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This may be trivial, but it is not clear to me.

Let $F=\mathbb F_q$ be a finite field and $L=\mathbb F_{q^m}$ be its finite extension of degree $m$. Let $Gal(L,F)$ denote Galois group of $L$ over $F$. Now, extend the standard norm to a map $N_{L/F}$ from $L[x]$ to $F[x]$ as below: $$ N_{L/F}: L[x] \mapsto F[x]$$ such that $$ N_{L/F}(h(x))= \prod_{\sigma \in Gal(L,F)} \sigma(h(x))$$, where $\sigma (x) =x$ for all $\sigma \in Gal(L,F)$.

Why $N_{L/F}(h(x)) \in F[x]$?

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It's trivial enough: the norm polynomial is fixed under the ring automorphisms from the Galois group, so the coefficients are in F. – Charles Matthews Feb 8 2011 at 12:26
Thanks Charles. It's clear to me now. – Sartaj Ul Hasan Feb 8 2011 at 18:03

closed as too localized by Charles Matthews, Alex Bartel, Franz Lemmermeyer, Pete L. Clark, George Lowther Feb 8 2011 at 20:22

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