Proof of a Lemma for a local field extension with certain properties












3














The following result is from "Euler Factors determine local Weil Representations" by Tim and Vladimir Dokchitser:




Lemma 1: Let $F/K$ be a cyclic extension of degree $n$ and ramification degree $e$. Then there exists a cyclic totally ramified extension $L/K$ of degree $e$ such that $FL/L$ is unramified of degree $n$.




The authors argue that it is enough to prove this Lemma for the case of $F/K$ having prime power degree and now I am trying to understand that.



I was able to prove the following result which should be helpful:




Lemma 2: Let $F/K$ be a finite extension with degree $n$ and let $p_1^{n_1} cdots p_r^{n_r}$ be the prime decomposition of $n$. Then there exist finite extensions $F_i/K$ with $[F_i:K]=p_i^{n_i}$ such that $F$ is the compositum of $F_1,dots,F_r$.




For simplicity, let $n = p_1^{n_1} p_2^{n_2}$ be the prime decomposition of $F/K$ and let $F = F_1 F_2$ be the corresponding decomposition of $F$ into prime power degree extensions of $K$ obtained from Lemma 2.



If Lemma 1 is true for the case of prime power degree extensions, there exist cyclic and totally ramified extensions $L_1$ and $L_2$ of $K$ of degree $e(F_1/K)$ and $e(F_2/K)$ such that $F_1L_1/L_1$ and $F_2 L_2/L_2$ are totally ramified of degree $p_1^{n_1}$ and $p_2^{n_2}$, resprectively.



To show the general case, I assume that $L$ must be $L_1 L_2$. Then it is unproblematic to show that $L/K$ is unramified of degree $n$ due to the coprimeness of $p_1$ and $p_2$. But I have no idea how to show that $FL/L$ is unramified of degree $n$. It is especially hard because the ground field $L$ is a compositum itself and there are no general rules like $[FL:K] leq [F:K] [L:K]$ which I could apply here.



Could you please help me completing this proof? Any help is really appreciated!










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    3














    The following result is from "Euler Factors determine local Weil Representations" by Tim and Vladimir Dokchitser:




    Lemma 1: Let $F/K$ be a cyclic extension of degree $n$ and ramification degree $e$. Then there exists a cyclic totally ramified extension $L/K$ of degree $e$ such that $FL/L$ is unramified of degree $n$.




    The authors argue that it is enough to prove this Lemma for the case of $F/K$ having prime power degree and now I am trying to understand that.



    I was able to prove the following result which should be helpful:




    Lemma 2: Let $F/K$ be a finite extension with degree $n$ and let $p_1^{n_1} cdots p_r^{n_r}$ be the prime decomposition of $n$. Then there exist finite extensions $F_i/K$ with $[F_i:K]=p_i^{n_i}$ such that $F$ is the compositum of $F_1,dots,F_r$.




    For simplicity, let $n = p_1^{n_1} p_2^{n_2}$ be the prime decomposition of $F/K$ and let $F = F_1 F_2$ be the corresponding decomposition of $F$ into prime power degree extensions of $K$ obtained from Lemma 2.



    If Lemma 1 is true for the case of prime power degree extensions, there exist cyclic and totally ramified extensions $L_1$ and $L_2$ of $K$ of degree $e(F_1/K)$ and $e(F_2/K)$ such that $F_1L_1/L_1$ and $F_2 L_2/L_2$ are totally ramified of degree $p_1^{n_1}$ and $p_2^{n_2}$, resprectively.



    To show the general case, I assume that $L$ must be $L_1 L_2$. Then it is unproblematic to show that $L/K$ is unramified of degree $n$ due to the coprimeness of $p_1$ and $p_2$. But I have no idea how to show that $FL/L$ is unramified of degree $n$. It is especially hard because the ground field $L$ is a compositum itself and there are no general rules like $[FL:K] leq [F:K] [L:K]$ which I could apply here.



