpha^3 = 2$.Let $K_1 = K(zeta_3)$, $K_2 = K(alpha)$ and $L=K_1(beta)$ where $beta$ is defined by $beta^3 = zeta_3 alpha$.



Now I would like to compute the degrees and ramification indices of $L/K_1$.



Discoveries and attempts:




  • I noticed that $L$ contains $F$ since $beta^3/zeta_3 = alpha$.

  • I know that $F/K_1$ is a totally ramified extension of degree $3$. This can be seen by taking a look at the extensions $F/K_1/K$ and $F/K_2/K$ and noticing that $K_1/K$ is unramified of degree $2$ and $K_2/K$ is totally ramified of degree $3$.


Could you please explain what can I do now to get information about $L/K_1$ resp. $L/F$? Thank you!






abstract-algebra algebraic-number-theory extension-field local-field ramification




abstract-algebra algebraic-number-theory extension-field local-field ramification






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Dec 22 at 21:01









Diglett

879520




879520












  • $mathbb{Z}_2[zeta_3]/mathbb{Z}_2$ is unramified of degree $2$ (that is $(2)$ is the unique prime ideal of $mathbb{Z}_2[zeta_3]$ so $[mathbb{Z}_2[zeta_3]/(2) : mathbb{Z}_2/(2)]=[mathbb{Z}_2[zeta_3]:mathbb{Z}_2]$). $ mathbb{Z}_2[zeta_3,2^{1/3}]/mathbb{Z}_2[zeta_3]$ is totally ramified of degree $3$ ($pi=2^{1/3}$ then $(pi)$ is the unique prime ideal of $mathbb{Z}_2[zeta_3,2^{1/3}]$ and $pi^3 / 2 in mathbb{Z}_2[zeta_3,2^{1/3}]^times$). $mathbb{Z}_2[zeta_32^{1/3}]$ is isomorphic to $mathbb{Z}_2[2^{1/3}]$ since $2^{1/3},zeta_32^{1/3}$ have the same minimal polynomial
    – reuns
    Dec 23 at 1:43












  • @reuns: Thank you for your response! Could you please explain how your observation can be used for this problem? I am not able to see any relation.
    – Diglett
    Dec 23 at 11:52


















  • $mathbb{Z}_2[zeta_3]/mathbb{Z}_2$ is unramified of degree $2$ (that is $(2)$ is the unique prime ideal of $mathbb{Z}_2[zeta_3]$ so $[mathbb{Z}_2[zeta_3]/(2) : mathbb{Z}_2/(2)]=[mathbb{Z}_2[zeta_3]:mathbb{Z}_2]$). $ mathbb{Z}_2[zeta_3,2^{1/3}]/mathbb{Z}_2[zeta_3]$ is totally ramified of degree $3$ ($pi=2^{1/3}$ then $(pi)$ is the unique prime ideal of $mathbb{Z}_2[zeta_3,2^{1/3}]$ and $pi^3 / 2 in mathbb{Z}_2[zeta_3,2^{1/3}]^times$). $mathbb{Z}_2[zeta_32^{1/3}]$ is isomorphic to $mathbb{Z}_2[2^{1/3}]$ since $2^{1/3},zeta_32^{1/3}$ have the same minimal polynomial
    – reuns
    Dec 23 at 1:43












  • @reuns: Thank you for your response! Could you please explain how your observation can be used for this problem? I am not able to see any relation.
    – Diglett
    Dec 23 at 11:52
















$mathbb{Z}_2[zeta_3]/mathbb{Z}_2$ is unramified of degree $2$ (that is $(2)$ is the unique prime ideal of $mathbb{Z}_2[zeta_3]$ so $[mathbb{Z}_2[zeta_3]/(2) : mathbb{Z}_2/(2)]=[mathbb{Z}_2[zeta_3]:mathbb{Z}_2]$). $ mathbb{Z}_2[zeta_3,2^{1/3}]/mathbb{Z}_2[zeta_3]$ is totally ramified of degree $3$ ($pi=2^{1/3}$ then $(pi)$ is the unique prime ideal of $mathbb{Z}_2[zeta_3,2^{1/3}]$ and $pi^3 / 2 in mathbb{Z}_2[zeta_3,2^{1/3}]^times$). $mathbb{Z}_2[zeta_32^{1/3}]$ is isomorphic to $mathbb{Z}_2[2^{1/3}]$ since $2^{1/3},zeta_32^{1/3}$ have the same minimal polynomial
– reuns
Dec 23 at 1:43






