An example that first comes to my mind deals with quadratic equations over a finite field $\Bbb{F}_{2^n}$ of characteristic two. There the additive Hilbert 90 says that
$$
x^2+x=a
$$
with $a\in \Bbb{F}_{2^n}$ has a solution (obviously then two solutions) in $\Bbb{F}_{2^n}$, if and only if $tr(a)=0$. This reinterpretation comes from the following observations:
- Squaring, i.e. the Frobenius automorphism $F$, is a generator of the Galois group $Gal(\Bbb{F}_{2^n}/\Bbb{F}_2)$.
- $F(x)-x=x^2-x=x^2+x$ for all $x\in \Bbb{F}_{2^n}$.
This leads to solvability criteria of a general quadratic over $\Bbb{F}_{2^n}$:
$$
x^2+bx+a=0\qquad(*)
$$
with $a,b\in\Bbb{F}_{2^n}$ has solutions in $\Bbb{F}_{2^n}$, iff $tr(a/b^2)=0$ -
divide $(*)$ by $b^2$, and write it in terms of the new variable $y=x/b$). Note that
the usual trick of completing the square is unavailable in characteristic two. Also note that $(*)$ has a double root in $\Bbb{F}_{2^n}$, if $b=0$.
Admittedly this example is not very satisfying here, because:
- we can derive this result without knowing about Hilbert 90, and
- it's not about algebraic number fields.