    Could you please help me completing this proof? Any help is really appreciated!










    share|cite|improve this question



























      3












      3








      3


      1





      The following result is from "Euler Factors determine local Weil Representations" by Tim and Vladimir Dokchitser:




      Lemma 1: Let $F/K$ be a cyclic extension of degree $n$ and ramification degree $e$. Then there exists a cyclic totally ramified extension $L/K$ of degree $e$ such that $FL/L$ is unramified of degree $n$.




      The authors argue that it is enough to prove this Lemma for the case of $F/K$ having prime power degree and now I am trying to understand that.



      I was able to prove the following result which should be helpful:




      Lemma 2: Let $F/K$ be a finite extension with degree $n$ and let $p_1^{n_1} cdots p_r^{n_r}$ be the prime decomposition of $n$. Then there exist finite extensions $F_i/K$ with $[F_i:K]=p_i^{n_i}$ such that $F$ is the compositum of $F_1,dots,F_r$.




      For simplicity, let $n = p_1^{n_1} p_2^{n_2}$ be the prime decomposition of $F/K$ and let $F = F_1 F_2$ be the corresponding decomposition of $F$ into prime power degree extensions of $K$ obtained from Lemma 2.



      If Lemma 1 is true for the case of prime power degree extensions, there exist cyclic and totally ramified extensions $L_1$ and $L_2$ of $K$ of degree $e(F_1/K)$ and $e(F_2/K)$ such that $F_1L_1/L_1$ and $F_2 L_2/L_2$ are totally ramified of degree $p_1^{n_1}$ and $p_2^{n_2}$, resprectively.



      To show the general case, I assume that $L$ must be $L_1 L_2$. Then it is unproblematic to show that $L/K$ is unramified of degree $n$ due to the coprimeness of $p_1$ and $p_2$. But I have no idea how to show that $FL/L$ is unramified of degree $n$. It is especially hard because the ground field $L$ is a compositum itself and there are no general rules like $[FL:K] leq [F:K] [L:K]$ which I could apply here.



      Could you please help me completing this proof? Any help is really appreciated!










      share|cite|improve this question















      The following result is from "Euler Factors determine local Weil Representations" by Tim and Vladimir Dokchitser:




      Lemma 1: Let $F/K$ be a cyclic extension of degree $n$ and ramification degree $e$. Then there exists a cyclic totally ramified extension $L/K$ of degree $e$ such that $FL/L$ is unramified of degree $n$.




      The authors argue that it is enough to prove this Lemma for the case of $F/K$ having prime power degree and now I am trying to understand that.



      I was able to prove the following result which should be helpful:




      Lemma 2: Let $F/K$ be a finite extension with degree $n$ and let $p_1^{n_1} cdots p_r^{n_r}$ be the prime decomposition of $n$. Then there exist finite extensions $F_i/K$ with $[F_i:K]=p_i^{n_i}$ such that $F$ is the compositum of $F_1,dots,F_r$.




      For simplicity, let $n = p_1^{n_1} p_2^{n_2}$ be the prime decomposition of $F/K$ and let $F = F_1 F_2$ be the corresponding decomposition of $F$ into prime power degree extensions of $K$ obtained from Lemma 2.



      If Lemma 1 is true for the case of prime power degree extensions, there exist cyclic and totally ramified extensions $L_1$ and $L_2$ of $K$ of degree $e(F_1/K)$ and $e(F_2/K)$ such that $F_1L_1/L_1$ and $F_2 L_2/L_2$ are totally ramified of degree $p_1^{n_1}$ and $p_2^{n_2}$, resprectively.



      To show the general case, I assume that $L$ must be $L_1 L_2$. Then it is unproblematic to show that $L/K$ is unramified of degree $n$ due to the coprimeness of $p_1$ and $p_2$. But I have no idea how to show that $FL/L$ is unramified of degree $n$. It is especially hard because the ground field $L$ is a compositum itself and there are no general rules like $[FL:K] leq [F:K] [L:K]$ which I could apply here.



      Could you please help me completing this proof? Any help is really appreciated!