$mathbb{Z}_2[zeta_3]/mathbb{Z}_2$ is unramified of degree $2$ (that is $(2)$ is the unique prime ideal of $mathbb{Z}_2[zeta_3]$ so $[mathbb{Z}_2[zeta_3]/(2) : mathbb{Z}_2/(2)]=[mathbb{Z}_2[zeta_3]:mathbb{Z}_2]$). $ mathbb{Z}_2[zeta_3,2^{1/3}]/mathbb{Z}_2[zeta_3]$ is totally ramified of degree $3$ ($pi=2^{1/3}$ then $(pi)$ is the unique prime ideal of $mathbb{Z}_2[zeta_3,2^{1/3}]$ and $pi^3 / 2 in mathbb{Z}_2[zeta_3,2^{1/3}]^times$). $mathbb{Z}_2[zeta_32^{1/3}]$ is isomorphic to $mathbb{Z}_2[2^{1/3}]$ since $2^{1/3},zeta_32^{1/3}$ have the same minimal polynomial
– reuns
Dec 23 at 1:43














@reuns: Thank you for your response! Could you please explain how your observation can be used for this problem? I am not able to see any relation.
– Diglett
Dec 23 at 11:52




@reuns: Thank you for your response! Could you please explain how your observation can be used for this problem? I am not able to see any relation.
– Diglett
Dec 23 at 11:52










1 Answer
1






active

oldest

votes


















0














Let $L_{1}=mathbb{Q}_{2}(beta)$ and note that $beta$ is a root of $x^{9}-2$ over $K$, which is an Eisenstein polynomial. Therefore $L_{1}/K$ is totally ramified of degree 9 (see e.g. Proposition 3.6 of Local Fields and Their Extensions by Fesenko & Vostokov for a proof if you don't know this result). Now, $K_{1}/K$ is Galois of degree 2, therefore we have by Galois theory
$$[L:L_{1}]=[K_{1}:K],$$
because $L=L_{1}K_{1}$. Therefore we conclude that $[L:L_{1}]=2$. So $[L:K]=18$ by the tower formula, whence $[L:K_{1}]=9$.



Now, note that
$$f(L/K)=f(L/K_{1})f(K_{1}/K)=2f(L/K_{1}),$$
because $K_{1}/K$ is unramified (where $f(L/K)$ means the inertia degree of the extension $L/K$). Also
$$e(L/K)=e(L/L_{1})e(L_{1}/K)=9e(L/L_{1}),$$
because $L_{1}/K$ is totally ramified (where $e(L/K)$ means the ramification index of the extension $L/K$). Thus we see that $2mid f(L/K)$ and $9mid e(L/K)$. By the fundamental identity we have $$18=[L:K]=e(L/K)f(L/K),$$
whence $f(L/K)=2$ and $e(L/K)=9$. This implies that $f(L/K_{1})=1$ because
$$2=f(L/K)=f(L/K_{1})f(K_{1}/K)=2f(L/K_{1}).$$
It also implies that $e(L/K_{1})=9$ because
$$9=e(L/K)=e(L/K_{1})e(K_{1}/K)=e(L/K_{1}).$$



Now for the extension $L/F$ we have
$$2=f(L/K)=f(L/F)f(F/K)=f(L/F)f(F/K_{1})f(K_{1}/K)=2f(L/F)f(F/K_{1}),$$
whence $f(L/F)=1$. Furthermore
$$9=e(L/K)=e(L/F)e(F/K)=e(L/F)e(F/K_{1})e(K_{1}/K)=3e(L/F),$$
whence $e(L/F)=3$ and we conclude that $[L:F]=e(L/F)f(L/F)=3$.






share|cite|improve this answer























    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%2f3049847%2fcomputing-degrees-and-ramification-indices-of-some-extensions-of-mathbbq-2%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









    0














    Let $L_{1}=mathbb{Q}_{2}(beta)$ and note that $beta$ is a root of $x^{9}-2$ over $K$, which is an Eisenstein polynomial. Therefore $L_{1}/K$ is totally ramified of degree 9 (see e.g. Proposition 3.6 of Local Fields and Their Extensions by Fesenko & Vostokov for a proof if you don't know this result). Now, $K_{1}/K$ is Galois of degree 2, therefore we have by Galois theory
    $$[L:L_{1}]=[K_{1}:K],$$
    because $L=L_{1}K_{1}$. Therefore we conclude that $[L:L_{1}]=2$. So $[L:K]=18$ by the tower formula, whence $[L:K_{1}]=9$.



