Here is a nice *characterization* of the norm mapping on a finite extension of fields $K/k$.

If $K/k$ is any finite extension of fields with degree $n$, then the norm mapping from $K$ to $k$ is the unique function $f \colon K \rightarrow k$ satisfying the following three conditions:

1) $f(xy) = f(x)f(y)$ for all $x$ and $y$ in $K$.

2) $f(c) = c^n$ for all $c$ in $k$.

3) $f$ is a polynomial function over $k$ of degree at most $n$, by which I mean there is a basis $\{e_1,\dots,e_n\}$ of $K/k$ relative to which $f$ can be described by a polynomial: there's a polynomial $P(x_1,\dots,x_n)$ in $k[x_1,\dots,x_n]$ such that $f(\sum_{i=1}^n c_ie_i) = P(c_1,\dots,c_n)$ for all $c_1,\dots,c_n$ in $k$. (Being a polynomial function is independent of the choice of basis.)

This is due to Harley Flanders. See the following two articles of his:

The Norm Function of an Algebraic Field Extension, Pacific J. Math 3 (1953), 103--113.

The Norm Function of an Algebraic Field Extension, II, Pacific J. Math 5 (1955), 519--528.

One nice consequence of this characterization of the norm, which Flanders points out, is that it gives a slick proof of the transitivity of the norm: if $K \supset F \supset k$ then the function
${\rm N}_{F/k} \circ {\rm N}_{K/F}$ satisfies the three conditions that characterize ${\rm N}_{K/k}$.