Field norm - Imported Knowlpedia Content

Let $L/K$ be a finite field extension of degree $n=[L:K]$ . For $\alpha\in L$ , consider the $K$ -linear map $m_\alpha:L\to L$ , $x\mapsto \alpha x$ . The (field) norm of $\alpha$ from $L$ to $K$ is

\mathrm{N}_{L/K}(\alpha) := \det(m_\alpha)\in K.

Remarks

If $L/K$ is separable (see separable extension) and $\Omega$ contains $L$ , then

\mathrm{N}_{L/K}(\alpha)=\prod_{\sigma} \sigma(\alpha),

where the product runs over all $K$ -embeddings $\sigma:L\hookrightarrow \Omega$ . The norm is multiplicative, $\mathrm{N}_{L/K}(\alpha\beta)=\mathrm{N}_{L/K}(\alpha)\mathrm{N}_{L/K}(\beta)$ , and satisfies the tower property in trace/norm in towers for a tower of fields $K\subseteq M\subseteq L$ .

Examples

Quadratic extension. Let $L=K(\sqrt{d})$ with $\mathrm{char}(K)\neq 2$ . For $\alpha=a+b\sqrt{d}$ ,

\mathrm{N}_{L/K}(\alpha)=(a+b\sqrt{d})(a-b\sqrt{d})=a^2-b^2d.

Norm via minimal polynomial. If $L=K(\alpha)$ is a simple extension and the minimal polynomial of $\alpha$ over $K$ is $m_\alpha(x)=x^n+c_{n-1}x^{n-1}+\cdots+c_0$ , then

\mathrm{N}_{L/K}(\alpha)=(-1)^n c_0.

(Here $c_0$ is the constant term.)

Finite fields. For $L=\mathbb{F}_{q^n}$ over $K=\mathbb{F}_q$ ,

\mathrm{N}_{L/K}(\alpha)=\alpha\cdot \alpha^{q}\cdot \alpha^{q^2}\cdots \alpha^{q^{n-1}}=\alpha^{(q^n-1)/(q-1)}.

This is compatible with the cyclic structure of the multiplicative group (see finite-field multiplicative group is cyclic).