Galois Groups in Ramification Theory












4












$begingroup$


I had a slight confusion about Galois groups over a base field which is complete with respect to a discrete valuation.



We know that there are irreducible polynomials such as $X^3+X^2+2X-8$ where the roots have different valuation in the splitting field. In this example, take $K=mathbb{Q}_2$ to be the base field and consider its splitting field. By Newton's Polygon, the three roots have valuations $0$, $1$ and $2$. (See the algebraic number theory notes at www.jmilne.org/math/)



However, it is also noted that Galois groups preserve the valuation regardless of what kind of extension it is (unramified, totally ramified, etc). This is by the uniqueness of extension of an absolute value in a complete field. (Analogous to the result, all norms of a finite real dimensional vector space are Lipschitz equivalent.)



To be more specific, if $v$ is a valuation then $v':xmapsto v(sigma(x))$ for $sigmain Gal(L/K)$ is another valuation and $Im(v)=Im(v')$, they must be the same valuation. (i.e. no nonsense about scale factors)



My question is that these two results seem to be contradicting each other. If we take $K=mathbb{Q}_2$ and $L$ to be the splitting field of $X^3+X^2+2X-8$, then a field automorphism in the Galois group needs to be able to send an element $alpha_1$, a root of $X^3+X^2+2X-8$ to another root $alpha_2$ of a different valuation.



So where did I reason wrongly?






galois-theory algebraic-number-theory valuation-theory ramification







share|cite|improve this question









$endgroup$

















    4












    $begingroup$


    I had a slight confusion about Galois groups over a base field which is complete with respect to a discrete valuation.



    We know that there are irreducible polynomials such as $X^3+X^2+2X-8$ where the roots have different valuation in the splitting field. In this example, take $K=mathbb{Q}_2$ to be the base field and consider its splitting field. By Newton's Polygon, the three roots have valuations $0$, $1$ and $2$. (See the algebraic number theory notes at www.jmilne.org/math/)



    However, it is also noted that Galois groups preserve the valuation regardless of what kind of extension it is (unramified, totally ramified, etc). This is by the uniqueness of extension of an absolute value in a complete field. (Analogous to the result, all norms of a finite real dimensional vector space are Lipschitz equivalent.)



    To be more specific, if $v$ is a valuation then $v':xmapsto v(sigma(x))$ for $sigmain Gal(L/K)$ is another valuation and $Im(v)=Im(v')$, they must be the same valuation. (i.e. no nonsense about scale factors)



    My question is that these two results seem to be contradicting each other. If we take $K=mathbb{Q}_2$ and $L$ to be the splitting field of $X^3+X^2+2X-8$, then a field automorphism in the Galois group needs to be able to send an element $alpha_1$, a root of $X^3+X^2+2X-8$ to another root $alpha_2$ of a different valuation.



    So where did I reason wrongly?






    galois-theory algebraic-number-theory valuation-theory ramification







    share|cite|improve this question









    $endgroup$















      4












      4








      4


      1



      $begingroup$


      I had a slight confusion about Galois groups over a base field which is complete with respect to a discrete valuation.



      We know that there are irreducible polynomials such as $X^3+X^2+2X-8$ where the roots have different valuation in the splitting field. In this example, take $K=mathbb{Q}_2$ to be the base field and consider its splitting field. By Newton's Polygon, the three roots have valuations $0$, $1$ and $2$. (See the algebraic number theory notes at www.jmilne.org/math/)



      However, it is also noted that Galois groups preserve the valuation regardless of what kind of extension it is (unramified, totally ramified, etc). This is by the uniqueness of extension of an absolute value in a complete field. (Analogous to the result, all norms of a finite real dimensional vector space are Lipschitz equivalent.)



      To be more specific, if $v$ is a valuation then $v':xmapsto v(sigma(x))$ for $sigmain Gal(L/K)$ is another valuation and $Im(v)=Im(v')$, they must be the same valuation. (i.e. no nonsense about scale factors)



      My question is that these two results seem to be contradicting each other. If we take $K=mathbb{Q}_2$ and $L$ to be the splitting field of $X^3+X^2+2X-8$, then a field automorphism in the Galois group needs to be able to send an element $alpha_1$, a root of $X^3+X^2+2X-8$ to another root $alpha_2$ of a different valuation.



