= Projective transitivity of a trinomial linearized polynomial
{title2=$L(X)=X^{q^d}+X^q+TX$}
For $d\geq2$, 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.
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