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Put $N=(q^d-1)/(q-1)$, so $N>1$ and $N\equiv 1$ in the characteristic. For a nonzero root $x$, division by $x$ givesThe projective point ${\mathbb{F}}_{q}x$ is therefore represented by $s=x^{q-1}$, satisfyingTwo nonzero roots give the same $s$ exactly when their ratio lies in ${\mathbb{F}}_q^{\times }$. Thus the $N$ projective points are precisely the roots of $P_t$. The polynomial $s^N+s+t$ is irreducible in ${\mathbb{F}}_q[s,t]$, since it is linear in $t$ and its quotient ring is ${\mathbb{F}}_q[s]$. Gauss's lemma makes it irreducible over ${\mathbb{F}}_q(t)$, so the action on projective points is transitive.

Fix one such coordinate $s$, so $t=-s^N-s$. The other coordinates satisfyWe prove its irreducibility even over ${\overline{\mathbb{F}}}_q(s)$. Make the birational change $z=Y/s$, $h=1/s$. Multiplication by $h^{N-1}$ transforms the equation intoChoose a nontrivial $N$th root of unity $\zeta$ over the algebraic closure. Since the characteristic does not divide $N$, $B$ has a simple zero at $\zeta$. Viewed as a polynomial in $h$ over ${\overline{\mathbb{F}}}_q[z]$, $h^{N-1}+B(z)$ is Eisenstein at the prime $z-\zeta$: the leading coefficient is one, every intermediate coefficient is zero, and the constant term is divisible by that prime exactly once. It is therefore irreducible.

The birational substitution preserves irreducibility after inverting $s$ or $h$. No factor supported solely on $s=0$ is lost, because $R(Y,0)=Y^{N-1}+1$ is nonzero. Gauss's lemma then proves that $R(Y,s)$ is irreducible over ${\overline{\mathbb{F}}}_q(s)$ and hence over ${\mathbb{F}}_q(s)$. Its roots are distinct: a repeated root would have $Y^{N-1}+1=0$ as well as $Y^N+Y+t=0$, forcing $t=0$, impossible generically.

The stabilizer of the chosen projective point is $\mathrm{Gal}(E/{\mathbb{F}}_q(s))$, and the irreducibility of $R$ makes it transitive on all the remaining projective points. This proves projective transitivity of a trinomial linearized polynomial, namely two-transitivity on the one-dimensional subspaces. The supplied group-theoretic result now applies to the linear subgroup from part (a), givingThe Eisenstein argument remains valid even though $N-1$ is divisible by the characteristic; it concerns irreducibility in $h$, not separability of that projection. Separability of the projective polynomial has been checked independently.