A note on Galois groups of linearized polynomials

Let $L(X)$ be a monic $q$-linearized polynomial over $F_q$ of degree $q^n$, where $n$ is an odd prime. Recently Gow and McGuire showed that the Galois group of $L(X)/X-t$ over the field of rational functions $F_q(t)$ is $GL_n(q)$ unless $L(X)=X^{q^n}$. The case of even $q$ remained open, but it was conjectured that the result holds too and partial results were given. In this note we settle this conjecture. In fact we use Hensel's Lemma to give a unified proof for all prime powers $q$.


Introduction
In this note we extend the main result [2, Theorem 2] from odd prime powers q to all prime powers.A q-linearized polynomial over F q of q-degree n is defined as L(X) = n i=0 a i X q i where a i ∈ F q and a n = 0.
Theorem 1.Let q be a prime power and n be an odd prime.Let L(X) be a monic q-linearized polynomial over F q of q-degree n.Then the Galois group of L(X)/X − t over F q (t) is GL n (q) unless L(X) = X q n .Using Hensel's Lemma we prove the following divisibility result for the order of the Galois group.Note that n need neither be odd, nor be a prime.Proposition 2. Let q be a prime power and n be a positive integer.Set L(X) = n i=m a i X q i where a i ∈ F q , a m , a n = 0 and 1 ≤ m ≤ n − 1.Then q m • (q m − 1) • (q n − 1) divides the order of the Galois group of L(X)/X − t over F q (t).
We show how Theorem 1 follows from Proposition 2. Upon replacing t with t − a 0 one may assume that a 0 = 0. Let G be the Galois group of L(X)/X − t over F q (t).It is well known and easy to prove that G ≤ GL n (q).Furthermore, ramification at t = ∞ tells that G contains a cyclic regular subgroup of order q n − 1. (See e.g.[5,Lemma 3.3] for a short proof.)A group theoretic classification result (see Remark 6) shows that G = GL n (q) or G ≤ ΓL 1 (q n ).The latter group has order n • (q n − 1).Suppose that G = GL n (q).Proposition 2 yields q m • (q m − 1) • (q n − 1) | n • (q n − 1).This implies q m | n, hence m = 1 and q = n because n is a prime.The contradiction n − 1 | 1 shows that there is no m with 1 ≤ m ≤ n − 1 such that a m = 0, hence we have the excluded case L(X) = X q n .

Proof of Proposition 2
In the following we let y be a variable over the field K and X be a variable over the field of rational functions K(y).
The proof of the following lemma uses Hensel's Lemma and the Eisenstein irreducibility criterion for polynomials with coefficients in a power series ring over a field.Lemma 3. Let f and g be non-constant polynomials over K. Suppose that (i) f has a root of multiplicity a, (ii) g has a root of multiplicity b, (iii) a and b are relatively prime, and Then a divides the order of the Galois group of f (X) − g(y) over K(y).
Proof.Let E be a common splitting field of f and g over K, and α and β be the roots of f and g of multiplicity a and b, respectively.As a and b are relatively prime, there are positive integers r and s such that sb − ra = 1.
Choose z with z s = y.Let E[[z]] be the ring of formal power series in z over E and E((z)) be its quotient field, which is the field of formal Laurent series in z.
By definition, K(y) is a subfield of E((z)), so the Galois group of f (X) − g(y) over E((z)) is a subgroup of the Galois group of f (X) − g(y) over K(y).
Thus we are done once we know that f (X)−g(y) has an irreducible factor over E((z)) of degree a. Upon replacing X and y with X + α and y + β we may and do assume that α = β = 0.
Thus f (X) = X a u(X) and g(y) = y b v(y) = z sb v(z s ) with u(0) = 0 = v(0).The degrees of the irreducible factors of f (X) − g(y) over E((z)) don't change if we replace X with Xz r .Recall that sb−ra = 1, so we are concerned with the factorization of As X a and u(Xz r ) are relatively prime, Hensel's Lemma provides a factorization H(X) = A(X)U(X) over E[[z]] such that A(X) ≡ X a (mod z).Furthermore, A(0)U(0) = H(0) = −zv(z s ), so z 2 does not divide A(0).Therefore A(X) is an Eisenstein polynomial of degree a with respect to z, and we are done.Remark 4. A variant of the proof would use standard facts about Newton polygons.Assume as above that f (X) = Z a u(X) and g(y) = y b v(y).Let ν be a valuation of the splitting field of f (X) − g(y) over E((y)) (E as above) which extends the y-adic valuation of E((y)).
The Newton polygon of f (X) − g(y) with respect to ν consists of two line segments.The first segment connects (0, b) with (a, 0), and the second segment connects (a, 0) with (n, 0).
From known properties of the Newton polygon, we get that f (X) − g(y) has exactly a roots γ with ν(γ) = b a .Let Γ be the set of these roots.Then γ∈Γ (X − γ) has coefficients in E((y)) and is irreducible over E((y)).Corollary 5. Let f ∈ K[X] be a polynomial of degree n for which two of its roots (in some extension of K) have relatively prime multiplicities a and b.Furthermore, assume that f (X)−t is separable over the field K(t) of rational functions in t.Then a • b • n divides the order of the Galois group of f (X) − t over K(t).
Proof.Let G be the Galois group of f (X) − t over K(t).Define y in an extension of K(t) by f (y Thus we need to show that a • b divides |H|. But this follows from Lemma 3 with f = g.First we use it with a root of multiplicity a of f (X) and a root of multiplicity b of f (y), and secondly with a root of multiplicity b of f (X) and a root of multiplicity a of f (y).
In order to prove Proposition 2, we are left to show that L(X)/X has roots of multiplicities q m − 1 and q m .We compute As a m = 0, we see that h is separable and h(0) = 0. From that we obtain the assertion about the multiplicities of the roots of L(X)/X.Remark 6.The main tool from group theory in the proof of Theorem 1 is [2, Theorem 7]: If a subgroup G of GL n (q) contains a Singer cycle (an element which permutes the nonzero elements of F n q cyclically), then GL n/d (q d ) ≤ G ≤ ΓL n/d (q d ) for some divisor d of n.
For this result Gow and McGuire refer to several much stronger results about transitive linear groups.In particular, these sources depend on the classification of the finite simple groups.
In [3], Kantor gives a short proof of this result without using the classification of the finite simple groups.His proof is based on the classification-free paper [1] where Cameron and Kantor determine the doubly transitive subgroups of PGL n (q), which later turned out to contain mistakes.These were fixed by Kantor in the revision [4].