Norm map over Galois extension

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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• 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...
– lee
Aug 7 '14 at 5:56
• "is of the form" is unappropriate here, since your $N$ depends on $L$. You should simply say something like "any intermediate $L$ satisfies $L=F(\lbrace N_{K/L}(a) |a \in K\rbrace)$". Aug 7 '14 at 8:11

1 Answer

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.