Continuity of trace and norm

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Let $L|K$ be a finite extension of discrete valuation fields (not necessarily complete). Consider the classical maps:



$$operatornameTr_K:Lto K$$
$$N_K:L^timesto K^times$$



Are such functions continuous with respect to the valuation topologies?




abstract-algebra general-topology field-theory valuation-theory




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  • 2




    Finite fields only have the trivial valuation, so I think that tag was not appropriate, and deleted it.
    – Jyrki Lahtonen
    Aug 15 at 11:23















up vote
2
down vote

favorite
1












Let $L|K$ be a finite extension of discrete valuation fields (not necessarily complete). Consider the classical maps:



$$operatornameTr_K:Lto K$$
$$N_K:L^timesto K^times$$



Are such functions continuous with respect to the valuation topologies?




abstract-algebra general-topology field-theory valuation-theory




share|cite|improve this question


















  • 2




    Finite fields only have the trivial valuation, so I think that tag was not appropriate, and deleted it.
    – Jyrki Lahtonen
    Aug 15 at 11:23













up vote
2
down vote

favorite
1









up vote
2
down vote

favorite
1






1





Let $L|K$ be a finite extension of discrete valuation fields (not necessarily complete). Consider the classical maps:



$$operatornameTr_K:Lto K$$
$$N_K:L^timesto K^times$$



Are such functions continuous with respect to the valuation topologies?




abstract-algebra general-topology field-theory valuation-theory




share|cite|improve this question














Let $L|K$ be a finite extension of discrete valuation fields (not necessarily complete). Consider the classical maps:



$$operatornameTr_K:Lto K$$
$$N_K:L^timesto K^times$$



Are such functions continuous with respect to the valuation topologies?





abstract-algebra general-topology field-theory valuation-theory





share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Aug 15 at 11:23









Jyrki Lahtonen

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105k12161357










asked Aug 15 at 8:25









manifold

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260213







  • 2




    Finite fields only have the trivial valuation, so I think that tag was not appropriate, and deleted it.
    – Jyrki Lahtonen
    Aug 15 at 11:23













  • 2




    Finite fields only have the trivial valuation, so I think that tag was not appropriate, and deleted it.
    – Jyrki Lahtonen
    Aug 15 at 11:23








2




2




Finite fields only have the trivial valuation, so I think that tag was not appropriate, and deleted it.
– Jyrki Lahtonen
Aug 15 at 11:23





Finite fields only have the trivial valuation, so I think that tag was not appropriate, and deleted it.
– Jyrki Lahtonen
Aug 15 at 11:23











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Yes.



Let $N$ be a normal closure of $L$ over $K$, the valuation on $L$ can be extended to $N$ (not necessarily unique, fix one). It suffices to show that each $K$-embedding $sigma: L to N$ is continuous, we can extend $sigma$ to $N$. Let $mathfrakP$ be the prime ideal of $N$.



Note that $sigma(mathfrakP)= mathfrakP$. Since $sigma:Nto N$ is additive, it suffices to show continuity at $0$, this is obvious: for $xin N$, $$x in mathfrakP^n iffsigma(x)in mathfrakP^n$$






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    1 Answer
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    active

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    1 Answer
    1






    active

    oldest

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    active

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    active

    oldest

    votes








    up vote
    2
    down vote



    accepted










    Yes.



    Let $N$ be a normal closure of $L$ over $K$, the valuation on $L$ can be extended to $N$ (not necessarily unique, fix one). It suffices to show that each $K$-embedding $sigma: L to N$ is continuous, we can extend $sigma$ to $N$. Let $mathfrakP$ be the prime ideal of $N$.



    Note that $sigma(mathfrakP)= mathfrakP$. Since $sigma:Nto N$ is additive, it suffices to show continuity at $0$, this is obvious: for $xin N$, $$x in mathfrakP^n iffsigma(x)in mathfrakP^n$$






    share|cite|improve this answer


























      up vote
      2
      down vote



      accepted










      Yes.



      Let $N$ be a normal closure of $L$ over $K$, the valuation on $L$ can be extended to $N$ (not necessarily unique, fix one). It suffices to show that each $K$-embedding $sigma: L to N$ is continuous, we can extend $sigma$ to $N$. Let $mathfrakP$ be the prime ideal of $N$.



      Note that $sigma(mathfrakP)= mathfrakP$. Since $sigma:Nto N$ is additive, it suffices to show continuity at $0$, this is obvious: for $xin N$, $$x in mathfrakP^n iffsigma(x)in mathfrakP^n$$






      share|cite|improve this answer
























        up vote
        2
        down vote



        accepted







        up vote
        2
        down vote



        accepted






        Yes.



        Let $N$ be a normal closure of $L$ over $K$, the valuation on $L$ can be extended to $N$ (not necessarily unique, fix one). It suffices to show that each $K$-embedding $sigma: L to N$ is continuous, we can extend $sigma$ to $N$. Let $mathfrakP$ be the prime ideal of $N$.



        Note that $sigma(mathfrakP)= mathfrakP$. Since $sigma:Nto N$ is additive, it suffices to show continuity at $0$, this is obvious: for $xin N$, $$x in mathfrakP^n iffsigma(x)in mathfrakP^n$$






        share|cite|improve this answer














        Yes.



        Let $N$ be a normal closure of $L$ over $K$, the valuation on $L$ can be extended to $N$ (not necessarily unique, fix one). It suffices to show that each $K$-embedding $sigma: L to N$ is continuous, we can extend $sigma$ to $N$. Let $mathfrakP$ be the prime ideal of $N$.



        Note that $sigma(mathfrakP)= mathfrakP$. Since $sigma:Nto N$ is additive, it suffices to show continuity at $0$, this is obvious: for $xin N$, $$x in mathfrakP^n iffsigma(x)in mathfrakP^n$$







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited Aug 15 at 11:09

























        answered Aug 15 at 10:25









        pisco

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