    Now, note that
    $$f(L/K)=f(L/K_{1})f(K_{1}/K)=2f(L/K_{1}),$$
    because $K_{1}/K$ is unramified (where $f(L/K)$ means the inertia degree of the extension $L/K$). Also
    $$e(L/K)=e(L/L_{1})e(L_{1}/K)=9e(L/L_{1}),$$
    because $L_{1}/K$ is totally ramified (where $e(L/K)$ means the ramification index of the extension $L/K$). Thus we see that $2mid f(L/K)$ and $9mid e(L/K)$. By the fundamental identity we have $$18=[L:K]=e(L/K)f(L/K),$$
    whence $f(L/K)=2$ and $e(L/K)=9$. This implies that $f(L/K_{1})=1$ because
    $$2=f(L/K)=f(L/K_{1})f(K_{1}/K)=2f(L/K_{1}).$$
    It also implies that $e(L/K_{1})=9$ because
    $$9=e(L/K)=e(L/K_{1})e(K_{1}/K)=e(L/K_{1}).$$



    Now for the extension $L/F$ we have
    $$2=f(L/K)=f(L/F)f(F/K)=f(L/F)f(F/K_{1})f(K_{1}/K)=2f(L/F)f(F/K_{1}),$$
    whence $f(L/F)=1$. Furthermore
    $$9=e(L/K)=e(L/F)e(F/K)=e(L/F)e(F/K_{1})e(K_{1}/K)=3e(L/F),$$
    whence $e(L/F)=3$ and we conclude that $[L:F]=e(L/F)f(L/F)=3$.






    share|cite|improve this answer




























      0














      Let $L_{1}=mathbb{Q}_{2}(beta)$ and note that $beta$ is a root of $x^{9}-2$ over $K$, which is an Eisenstein polynomial. Therefore $L_{1}/K$ is totally ramified of degree 9 (see e.g. Proposition 3.6 of Local Fields and Their Extensions by Fesenko & Vostokov for a proof if you don't know this result). Now, $K_{1}/K$ is Galois of degree 2, therefore we have by Galois theory
      $$[L:L_{1}]=[K_{1}:K],$$
      because $L=L_{1}K_{1}$. Therefore we conclude that $[L:L_{1}]=2$. So $[L:K]=18$ by the tower formula, whence $[L:K_{1}]=9$.



      Now, note that
      $$f(L/K)=f(L/K_{1})f(K_{1}/K)=2f(L/K_{1}),$$
      because $K_{1}/K$ is unramified (where $f(L/K)$ means the inertia degree of the extension $L/K$). Also
      $$e(L/K)=e(L/L_{1})e(L_{1}/K)=9e(L/L_{1}),$$
      because $L_{1}/K$ is totally ramified (where $e(L/K)$ means the ramification index of the extension $L/K$). Thus we see that $2mid f(L/K)$ and $9mid e(L/K)$. By the fundamental identity we have $$18=[L:K]=e(L/K)f(L/K),$$
      whence $f(L/K)=2$ and $e(L/K)=9$. This implies that $f(L/K_{1})=1$ because
      $$2=f(L/K)=f(L/K_{1})f(K_{1}/K)=2f(L/K_{1}).$$
      It also implies that $e(L/K_{1})=9$ because
      $$9=e(L/K)=e(L/K_{1})e(K_{1}/K)=e(L/K_{1}).$$



      Now for the extension $L/F$ we have
      $$2=f(L/K)=f(L/F)f(F/K)=f(L/F)f(F/K_{1})f(K_{1}/K)=2f(L/F)f(F/K_{1}),$$
      whence $f(L/F)=1$. Furthermore
      $$9=e(L/K)=e(L/F)e(F/K)=e(L/F)e(F/K_{1})e(K_{1}/K)=3e(L/F),$$
      whence $e(L/F)=3$ and we conclude that $[L:F]=e(L/F)f(L/F)=3$.






      share|cite|improve this answer


























        0












        0








        0






        Let $L_{1}=mathbb{Q}_{2}(beta)$ and note that $beta$ is a root of $x^{9}-2$ over $K$, which is an Eisenstein polynomial. Therefore $L_{1}/K$ is totally ramified of degree 9 (see e.g. Proposition 3.6 of Local Fields and Their Extensions by Fesenko & Vostokov for a proof if you don't know this result). Now, $K_{1}/K$ is Galois of degree 2, therefore we have by Galois theory
        $$[L:L_{1}]=[K_{1}:K],$$
        because $L=L_{1}K_{1}$. Therefore we conclude that $[L:L_{1}]=2$. So $[L:K]=18$ by the tower formula, whence $[L:K_{1}]=9$.



