Norm map

The mapping $\def\N{N_{K/k}}\N$ of a field $K$ into a field $k$, where $K$ is a finite extension of $k$ (cf. Extension of a field), that sends an element $\def\a{\alpha}\a\in K$ to the element $\N(\a)$ that is the determinant of the matrix of the $k$-linear mapping $K\to K$ that takes $x\in K$ to $\a x$. The element $\N(\a)$ is called the norm of the element $\a$.

One has $\N(\a) = 0$ if and only if $\a = 0$. For any $\def\b{\beta}\a,\b\in K$,

$$\N(\a\b) = \N(\a)\N(\b),$$ that is, $\N$ induces a homomorphism of the multiplicative groups $K^*\to k^*$, which is also called the norm map. For any $\a\in k$,

$$\N(\a)=\a^n,\ \ \textrm{ where } n=[K:k].$$ The group $\N(K^*)$ is called the norm subgroup of $k^*$, or the group of norms (from $K$ into $k$). If $f(x)=x^n+a_{n-1}x^{n-1}+\cdots+a_0$ is the characteristic polynomial of $\a\in K$ relative to $k$, then

$$\N(\a) = (-1)^n\a_0.$$ Suppose that $K/k$ is separable (cf. Separable extension). Then for any $\a\in K$,

$$\N(\a) = \prod_{i=1}^n\sigma_i(\a),$$ where the $\sigma_i$ are all the isomorphisms of $K$ into the algebraic closure $\bar k$ of $k$ fixing the elements of $k$.

The norm map is transitive. If $L/K$ and $K/k$ are finite extensions, then

$$N_{L/k}(\a)=\N(N_{L/K}(\a))$$ for any $\a\in L$.

References

[BoSh]	Z.I. Borevich, I.R. Shafarevich, "Number theory", Acad. Press (1966) (Translated from Russian) (German translation: Birkhäuser, 1966) MR0195803 Zbl 0145.04902
[La]	S. Lang, "Algebra", Addison-Wesley (1984) MR0799862 MR0783636 MR0760079 Zbl 0712.00001