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






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      edited Dec 20 '18 at 0:26

























      asked Dec 19 '18 at 22:07









      Diglett

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          I doubt there could be a purely Galois theoretic answer to your problem since it deals specifically with local fields. By the way, you don't recall what you mean by local fields, so I'll assume that all residual fields are finite, in order to apply the most powerful tool at our disposal, namely local CFT. My reference throughout will be Cassels-Fröhlich, chap. VI by Serre. I keep your notations $F/K$ of degree $n$, etc., adding $n=ef$ as usual, so that we have a tower of extensions $F/N/K$, where $N$ is the maximal unramified subextension, of degree $f$ over $K$. The question is to construct a totally ramified extension $L/K$ of degree $e$ s.t. $FL/L$ is unramified of degree $n$. It could be possible to proceed directly using the classical norm and ramification properties in extensions of local fields, but this would be rather technical and complicated (see Serre's comment on the local Kronecker-Weber theorem, loc. cit., top of p.147).



          We'll appeal instead to Lubin-Tate's theory, which provides us with a diect product decomposition of $G(K_{ab}/K)$, where $K_{ab}$ is the maximal abelian extension of $K$. Fix a uniformizer $pi$ of $K$. Acording to loc. cit. thm.3, p.143, there exists an extension $K_{pi}/K$ (independent of the choice of $pi$), linearly disjoint from the maximal unramified extension $K_{nr}/K$, s.t. $K_{ab}=K_{nr}.K_{pi}$ and we have a direct product decomposition $G(K_{ab}/K)cong G(K_{ab}/K_{nr}) times G(K_{ab}/K_{pi}) cong G(K_{nr}/K) times G(K_{pi}/K)$. Recall that the norm residue symbol $(pi,K_{nr}/K)$ gives the Frobenius automorphism which topologically generates $G(K_{nr}/K)cong hat {mathbf Z}$ (the profinite completion of $mathbf Z$). On the other "branch", the norm residue map $uin U_K to (u,K_{pi}/K)$ realizes an isomorphism $U_K cong G(K_{pi}/K)$.



          Back to your problem (it would be convenient to draw a Galois diagram): $N/K$ is a finite subextension degree $f$ of $K_{nr}/K$, and the totally ramified extension $F/N$ lifts to an extension $F':=F.K_{nr}$ of $K_{nr}$ of the same degree $e$, which in turn goes down to a totally ramified extension (because contained in $K_{pi})$ $L/K$ of degree $e$ s.t. $G(F'/K_{nr})cong G(L/K)$. Note that $F'$ is no other than $L_{nr}$, but $N.L neq F$ because $F/K$ is cyclic. It remains only to show that $F.L/L$ is an unramified extension of degree $n$. Because $F/K$ is cyclic, $N.L neq F$ and the extension $F.L.N/N$ is bicyclic, with Galois group $mathbf Z/e times mathbf Z/e$, the three subextensions of degree $e$ over $N$ being $F, N.L$ and $F.L$. Let us show that $F.L/N$ is unramified. By the construction of $L$ and $F'$ via inductive limits, there exists a finite subextension $M/N$ of $K_{nr}/N$ s.t. $F'=M.L.K_{nr}$, so $M.L$ contains $F.L$ and $F.L/N$ is unramified of degree $e$, and finally $F.L/L$ is unramified of degree $n$, as desired.






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            I doubt there could be a purely Galois theoretic answer to your problem since it deals specifically with local fields. By the way, you don't recall what you mean by local fields, so I'll assume that all residual fields are finite, in order to apply the most powerful tool at our disposal, namely local CFT. My reference throughout will be Cassels-Fröhlich, chap. VI by Serre. I keep your notations $F/K$ of degree $n$, etc., adding $n=ef$ as usual, so that we have a tower of extensions $F/N/K$, where $N$ is the maximal unramified subextension, of degree $f$ over $K$. The question is to construct a totally ramified extension $L/K$ of degree $e$ s.t. $FL/L$ is unramified of degree $n$. It could be possible to proceed directly using the classical norm and ramification properties in extensions of local fields, but this would be rather technical and complicated (see Serre's comment on the local Kronecker-Weber theorem, loc. cit., top of p.147).



