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Let $K$ be a Galois extension of $F$. Prove or disprove that any intermediate field $L$ of $K/F$ is of the form $L=F(\{N(a)\mid a \in K\})$, where $N$ the is norm map of $K/L$.

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    $\begingroup$ Welcome to mathSE: since you are a new user, I wanted to let you know a few things about the site. In order to get the best possible answers, it is helpful if you say in what context you encountered the problem, and what your thoughts on it are so far; this will prevent people from telling you things you already know, and help them write their answers at an appropriate level. Also, many find the use of imperative ("Prove or disprove") to be rude when asking for help; please consider rewriting your post. $\endgroup$ Apr 28 '12 at 7:22
  • $\begingroup$ I can prove it if F is a finite field. But in general I guess it's not right and should have some counterexample. And I just can't find one... $\endgroup$
    – lee
    Aug 7 '14 at 5:56
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    $\begingroup$ "is of the form" is unappropriate here, since your $N$ depends on $L$. You should simply say something like "any intermediate $L$ satisfies $L=F(\lbrace N_{K/L}(a) |a \in K\rbrace)$". $\endgroup$ Aug 7 '14 at 8:11
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This assertion is right. We only prove the case when L is an infinite field, the proof when L is just a finite field is much easier because the Galois group is generated by the Frobenius map and the norm is easy to compute.

Proof:Suppose that $K=F(\beta)$ with $f(x)=min(\beta,L)=\prod_{\sigma\in Gal(K/L)}{(x-\sigma(\beta))}$ $=x^n+a_{n-1}x^{n-1}+...+a_{1}x+a_{0}$. A simple observation here is that $F(\{a_{i}\})=L$, so we only need to show that $F(\{N_{K/L}(\alpha)\mid\alpha \in K\})=F(\{a_{i}\})$. Using the definition of Norm, we have $N_{K/L}(\alpha)=\prod_{\sigma\in Gal(K/L)}\sigma(\alpha)$, for any $\alpha$ in K. Let $\alpha=a-\beta$, where $a \in F$, we have $N_{K/L}(\alpha)=f(a)$. At last, using a little linear algebra, we can show that $F(\{f(a)\mid a\in F\})=F(\{a_{i}\})$(hint: using Vandermonde Determinant). Now the proof is complete.

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