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Let $L/K$ be extension of complete discrete valued fields, $S,R$ be their ring of integers, if $L/K$ is tamely ramified, then $Tr_{L/K}(S)=R$. How does this imply $v_L(\mathfrak{D})=e-1$? Here $\mathfrak{D}$ is the different ideal in $S$. In Cassels and Frolich's Algebraic number theory p.13,Theorem 2, we first note that $Tr_{L/K}(S)^{-1}=\mathfrak{D}^{-1}\cap K$, then I was not clear how they deduced the inequality: $v_k(Tr_{L/K}(S))\leq \frac{v_L(\mathfrak{D})}{e}<v_k(Tr_{L/K}(S))+1$?

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  • $\begingroup$ Your first sentence deals with the unramified case, but it looks to me as if everything else deals with the tamely ramified case. If this is right, I think you should make that clear in the text. $\endgroup$ – Lubin Dec 18 '15 at 5:15
  • $\begingroup$ @Lubin You are right, it should be tamely ramified..(not unramified), I've corrected it $\endgroup$ – Qixiao Dec 18 '15 at 16:40

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