            We'll appeal instead to Lubin-Tate's theory, which provides us with a diect product decomposition of $G(K_{ab}/K)$, where $K_{ab}$ is the maximal abelian extension of $K$. Fix a uniformizer $pi$ of $K$. Acording to loc. cit. thm.3, p.143, there exists an extension $K_{pi}/K$ (independent of the choice of $pi$), linearly disjoint from the maximal unramified extension $K_{nr}/K$, s.t. $K_{ab}=K_{nr}.K_{pi}$ and we have a direct product decomposition $G(K_{ab}/K)cong G(K_{ab}/K_{nr}) times G(K_{ab}/K_{pi}) cong G(K_{nr}/K) times G(K_{pi}/K)$. Recall that the norm residue symbol $(pi,K_{nr}/K)$ gives the Frobenius automorphism which topologically generates $G(K_{nr}/K)cong hat {mathbf Z}$ (the profinite completion of $mathbf Z$). On the other "branch", the norm residue map $uin U_K to (u,K_{pi}/K)$ realizes an isomorphism $U_K cong G(K_{pi}/K)$.



            Back to your problem (it would be convenient to draw a Galois diagram): $N/K$ is a finite subextension degree $f$ of $K_{nr}/K$, and the totally ramified extension $F/N$ lifts to an extension $F':=F.K_{nr}$ of $K_{nr}$ of the same degree $e$, which in turn goes down to a totally ramified extension (because contained in $K_{pi})$ $L/K$ of degree $e$ s.t. $G(F'/K_{nr})cong G(L/K)$. Note that $F'$ is no other than $L_{nr}$, but $N.L neq F$ because $F/K$ is cyclic. It remains only to show that $F.L/L$ is an unramified extension of degree $n$. Because $F/K$ is cyclic, $N.L neq F$ and the extension $F.L.N/N$ is bicyclic, with Galois group $mathbf Z/e times mathbf Z/e$, the three subextensions of degree $e$ over $N$ being $F, N.L$ and $F.L$. Let us show that $F.L/N$ is unramified. By the construction of $L$ and $F'$ via inductive limits, there exists a finite subextension $M/N$ of $K_{nr}/N$ s.t. $F'=M.L.K_{nr}$, so $M.L$ contains $F.L$ and $F.L/N$ is unramified of degree $e$, and finally $F.L/L$ is unramified of degree $n$, as desired.






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              0














              I doubt there could be a purely Galois theoretic answer to your problem since it deals specifically with local fields. By the way, you don't recall what you mean by local fields, so I'll assume that all residual fields are finite, in order to apply the most powerful tool at our disposal, namely local CFT. My reference throughout will be Cassels-Fröhlich, chap. VI by Serre. I keep your notations $F/K$ of degree $n$, etc., adding $n=ef$ as usual, so that we have a tower of extensions $F/N/K$, where $N$ is the maximal unramified subextension, of degree $f$ over $K$. The question is to construct a totally ramified extension $L/K$ of degree $e$ s.t. $FL/L$ is unramified of degree $n$. It could be possible to proceed directly using the classical norm and ramification properties in extensions of local fields, but this would be rather technical and complicated (see Serre's comment on the local Kronecker-Weber theorem, loc. cit., top of p.147).



              We'll appeal instead to Lubin-Tate's theory, which provides us with a diect product decomposition of $G(K_{ab}/K)$, where $K_{ab}$ is the maximal abelian extension of $K$. Fix a uniformizer $pi$ of $K$. Acording to loc. cit. thm.3, p.143, there exists an extension $K_{pi}/K$ (independent of the choice of $pi$), linearly disjoint from the maximal unramified extension $K_{nr}/K$, s.t. $K_{ab}=K_{nr}.K_{pi}$ and we have a direct product decomposition $G(K_{ab}/K)cong G(K_{ab}/K_{nr}) times G(K_{ab}/K_{pi}) cong G(K_{nr}/K) times G(K_{pi}/K)$. Recall that the norm residue symbol $(pi,K_{nr}/K)$ gives the Frobenius automorphism which topologically generates $G(K_{nr}/K)cong hat {mathbf Z}$ (the profinite completion of $mathbf Z$). On the other "branch", the norm residue map $uin U_K to (u,K_{pi}/K)$ realizes an isomorphism $U_K cong G(K_{pi}/K)$.