        Now, note that
        $$f(L/K)=f(L/K_{1})f(K_{1}/K)=2f(L/K_{1}),$$
        because $K_{1}/K$ is unramified (where $f(L/K)$ means the inertia degree of the extension $L/K$). Also
        $$e(L/K)=e(L/L_{1})e(L_{1}/K)=9e(L/L_{1}),$$
        because $L_{1}/K$ is totally ramified (where $e(L/K)$ means the ramification index of the extension $L/K$). Thus we see that $2mid f(L/K)$ and $9mid e(L/K)$. By the fundamental identity we have $$18=[L:K]=e(L/K)f(L/K),$$
        whence $f(L/K)=2$ and $e(L/K)=9$. This implies that $f(L/K_{1})=1$ because
        $$2=f(L/K)=f(L/K_{1})f(K_{1}/K)=2f(L/K_{1}).$$
        It also implies that $e(L/K_{1})=9$ because
        $$9=e(L/K)=e(L/K_{1})e(K_{1}/K)=e(L/K_{1}).$$



        Now for the extension $L/F$ we have
        $$2=f(L/K)=f(L/F)f(F/K)=f(L/F)f(F/K_{1})f(K_{1}/K)=2f(L/F)f(F/K_{1}),$$
        whence $f(L/F)=1$. Furthermore
        $$9=e(L/K)=e(L/F)e(F/K)=e(L/F)e(F/K_{1})e(K_{1}/K)=3e(L/F),$$
        whence $e(L/F)=3$ and we conclude that $[L:F]=e(L/F)f(L/F)=3$.






        share|cite|improve this answer














        Let $L_{1}=mathbb{Q}_{2}(beta)$ and note that $beta$ is a root of $x^{9}-2$ over $K$, which is an Eisenstein polynomial. Therefore $L_{1}/K$ is totally ramified of degree 9 (see e.g. Proposition 3.6 of Local Fields and Their Extensions by Fesenko & Vostokov for a proof if you don't know this result). Now, $K_{1}/K$ is Galois of degree 2, therefore we have by Galois theory
        $$[L:L_{1}]=[K_{1}:K],$$
        because $L=L_{1}K_{1}$. Therefore we conclude that $[L:L_{1}]=2$. So $[L:K]=18$ by the tower formula, whence $[L:K_{1}]=9$.



        Now, note that
        $$f(L/K)=f(L/K_{1})f(K_{1}/K)=2f(L/K_{1}),$$
        because $K_{1}/K$ is unramified (where $f(L/K)$ means the inertia degree of the extension $L/K$). Also
        $$e(L/K)=e(L/L_{1})e(L_{1}/K)=9e(L/L_{1}),$$
        because $L_{1}/K$ is totally ramified (where $e(L/K)$ means the ramification index of the extension $L/K$). Thus we see that $2mid f(L/K)$ and $9mid e(L/K)$. By the fundamental identity we have $$18=[L:K]=e(L/K)f(L/K),$$
        whence $f(L/K)=2$ and $e(L/K)=9$. This implies that $f(L/K_{1})=1$ because
        $$2=f(L/K)=f(L/K_{1})f(K_{1}/K)=2f(L/K_{1}).$$
        It also implies that $e(L/K_{1})=9$ because
        $$9=e(L/K)=e(L/K_{1})e(K_{1}/K)=e(L/K_{1}).$$



        Now for the extension $L/F$ we have
        $$2=f(L/K)=f(L/F)f(F/K)=f(L/F)f(F/K_{1})f(K_{1}/K)=2f(L/F)f(F/K_{1}),$$
        whence $f(L/F)=1$. Furthermore
        $$9=e(L/K)=e(L/F)e(F/K)=e(L/F)e(F/K_{1})e(K_{1}/K)=3e(L/F),$$
        whence $e(L/F)=3$ and we conclude that $[L:F]=e(L/F)f(L/F)=3$.







