Cohn's Irreducibility criterion in function fields over finite field

Question

Generalizing Cohn's irreducibility criterion for Function fields over finite fields.

Let $\mathbb{F}_q$ denote the field with $q$ elements. Then we denote its field of rational functions by $\mathbb{F}_q(t)=K$. We define the norm in this field by setting $$ \left| \frac{f(t)}{g(t)} \right| = q^{\operatorname{deg} \ f \ - \ \operatorname{deg} \ g} $$ where $f(t)$ and $g(t)$ are polynomials with coefficients in $\mathbb{F}_q$. The defined norm has the following properties:

1. For all $\beta \in K$; $|\beta| \ge 0$.

2. For all $\beta, \gamma \in K$; $|\beta\gamma|=|\beta| \ |\gamma|$

3. For all $\beta, \gamma \in K$; $|\beta+\gamma| \le |\beta| + |\gamma|$

Proof for these are given here.

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Here is lemma which tells something about the roots of a polynomial in $K[X]$.

Lemma 1: Let $f(x)=a_m(t)x^m+a_{m-1}(t)x^{m-1}+\cdots+a_1(t)x+a_0(t)$ be a polynomial in $K[X]$ of degree $m$ and have $\alpha(t) \in \bar{K}$ as a root. Then $$ |\alpha(t)|<H+1 \ \ \text{where} \ \ H:=\max\limits_{0 \le i \le m-1} \left| \frac{a_i(t)}{a_m(t)}\right| $$

Proof: Since $\alpha(t)$ is a root of $f(x)$ we have $$ -a_m(t)(\alpha(t))^m= a_{m-1}(t)(\alpha(t))^{m-1}+\cdots+a_1(t)\alpha(t)+a_0(t) $$ $$ \implies - (\alpha(t))^m= \frac{a_{m-1}(t)}{a_m(t)}\alpha(t)^{m-1}+\cdots+ \frac{a_1(t)}{a_m(t)}\alpha(t)+\frac{a_0(t)}{a_m(t)} $$ $$ \implies |\alpha(t)|^m \le H \ \left(|\alpha(t)|^{m-1}+\cdots+|\alpha(t)|+1 \right) = H \left(\frac{|\alpha(t)|^m-1}{|\alpha(t)|-1} \right) $$

If $|\alpha(t)| \le 1$ then $|\alpha(t)| < H+1$. If $|\alpha(t)| > 1$ then we have $$ |\alpha(t)|^m \left( |\alpha(t)|-1\right) \le H \left(|\alpha(t)|^m\right) - H < H \left(|\alpha(t)|^m\right) $$ $$ \implies |\alpha(t)| < H+1 $$

Please let me know if the proof has any flaws?

Second Part:

If we fix a polynomial $b(t) \in \mathbb{F}_q[t]$, of positive degree, then an irreducible polynomial $p(t)$ can be "written in base b(t)" via the Euclidean algorithm: $$ p(t)=a_m(t)b(t)^m+\cdots+a_1(t)b(t)+a_0(t) $$ where $a_i(t) \in \mathbb{F}_q[t] \subset K$.

Note-1: Because of the division algorithm we can tell that $$ 0 \le \operatorname{deg} \ a_i(t) \le \operatorname{deg} \ b(t)-1 $$ $$ \implies \ \ \text{the maximum value of H in this polynomial is} \ \ q^{\operatorname{deg} \ b(t)-1} $$

In the analogy with the theorem of Cohn. We can inquire if the polynomial $$ f(x)=a_m(t)x^m+\cdots+a_1(t)x+a_0(t) $$ is irreducible in $K[X]$.

Let $g(x)$ be a nontrivial factor of $f(x)$.

Note-2: If $g(x)$ be a nontrivial factor of $f(x)$, then $g(b(t))$ is also a nontrivial factor of $f(b(t))=p(t)$.

By factoring the polynomial over the algebraic closure $\bar{K}$ of $K$ we get $$ g(x)=c(t)\prod_i(x-\alpha_i(t)) $$ where $\alpha_i(t) \in \bar{K} \ , \ c(t) \in \mathbb{F}_q[t]$; since $c(t)=a_m(t)$ (the leading coefficient of $f(x)$) $$ \implies |c(t)| = |a_m(t)| = q^{\operatorname{deg} \ a_m(t)} \ge 1 $$

Therefore

$$ |g(x)|=\left|c(t)\prod_i(x-\alpha_i(t))\right| \ge \left| \prod_i(x-\alpha_i(t)) \right| \ge \prod_i(|x|-|\alpha_i(t)|) $$

$$ \implies |g(b(t))| \ge \prod_i(|b(t)|-|\alpha_i(t)|) > q^{deg \ b(t)} - (H+1) \ge q^{deg \ b(t)} - (q^{deg \ b(t)-1}+1) \ge 1 $$

Unless $q=2$ and $deg \ b(t)=1$.

which tells that $g(b(t))$ is a nontrivial factor of $f(b(t))=p(t)$. Contradiction!

If $deg \ b =1$ then $f(x)$ is the same as $p(t)$ after a linear change of variable.

Hence $f(x)$ must be irreducible.

My Questions:

1. Here the norm is defined on $K$, how do I extend this norm to $\bar{K}?$

2. Are the above arguments correct?

3. Where are we using the condition "finiteness of field"?