              Back to your problem (it would be convenient to draw a Galois diagram): $N/K$ is a finite subextension degree $f$ of $K_{nr}/K$, and the totally ramified extension $F/N$ lifts to an extension $F':=F.K_{nr}$ of $K_{nr}$ of the same degree $e$, which in turn goes down to a totally ramified extension (because contained in $K_{pi})$ $L/K$ of degree $e$ s.t. $G(F'/K_{nr})cong G(L/K)$. Note that $F'$ is no other than $L_{nr}$, but $N.L neq F$ because $F/K$ is cyclic. It remains only to show that $F.L/L$ is an unramified extension of degree $n$. Because $F/K$ is cyclic, $N.L neq F$ and the extension $F.L.N/N$ is bicyclic, with Galois group $mathbf Z/e times mathbf Z/e$, the three subextensions of degree $e$ over $N$ being $F, N.L$ and $F.L$. Let us show that $F.L/N$ is unramified. By the construction of $L$ and $F'$ via inductive limits, there exists a finite subextension $M/N$ of $K_{nr}/N$ s.t. $F'=M.L.K_{nr}$, so $M.L$ contains $F.L$ and $F.L/N$ is unramified of degree $e$, and finally $F.L/L$ is unramified of degree $n$, as desired.






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                0












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                I doubt there could be a purely Galois theoretic answer to your problem since it deals specifically with local fields. By the way, you don't recall what you mean by local fields, so I'll assume that all residual fields are finite, in order to apply the most powerful tool at our disposal, namely local CFT. My reference throughout will be Cassels-Fröhlich, chap. VI by Serre. I keep your notations $F/K$ of degree $n$, etc., adding $n=ef$ as usual, so that we have a tower of extensions $F/N/K$, where $N$ is the maximal unramified subextension, of degree $f$ over $K$. The question is to construct a totally ramified extension $L/K$ of degree $e$ s.t. $FL/L$ is unramified of degree $n$. It could be possible to proceed directly using the classical norm and ramification properties in extensions of local fields, but this would be rather technical and complicated (see Serre's comment on the local Kronecker-Weber theorem, loc. cit., top of p.147).



                We'll appeal instead to Lubin-Tate's theory, which provides us with a diect product decomposition of $G(K_{ab}/K)$, where $K_{ab}$ is the maximal abelian extension of $K$. Fix a uniformizer $pi$ of $K$. Acording to loc. cit. thm.3, p.143, there exists an extension $K_{pi}/K$ (independent of the choice of $pi$), linearly disjoint from the maximal unramified extension $K_{nr}/K$, s.t. $K_{ab}=K_{nr}.K_{pi}$ and we have a direct product decomposition $G(K_{ab}/K)cong G(K_{ab}/K_{nr}) times G(K_{ab}/K_{pi}) cong G(K_{nr}/K) times G(K_{pi}/K)$. Recall that the norm residue symbol $(pi,K_{nr}/K)$ gives the Frobenius automorphism which topologically generates $G(K_{nr}/K)cong hat {mathbf Z}$ (the profinite completion of $mathbf Z$). On the other "branch", the norm residue map $uin U_K to (u,K_{pi}/K)$ realizes an isomorphism $U_K cong G(K_{pi}/K)$.



