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












2












$begingroup$


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







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$endgroup$












  • $begingroup$
    $mathbb{Z}_2[zeta_3]/mathbb{Z}_2$ 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
    $endgroup$
    – reuns
    Dec 23 '18 at 1:43












  • $begingroup$
    @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.
    $endgroup$
    – Diglett
    Dec 23 '18 at 11:52
















2












$begingroup$


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









$endgroup$












  • $begingroup$
    $mathbb{Z}_2[zeta_3]/mathbb{Z}_2$ 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
    $endgroup$
    – reuns
    Dec 23 '18 at 1:43












  • $begingroup$
    @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.
    $endgroup$
    – Diglett
    Dec 23 '18 at 11:52














2












2








2





$begingroup$


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









$endgroup$




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













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asked Dec 22 '18 at 21:01









DiglettDiglett

9801521




9801521












  • $begingroup$
    $mathbb{Z}_2[zeta_3]/mathbb{Z}_2$ 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
    $endgroup$
    – reuns
    Dec 23 '18 at 1:43












  • $begingroup$
    @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.
    $endgroup$
    – Diglett
    Dec 23 '18 at 11:52


















  • $begingroup$
    $mathbb{Z}_2[zeta_3]/mathbb{Z}_2$ 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
    $endgroup$
    – reuns
    Dec 23 '18 at 1:43












  • $begingroup$
    @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.
    $endgroup$
    – Diglett
    Dec 23 '18 at 11:52
















$begingroup$
$mathbb{Z}_2[zeta_3]/mathbb{Z}_2$ 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
$endgroup$
– reuns
Dec 23 '18 at 1:43






$begingroup$
$mathbb{Z}_2[zeta_3]/mathbb{Z}_2$ 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
$endgroup$
– reuns
Dec 23 '18 at 1:43














$begingroup$
@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.
$endgroup$
– Diglett
Dec 23 '18 at 11:52




$begingroup$
@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.
$endgroup$
– Diglett
Dec 23 '18 at 11:52










1 Answer
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$begingroup$

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 $[L:F]=e(L/F)f(L/F)=3$.






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    $begingroup$

    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











    $endgroup$


















      0












      $begingroup$

      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











      $endgroup$
















        0












        0








        0





        $begingroup$

        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











        $endgroup$



        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 $[L:F]=e(L/F)f(L/F)=3$.







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        edited Dec 25 '18 at 18:44

























        answered Dec 25 '18 at 18:36









        YumekuiMathYumekuiMath

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