Applications of additive version of Hilbert's theorem 90 Additive version of Hilbert's theorem 90 says that whenever $k \subset F$ is cyclic Galois extension with Galois group generated by $g$,  and $a$ is element of $L$ with trace 0, there exists an element $b$ of $L$ such that $a = b - g(b)$.
The corresponding multiplicative version with norm instead of trace has many interesting applications, for instance determining Pythagorean triples, or solving Pell's equation. How about the additive version? Where it can be useful?
 A: Google Artin-Schreier theory, an additive analogue of Kummer theory. In particular, the proof of the Artin-Schreier theorem about non-algebraically closed fields whose algebraic closure is a finite extension is a beautiful application (when treating the case of fields with characteristic $p$).
Quite generally, these additive theorems are useful in characteristic $p$ when dealing with extensions of degree $p$ or more generally a power of $p$, depending on the situation. If you are not a fan of characteristic $p$ then the additive Hilbert Theorem 90 will not be your friend.
A: 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.

