TITLE: On a formula of the norm of an element of a finite extension of a field
QUESTION [7 upvotes]: Theorem
Let $F$ be a field.
Let $K$ be a finite extension of $F$.
Let $[K : F]_i$ be the inseparable degree of $K/F$.
Let $\bar{K}$ be an algebraic closure of $K$.
Let $S$ be the set of $F$-embeddings of $K$ into $\bar{K}$.
Let $\alpha \in K$.
Then $N_{K/F}(\alpha) = (\prod_{\sigma \in S}\sigma(\alpha))^{[K : F]_i}$
This is proved in Theorem 60 in page 39 of the lecture note written by Pete L. Clark.
I don't understand the proof.
Would any one please enlighten me?
EDIT
Why down votes??
Asking for help to understand a proof should be frowned upon?
EDIT
I understand the prerequisites for the proof, i.e. the content of section 6 and Corollary 58.
EDIT
Related question 1,
Related question 2
EDIT
Let $f(X)$ be the characteristic polynomial of $\alpha$.
Let $g(X)$ be the minimal polynomial of $\alpha$.
$By$ $ $ $Corollary$ $ $ 58, $f(X)$ = $g(X)^{[K:F(\alpha)]}$.
The set {$\sigma(\alpha); \sigma ∈ S$} is the set of the roots of g(X).
However, it is not clear to me that the equation $N_{K/F}(\alpha) = (\prod_{\sigma \in S}\sigma(\alpha))^{[K : F]_i}$ follows immediately.
Would anyone please explain why the equation follows other than just saying it is straightforward?
EDIT
It's amazing that some people regard questioning a proof as an attack to the author's credibilty. Everybody makes a mistake. Even Grothendieck made a non-trivial mistake(see Misconceptions About $K_{X}$ by Kleiman).
I think this attitude is harmful to healthy development of mathematics. 
I'm not claiming that the proof is wrong, though.
EDIT
To anyone who thinks the proof is straightforward, please explain to me in detail.
I am not as smart as you.
EDIT
It's surprising that no one explained the proof so far.
I believe every step of any proof can be reduced to (really) trivial statements.
If you think it's straightforward, please reduce it to more trivial statements that anybody who has basic knowledge of abstract algebra can understand.

REPLY [3 votes]: Well the theorem is a bit too hastily stated, for example if you happen to pick $\alpha \in F$, then $f(t) = (t-\alpha)^n$ and has only one distinct root.
The correct statement is :
Let $K/F$ be a field extension of degree $n < \infty$ and separable degree $m$. Let $\overline{K}$ be an algebraic closure of $K$. Let $\alpha \in K$ and let $f(t)$ be the characteristic polynomial of $\alpha \bullet \in End_F(K)$. Let $\sigma_1 \ldots \sigma_m$ be the distinct $F$-algebra embeddings of $K$ into $\overline{K}$. Then the factorisation of $f(t)$ on $\overline{K}$ is $f(t) = \prod_{i=1}^m (t- \sigma_i(\alpha))^{n/m}$.
First, let $K_0 = F(\alpha)$, $n_0 = [K_0 : F]$, $m_0$ be the separable degree of $[K_0 : F]$, $\tau_1 \ldots \tau_{m_0}$ be the embeddings of $K_0$ into $\overline{K}$, and $f_0(t)$ be the minimal polynomial of $\alpha$ over $F$.
Then, by corollary 58, $f(t) = f_0(t)^{[K : K_0]} = f_0(t)^{n/n_0}$.
According to the proof of theorem 38, each $\tau_i$ is the restriction to $K_0$ of $m/m_0$ many embeddings $\sigma_j$, therefore
$\prod_{i=1}^m (t- \sigma_i(\alpha))^{n/m} = \prod_{i=1}^{m_0} (t- \tau_i(\alpha))^{n/m_0} $, so we only need to prove that $f_0(t) = \prod_{i=1}^{m_0} (t- \tau_i(\alpha))^{n_0/m_0}$.
From now, suppose we are in the case $K = K(\alpha)$. Let $K_s$ be the separable closure of $F$ in $K$, so by Proposition 50, $K/K_s$ is purely inseparable and $K_s/F$ is separable. By corollary 47, there is an integer $a\ge 0$ such that $K_s = K(\alpha^{p^a})$.
Let $g(t)$ be the minimal polynomial of $\alpha^{p^a}$ over $F$.
Then $f(t) = g(t^{p^a})$, because $g(\alpha^{p^a}) = 0$ and they have the same degree $n = p^a m$.
Meanwhile, $(t - \sigma_i(\alpha))^{p^a} = (t^{p^a} - \sigma_i(\alpha^{p^a}))$ so we only need to prove that $g(t) = \prod_{i=1}^m (t - \sigma_i(\alpha^{p^a}))$.
So we only need to prove the theorem when $\alpha$ is separable.
But then this is clear : the $\sigma_i(\alpha)$ are roots of $f(t)$ because $\sigma_i$ are embeddings of $F$-algebras, and every factor in the product is distinct, so the product has to divide $f(t)$, and both polynomials have the same degree, so they must be equal.