Is $N_{k\subset K}$ the only *norm* on the field extension $k\subset K$? In several examples of field extensions the norm function is very useful. For instance, in $\mathbb{Q}\subset\mathbb{Q}(\sqrt{2})$, the norm is $N(x+y\sqrt{2})=x^2-2y^2$. In $\mathbb{R}\subset\mathbb{C}$, $N(x+iy)=x^2+y^2$.
These norms are useful mainly because they satisfy 1) $N(\alpha\beta)=N(\alpha)N(\beta)$ 2)$N(1)=1$. 
Let $k\subset K$ be a finite field extension of degree $n$, then it is not difficult to see that $K$ corresponds to a field in $M_{n}(k)$ via \begin{equation}
\alpha\in K\mapsto M_{\alpha}\in M_n(k),
\end{equation} where $M_\alpha$ is the multiplication-by-$\alpha$ operator. 
However it was a real surprise to see (e.g. VII.1 of Algebra: Chapter 0) that 
the determinant 
\begin{equation}
N_{k\subset K}(\alpha)=\det(M_\alpha)
\end{equation}  is exactly the familiar norms. 
Since we know determinant function is characterized by multiplicity, linearity in rows, and equalling $1$ at $I$, it seems natural to ask whether 1) and 2) would characterize $N_{k\subset K}$.   
So my question is

Is $\alpha\mapsto\det(M_\alpha)$ the only function on $F$ that satisfies both 1) and 2)?

On the one hand this seems wrong as we need three axioms to characterize $\det$ but now we only have two. On the other hand, however, we only need to define norm on an $n$ dimensional space while $\det$ is defined on an $n^2$ dimensional space.
Can someone give a hint?
Thanks very much!
 A: Let $k \subset K$ be an extension of fields. A function $F: K \rightarrow k$ satisfying your requirements is a homomorphism of $K^\times$ to $k^\times$ when restricted to $K^\times$. In particular given a homomorphism $\psi:K^\times\rightarrow k^\times$ we can extend it to a function satisfying 1) and 2) by letting $\psi(0)=0$. 
So for instance we could take the trivial homoorphism between $K^\times$ and $k^\times$. Or we could compose the norm map with a suitable automorphism of $k$ to get another such function. For instance if we considered the norm on the extension $\mathbb C(x,y) \subset \mathbb C(\sqrt{x},y)$ and composed the norm with the automorphism switching $x$ and $y$. 
A: Every element of ${\bf Z}[\sqrt2]$ has a unique expression as a product of primes. For each prime $p$ define $N(p)$ however you like, then extend to the whole ring by multiplicativity. Then every element of ${\bf Q}(\sqrt2)$ is $r\alpha$ for some rational $r$ and some $\alpha$ in ${\bf Z}[\sqrt2]$; define its norm to be $r^2N(\alpha)$. 