      So where did I reason wrongly?






      galois-theory algebraic-number-theory valuation-theory ramification







      share|cite|improve this question









      $endgroup$




      I had a slight confusion about Galois groups over a base field which is complete with respect to a discrete valuation.



      We know that there are irreducible polynomials such as $X^3+X^2+2X-8$ where the roots have different valuation in the splitting field. In this example, take $K=mathbb{Q}_2$ to be the base field and consider its splitting field. By Newton's Polygon, the three roots have valuations $0$, $1$ and $2$. (See the algebraic number theory notes at www.jmilne.org/math/)



      However, it is also noted that Galois groups preserve the valuation regardless of what kind of extension it is (unramified, totally ramified, etc). This is by the uniqueness of extension of an absolute value in a complete field. (Analogous to the result, all norms of a finite real dimensional vector space are Lipschitz equivalent.)



      To be more specific, if $v$ is a valuation then $v':xmapsto v(sigma(x))$ for $sigmain Gal(L/K)$ is another valuation and $Im(v)=Im(v')$, they must be the same valuation. (i.e. no nonsense about scale factors)



      My question is that these two results seem to be contradicting each other. If we take $K=mathbb{Q}_2$ and $L$ to be the splitting field of $X^3+X^2+2X-8$, then a field automorphism in the Galois group needs to be able to send an element $alpha_1$, a root of $X^3+X^2+2X-8$ to another root $alpha_2$ of a different valuation.



      So where did I reason wrongly?






      galois-theory algebraic-number-theory valuation-theory ramification




      galois-theory algebraic-number-theory valuation-theory ramification






      share|cite|improve this question













      share|cite|improve this question











      share|cite|improve this question




      share|cite|improve this question










      asked Dec 25 '18 at 3:41









      darumadaruma

      982612




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          $begingroup$

          Your problem is that you didn’t use the fact that Newton Polygon tells you that your cubic is by no means irreducible: in fact, it splits over $Bbb Q_2$ into three linear factors. In other words, as a $Bbb Q_2$ polynomial, its splitting field is $Bbb Q_2$.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Thank you, that makes sense.
            $endgroup$
            – daruma
            Dec 25 '18 at 5:29












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          1 Answer
          1






          active

          oldest

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          active

          oldest

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          active

          oldest

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          6












          $begingroup$

          Your problem is that you didn’t use the fact that Newton Polygon tells you that your cubic is by no means irreducible: in fact, it splits over $Bbb Q_2$ into three linear factors. In other words, as a $Bbb Q_2$ polynomial, its splitting field is $Bbb Q_2$.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Thank you, that makes sense.
            $endgroup$
            – daruma
            Dec 25 '18 at 5:29
















          6












          $begingroup$

          Your problem is that you didn’t use the fact that Newton Polygon tells you that your cubic is by no means irreducible: in fact, it splits over $Bbb Q_2$ into three linear factors. In other words, as a $Bbb Q_2$ polynomial, its splitting field is $Bbb Q_2$.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Thank you, that makes sense.
            $endgroup$
            – daruma
            Dec 25 '18 at 5:29














          6












          6








          6





          $begingroup$

          Your problem is that you didn’t use the fact that Newton Polygon tells you that your cubic is by no means irreducible: in fact, it splits over $Bbb Q_2$ into three linear factors. In other words, as a $Bbb Q_2$ polynomial, its splitting field is $Bbb Q_2$.






          share|cite|improve this answer









          $endgroup$



          Your problem is that you didn’t use the fact that Newton Polygon tells you that your cubic is by no means irreducible: in fact, it splits over $Bbb Q_2$ into three linear factors. In other words, as a $Bbb Q_2$ polynomial, its splitting field is $Bbb Q_2$.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Dec 25 '18 at 5:07









          LubinLubin

          45.7k44688




          45.7k44688












          • $begingroup$
            Thank you, that makes sense.
            $endgroup$
            – daruma
            Dec 25 '18 at 5:29


















          • $begingroup$
            Thank you, that makes sense.
            $endgroup$
            – daruma
            Dec 25 '18 at 5:29
















          $begingroup$
          Thank you, that makes sense.
          $endgroup$
          – daruma
          Dec 25 '18 at 5:29




          $begingroup$
          Thank you, that makes sense.
          $endgroup$
          – daruma
          Dec 25 '18 at 5:29


















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