TITLE: To prove that the norm of a tower of field extensions is the composition of norms
QUESTION [6 upvotes]: We know that if we set $K$, $F$ and $L$ fields, with $L$ a finite extension of $F$ and $F$ a finite extension of $K$. Then we have the norm equality $$N_{L/K}=N_{F/K}\circ N_{L/F}$$ The common proof is by using rational canonical forms and elementary matrices. I was wondering if there is a way only by using the tools of field extensions. Thank you.

REPLY [2 votes]: We prove the theorem for separable field extensions $F$ over $K$ and $L$ over $F$. By the theorem of the primitive element in field theory there exists a monic irreducible polynomial $f(x)=x^m+a_{m-1}x^{m-1}+\dots+a_0$ such that
$$F=K[x]:=K[X]/\langle f(X)\rangle.$$
Similarly there exists a monic irreducible polynomial $g(y)=y^n+b_{n-1}y^{n-1}+\dots+b_0$ with coefficients $b_k\in F$ such that
$$L=K[x,y]=F[y]:=F[Y]/\langle g(Y)\rangle.$$
The field $L$ has dimension $m\cdot n$ over the field $K$. By the theorem of the primitive element there exists a monic irreducible polynomial $h(z)=z^{mn}+c_{mn-1}z^{m-1}+\dots+c_0$ with coefficients $c_k\in K$ such that
$$L=K[z]:=K[Z]/\langle h(Z)\rangle.$$
Every zero $z_i$ of the polynomial $h(z)$ produces an isomorphic field
$$L\cong L_i=K[z_i]:=K[Z_i]/\langle h(Z_i)\rangle$$
and therefore there exists an isomorphism $\hat\sigma_i$ such that $L_i=\hat\sigma_i(L)$ with $z_i=\sigma_i(z)$. But because the field $L$ is also generated by adjoining the zeros $x\in K$ and $y\in F$
$$L=K[x,y]$$
this homomorphism $\hat\sigma_i$ must send the zero $x$ to a zero $x_i:=\hat\sigma_i(x)$ of the polynomial $f(x)$ and the zero $y$ to a zero $y_{ik}:=\hat\sigma_i(y)$ of the polynomial 
$$g_i(y):=\hat\sigma_i(g(y))=y^n+\hat\sigma_i(b_{n-1})y^{n-1}+\dots+\hat\sigma_i(b_0).$$
For every $i$ there are $n$ zeros $y_{ik}$ because the polynomial $g_i(y)$ remains irreducible over the field $L_i$. We define the homomorphism $\sigma_{ik}$ over the field $L$ to be the homomorphism that sends the zero $x$ to the zero $x_i$ and the zero $y$ to the zero $y_{ik}$ that was adjoined over the field $K[x_i]$:
$$L_{ik}:=K[x_i,y_{ik}]=\sigma_{ik}(K[x,y]).$$
In adjoining the zero $y_{ik}$ to the fields $K[x_i]$ there exists $n$ homomorphism $\tau_{ik}$ such that
$$L_{ik}:=K[x_i,y_{ik}]=\tau_{ik}(K[x_i,y_{i0}])$$
and the field $F=K[x_i]$ remains invariant. Now we have developed the means to prove the formula in the question. The norm of an element $\zeta\in L$ is defined to
$$\tag{1}N_{L/K}(\zeta):=\prod_{0\le i\lt mn}\hat\sigma_{i}(\zeta)=\prod_{0\le i\lt m}\prod_{0\le k\lt n}\sigma_{ik}(\zeta).$$
We reorder the zeros such that $x_0=x$. Let the homomorphism $\sigma_{0,k}$ be the identity on the zero $x$. Each factor in the product $\prod_{0\le k\lt n}\sigma_{0k}(\zeta)$ contains exactly one of the zeros $y_{0,k}$ of the polynomial $g(y)$ that can be adjoined above to the field $F$. But then this product must be a symmetrical polynomial in the zeros $y_{0,k}$ and we can write the product to
$$\tag{2}\prod_{0\le k\lt n}\sigma_{0k}(\zeta)=d_mx^m+d_{m-1}x^{n-1}+\dots+d_0$$
with coefficients $d_j\in K[x,S_{00},\dots,S_{0m}]$ with the symmetrical polynomials
$$S_{00}:=1, S_{01}:=y_{01}+\dots+y_{0n}, \dots, S_{0n}:=y_{01}\cdots y_{0n}$$
and $y_{0k}:=\tau_{0k}(y)$. But these symmetric polynomials are known to be $S_{0k}=(-1)^kb_k\in F=K[x]$ of the polynomial $g(y)$ above. Therefore the product $(2)$ gives an element in the field $F=K[x]$.
In taking the product $\prod_{0\le k\lt n}\sigma_{ik}(\zeta)$ for a fixed $i$ we get a similar  result to equation $(2)$ with 
$$\tag{3}\prod_{0\le k\lt n}\sigma_{ik}(\zeta)=\sigma_{ik}(d_m)x_i^m+\sigma_{ik}(d_{m-1})x_i^{n-1}+\dots+\sigma_{ik}(d_0)$$
with coefficients $\sigma_{ik}(d_j)\in K[x_i,\sigma_{ik}(S_0),\dots,\sigma_{ik}(S_m)]$. But those are just the coefficients of the poynomial $g_i$ or $S_{ik}=\sigma_{ik}(S_{0k})=(-1)^k\sigma_{ik}(b_k)\in K[x_i]$.
The homomorphism $\sigma_{ik}$ of the field $L$ over $K$ can be restricted to the homomorphism $\kappa_i$ of the field $K[x_i]$ over the field $K$. But these homomorphism $\kappa_i$ just map the the zero $x$ to the zero $x_i$ and therefore they are the field homomorphism of the field $F=K[x_i]$ over the field $K$. In the formulas $(1)$ and $(3)$ we have seen that the inner product over $k$ gives an element in the field $K[x_i]$. Because the formulas $(2)$ and $(3)$ give
$$\prod_{0\le k\lt n}\sigma_{ik}(\zeta)=\kappa_i\left(\prod_{0\le k\lt m}\sigma_{0k}(\zeta)\right)$$
we can rewrite the formula $(1)$  to
$$N_{L/K}(\zeta)=\prod_{0\le i\lt m}\kappa_i\left(\prod_{0\le k\lt n}\sigma_{0k}(\zeta)\right).$$
But the homomorphism $\sigma_{0k}$ are the homomorphism $\tau_k:=\tau_{0,k}=\sigma_{0k}$ of the field $L$ over the field $F$. This gives directly
$$N_{L/K}(\zeta)=N_{F/K}\circ N_{L/F}(\zeta)=\prod_{0\le i\lt m}\kappa_i\left(\prod_{0\le k\lt n}\tau_{k}(\zeta)\right).$$