Past exam of the mathematics course of the University of Cambridge / 2002 / iii / Paper 3 / 5

The PDF has $q$-power exponents; the polynomial is a linearized polynomial over a finite field, not the degree-$d$ polynomial produced by the converted TeX. Let $V$ be its root set in an algebraic closure. Because $(x+y{)}^{q^i}=x^{q^i}+y^{q^i}$ and $c^{q^i}=c$ for $c\in {\mathbb{F}}_q$,Thus $V$ is an ${\mathbb{F}}_q$-vector space. Its derivative is $a_0\ne 0$, and its degree is $q^d$, so it has exactly $q^d$ distinct roots. A finite vector space of that size has dimension $d$.

Every splitting-field automorphism fixes ${\mathbb{F}}_q$ and carries roots to roots. Hence it acts linearly on $V$. The action is faithful because the splitting field is generated by the roots. Choosing a basis identifies this injection withDifferent bases conjugate the matrix subgroup. The linear action itself is canonical, while a particular matrix realization is basis-dependent.

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.

Write $d=2n$ with $n\ge 2$, and let $E$ be the splitting field. The root set of $X^{q^{2n}}+X^{q^2}+tX$ is an $n$-dimensional vector space over ${\mathbb{F}}_{q^2}$, by the argument of part (a) with $q$ replaced by $q^2$.

The constants ${\mathbb{F}}_{q^2}$ really occur inside $E$, rather than only in the ambient algebraic closure. Choose a nonzero root $\alpha$. For every $c\in {\mathbb{F}}_{q^2}$, $c\alpha$ is also a root, so both elements lie in $E$ andLet $K={\mathbb{F}}_{q^2}(t)$. Since $K/{\mathbb{F}}_q(t)$ is Galois of degree two,This index statement does not require proving that $E$ has no still larger constant field.

As an extension of $K$, $E$ is the splitting field of the same polynomial. Applying part (b) with base constant size $q^2$ and dimension $n$ givesThe special linear group is the kernel of determinant in the general linear group, so every element of $H$ normalizes it. Therefore $H$ is the required normal index-two subgroup normalizing $SL(d/2,q^2)$.

In fact the full group is semilinear over ${\mathbb{F}}_{q^2}$: an element outside $H$ acts on constants as $c\mapsto c^q$ and satisfies $g(cv)=c^{q}g(v)$. Field automorphisms preserve determinant one, so this also explains the normalizer behavior geometrically. This is the quadratic constant extension of a q-squared linearized splitting field, not the polynomial of part (b) with its $X^q$ term unchanged.