                Back to your problem (it would be convenient to draw a Galois diagram): $N/K$ is a finite subextension degree $f$ of $K_{nr}/K$, and the totally ramified extension $F/N$ lifts to an extension $F':=F.K_{nr}$ of $K_{nr}$ of the same degree $e$, which in turn goes down to a totally ramified extension (because contained in $K_{pi})$ $L/K$ of degree $e$ s.t. $G(F'/K_{nr})cong G(L/K)$. Note that $F'$ is no other than $L_{nr}$, but $N.L neq F$ because $F/K$ is cyclic. It remains only to show that $F.L/L$ is an unramified extension of degree $n$. Because $F/K$ is cyclic, $N.L neq F$ and the extension $F.L.N/N$ is bicyclic, with Galois group $mathbf Z/e times mathbf Z/e$, the three subextensions of degree $e$ over $N$ being $F, N.L$ and $F.L$. Let us show that $F.L/N$ is unramified. By the construction of $L$ and $F'$ via inductive limits, there exists a finite subextension $M/N$ of $K_{nr}/N$ s.t. $F'=M.L.K_{nr}$, so $M.L$ contains $F.L$ and $F.L/N$ is unramified of degree $e$, and finally $F.L/L$ is unramified of degree $n$, as desired.






                share|cite|improve this answer












                I doubt there could be a purely Galois theoretic answer to your problem since it deals specifically with local fields. By the way, you don't recall what you mean by local fields, so I'll assume that all residual fields are finite, in order to apply the most powerful tool at our disposal, namely local CFT. My reference throughout will be Cassels-Fröhlich, chap. VI by Serre. I keep your notations $F/K$ of degree $n$, etc., adding $n=ef$ as usual, so that we have a tower of extensions $F/N/K$, where $N$ is the maximal unramified subextension, of degree $f$ over $K$. The question is to construct a totally ramified extension $L/K$ of degree $e$ s.t. $FL/L$ is unramified of degree $n$. It could be possible to proceed directly using the classical norm and ramification properties in extensions of local fields, but this would be rather technical and complicated (see Serre's comment on the local Kronecker-Weber theorem, loc. cit., top of p.147).



                We'll appeal instead to Lubin-Tate's theory, which provides us with a diect product decomposition of $G(K_{ab}/K)$, where $K_{ab}$ is the maximal abelian extension of $K$. Fix a uniformizer $pi$ of $K$. Acording to loc. cit. thm.3, p.143, there exists an extension $K_{pi}/K$ (independent of the choice of $pi$), linearly disjoint from the maximal unramified extension $K_{nr}/K$, s.t. $K_{ab}=K_{nr}.K_{pi}$ and we have a direct product decomposition $G(K_{ab}/K)cong G(K_{ab}/K_{nr}) times G(K_{ab}/K_{pi}) cong G(K_{nr}/K) times G(K_{pi}/K)$. Recall that the norm residue symbol $(pi,K_{nr}/K)$ gives the Frobenius automorphism which topologically generates $G(K_{nr}/K)cong hat {mathbf Z}$ (the profinite completion of $mathbf Z$). On the other "branch", the norm residue map $uin U_K to (u,K_{pi}/K)$ realizes an isomorphism $U_K cong G(K_{pi}/K)$.



                Back to your problem (it would be convenient to draw a Galois diagram): $N/K$ is a finite subextension degree $f$ of $K_{nr}/K$, and the totally ramified extension $F/N$ lifts to an extension $F':=F.K_{nr}$ of $K_{nr}$ of the same degree $e$, which in turn goes down to a totally ramified extension (because contained in $K_{pi})$ $L/K$ of degree $e$ s.t. $G(F'/K_{nr})cong G(L/K)$. Note that $F'$ is no other than $L_{nr}$, but $N.L neq F$ because $F/K$ is cyclic. It remains only to show that $F.L/L$ is an unramified extension of degree $n$. Because $F/K$ is cyclic, $N.L neq F$ and the extension $F.L.N/N$ is bicyclic, with Galois group $mathbf Z/e times mathbf Z/e$, the three subextensions of degree $e$ over $N$ being $F, N.L$ and $F.L$. Let us show that $F.L/N$ is unramified. By the construction of $L$ and $F'$ via inductive limits, there exists a finite subextension $M/N$ of $K_{nr}/N$ s.t. $F'=M.L.K_{nr}$, so $M.L$ contains $F.L$ and $F.L/N$ is unramified of degree $e$, and finally $F.L/L$ is unramified of degree $n$, as desired.







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                answered Dec 27 '18 at 9:04









                nguyen quang do

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