Linearized polynomial over a finite field ($L(X)={\sum }_{i=0}^d{a}_i{X}^{q^i}$)

Over a field containing ${\mathbb{F}}_q$, a sum of $q$-power monomials is ${\mathbb{F}}_q$-linear. If its leading coefficient and linear coefficient are nonzero, its root space has exactly $q^d$ elements and dimension $d$ over ${\mathbb{F}}_q$. Automorphisms of its splitting field act faithfully and linearly on that root space, embedding the Galois group in $GL(d,q)$.

For even $d=2n$, the roots of $X^{q^d}+X^{q^2}+TX$ form an $n$-dimensional vector space over ${\mathbb{F}}_{q^2}$. Ratios $(c\alpha )/\alpha$ for a nonzero root $\alpha$ show that the splitting field contains every $c\in {\mathbb{F}}_{q^2}$. The subgroup fixing that quadratic constant extension is normal of index two and lies in $GL(n,q^2)$; applying projective transitivity over ${\mathbb{F}}_{q^2}(T)$ makes it contain $SL(n,q^2)$. It therefore normalizes that special linear group, while the full group acts semilinearly.

For $d\ge 2$, nonzero roots define projective-point coordinates $s=X^{q-1}$, satisfying $s^N+s+T=0$, where $N=(q^d-1)/(q-1)$. This polynomial is irreducible over ${\mathbb{F}}_q(T)$. After fixing one root $s$, the remaining coordinates satisfy $R(Y,s)=(Y^N-s^N)/(Y-s)+1$. Substitute $z=Y/s$ and $h=1/s$: the equation becomes $h^{N-1}+(z^N-1)/(z-1)=0$. At a nontrivial $N$th root of unity the second term has a simple zero, so the polynomial is Eisenstein over the algebraic closure's rational function field. This proves irreducibility of $R$ and hence two-transitivity on projective points.