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Let $k = \mathbb{F}_{p^n} = \mathbb{F}_q$ finite field of $q = p^n$ and $[K:k]=2$ Galois extension of degree 2. Then $K = \mathbb{F}_{q^2} = \mathbb{F}_{(p^n)^2} = \mathbb{F}_{p^{2n}}$. It is generated by $\sigma : x \mapsto x^q$ the Frobenius automorphism. I want to calculate how many elements there are in $K^\times = K - \{ 0 \}$ such that $N_{K/k}(x) = \sigma(x)x = 1$.

As a numerical example let $p=2$ and $n=10$, then $k = \mathbb{F}_{2^{10}}$ and $K = \mathbb{F}_{2^{20}}$.

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By Hilbert 90, the kernel of $N_{K/k}: K^\times \to k^\times$ is the image of the homomorphism $\phi: K^\times \to K^\times, x \mapsto \sigma(x)/x$. The kernel of $\phi$ is $\{x \in K^\times : \sigma(x) = x \} = k^\times$, so we have an exact sequence $$1 \longrightarrow k^\times \longrightarrow K^\times \overset{\phi}{\longrightarrow} \operatorname{ker}N_{K/k} \longrightarrow 1.$$ Therefore, $$|\operatorname{ker}N_{K/k}| = \frac{|K^\times|}{|k^\times|} = \frac{q^2-1}{q-1} = q+1.$$

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  • $\begingroup$ And this shows that the norm is surjective, right? $\endgroup$ – awllower Feb 7 '13 at 15:29
  • $\begingroup$ Yes, by $|\operatorname{im} N_{K/k}| = |K^\times| / |\operatorname{ker} N_{K/k}| = (q^2-1)/(q+1) = q-1 = |k^\times|$. $\endgroup$ – marlu Feb 7 '13 at 15:48
  • $\begingroup$ Hmm. Wasn't surjectivity of the norm map already included in that exact sequence? $\endgroup$ – Jyrki Lahtonen Feb 7 '13 at 21:48
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    $\begingroup$ I don't think so. The exactness at the right only says that $\phi: K^\times \to \operatorname{ker} N_{K/k}$ is surjective (which we know from Hilbert 90). $\endgroup$ – marlu Feb 7 '13 at 23:03
  • $\begingroup$ But from the sequence can we infer that $|k^*|=|K^*|/|KerN_{K/k}|=|ImN_{K/k}|$, so the norm is surjective. $\endgroup$ – awllower Feb 8 '13 at 3:39
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A different approach is given by the following argument (not using Hilbert 90). Pick an element $\gamma\in K\setminus k$. Let $z$ be an arbitrary element of $k$. Consider $$ x=\frac{z+\gamma}{z+\gamma^q}\in K. $$ We have $$ \sigma(x)=\frac{\sigma(z)+\sigma(\gamma)}{\sigma(z)+\sigma(\gamma^q)}=\frac{z+\gamma^q}{z+\gamma}=\frac1x. $$ Therefore $N(x)=1$. It is easy to show that different choices of $z$ yield different elements $x$, so there are at least $q$ elements in $\mathrm{ker} N_{K/k}$. Furthermore, obviously $1\in \mathrm{ker} N_{K/k}$ and $x=x(z)\neq1$ for all $z\in k$. Therefore $|\mathrm{ker} N_{K/k}|\ge q+1$. On the other hand $$ N_{K/k}(x)=x^{q+1}, $$ so there cannot be more than $q+1$ solutions to the equation $N_{K/k}(x)=1$.

Note that the construction is a cousin of the parametrization of points on the complex unit circle (other than $z=1$) by the recipe $$ z=\frac{x+i}{x-i},\quad x\in\mathbb{R}. $$

Admittedly Hilbert 90 also has a lot of appeal here.

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  • $\begingroup$ And in fact this argument works for any extension, not necessarily of degree $2$. A very nice answer and analogy indeed. $\endgroup$ – awllower Feb 8 '13 at 3:42
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    $\begingroup$ Using cyclicity of the multiplicative group of a finite field is surely the simplest way to settle this claim. However, using that feels like a broken record sometimes... $\endgroup$ – Jyrki Lahtonen Feb 10 '13 at 9:52

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