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited 2 days ago

























        answered 2 days ago









        YumekuiMath

        34114




        34114






























            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.





            Some of your past answers have not been well-received, and you're in danger of being blocked from answering.


            Please pay close attention to the following guidance:


            • 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.


            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%2f3049847%2fcomputing-degrees-and-ramification-indices-of-some-extensions-of-mathbbq-2%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

            Human spaceflight

            Image
            Human spaceflight From Wikipedia, the free encyclopedia Jump to navigation Jump to search "Space traveler" redirects here. For other uses, see Space traveler (disambiguation). @media all and (max-width:720px){.mw-parser-output .tmulti>.thumbinner{width:100%!important;max-width:none!important}.mw-parser-output .tmulti .tsingle{float:none!important;max-width:none!important;width:100%!important;text-align:center}} Apollo 11 crewmember Buzz Aldrin on the Moon, 1969 International Space Station crewmember Tracy Caldwell Dyson views the Earth, 2010 Space Shuttle Discovery heads into space with a crew aboard, STS-121 in 2006 Inside a space suit on the Canadarm, 1993 Human spaceflight (also referred to as crewed spaceflight or manned spaceflight ) is space travel with a crew or passengers aboard the spacecraft. Spacecraft carrying people may be operated directly, by human crew, or it may be e...
            Read more

            Can not write log (Is /dev/pts mounted?) - openpty in Ubuntu-on-Windows?

            Image
            0 1 What exactly does the error message beginning with E: mean? I assume it has to do with the file structure on Ubuntu-on-Windows, but what exactly? The following package was automatically installed and is no longer required: os-prober Use 'apt-get autoremove' to remove it. The following extra packages will be installed: libxslt1.1 The following NEW packages will be installed: libxslt1.1 xmlstarlet 0 upgraded, 2 newly installed, 0 to remove and 0 not upgraded. Need to get 435 kB of archives. After this operation, 1023 kB of additional disk space will be used. Get:1 http://archive.ubuntu.com/ubuntu/ trusty/main libxslt1.1 amd64 1.1.28-2build1 [145 kB] Get:2 http://archive.ubuntu.com/ubuntu/ trusty/universe xmlstarlet amd64 1.5.0-1 [290 kB] ...
            Read more

            張江高科駅

            Image
            張江高科駅 ちょうこうこうか Zhangjiang Hi-Tech Park 所在地 中国上海市浦東新区 所属事業者 上海地鉄運営有限公司（上海軌道交通・駅詳細） 上海浦東現代有軌交通有限公司（駅詳細） テンプレートを表示 張江高科駅 各種表記 繁体字： 張江高科站 簡体字： 张江高科站 拼音 ： Zhàngjiāng Gāokē Zhàn 発音： ヂャンジャン ガオクー ヂャン 英文： Zhangjiang Hi-Tech Park Station テンプレートを表示 張江高科駅 （ちょうこうこうか-えき）は中華人民共和国上海市浦東新区に位置する、上海軌道交通2号線の駅である。 なお、張江高科駅前にある張江軌道電車の 張江地鉄駅駅 についてもここで述べる。 目次 1 利用可能な鉄道路線 2 沿革 3 駅構造 3.1 上海軌道交通 3.2 張江軌道電車 4 駅周辺 5 隣の駅 6 注釈 7 関連項目 利用可能な鉄道路線 上海軌道交通 ■ 2号線 上海浦東現代有軌交通有限公司 ■ 張江軌道電車 沿革 高架駅時代の姿 2000年12月27日 - 2号線の駅が高架駅として開業。 2010年 1月1日 - 張江軌道電車の開業に伴い、張江地鉄駅駅（停留所）を設置。 2月13日 - 駅地下化に伴い高架駅の運用停止。 2月24日 - 10日間の閉鎖を経て地下新駅の開業。 2017年2月9日 - リニューアル工事を経て再開業。乗降客の増加に対応し、ホームへのエレベーター、エスカレーターを増設、券売機エリアを拡張した [1] 。 駅構造 上海軌道交通 上海軌道交通 張江高科駅 5号口 ちょうこうこうか Zhangjiang Hi-Tech Park ◄ 竜陽路 (2.60km) (1.61km) 金科路 ► 所在地 中国上海市浦東新区松涛路 祖冲之路 （中国語版） 北緯31度12分11秒 東経121度34分50秒  /  北緯31.2...
            Read more