explicit hilbert irreducibility
Abstract.
Let be an irreducible polynomial in two variables with rational coefficients. It follows from Hilbert’s Irreducibility Theorem that for most rational numbers the specialized polynomial is irreducible and has the same Galois group as . We discuss here a method for obtaining an explicit description of the set of exceptional numbers , i.e., those for which is either reducible or has a different Galois group than . To illustrate the method we determine the exceptional specializations of two polynomials of degrees four and six.
1. Introduction
Let be an irreducible polynomial in two variables with rational coefficients. Regarding as an element of the ring , let be the Galois group of , i.e., the Galois group of a splitting field for over . For any rational number we may consider the specialized polynomial and its Galois group, which we denote by . The Hilbert Irreducibility Theorem (henceforth abbreviated HIT) implies that as varies over all rational numbers, most specializations remain irreducible and have Galois group isomorphic to . However, there may exist rational numbers for which either is reducible or is not isomorphic to ; we will call the set of all such numbers the exceptional set of , denoted . The main purpose of this article is to develop a method for obtaining an explicit description of this exceptional set.
A standard step^{1}^{1}1See Lang [8, Chap. 9, §1] or Serre [11, §3.3]. in the proof of HIT is to show that there exist a finite set and algebraic curves having the following property: if is such that either is reducible or is not isomorphic to , then is a coordinate of a rational point on one of the curves (or more generally, is in the image of a map ). Our approach to obtaining an explicit description of the set is based on a method for finding such a set and curves . Theorem 1.1 below, which was motivated by Serre’s treatment of HIT in [11, §3.3] and by results of DèbesWalkowiak [4, §3.1], allows us to reduce the problem of finding both a set and defining equations for curves to problems in computational group theory and Galois theory.
Theorem 1.1.
Let and be the discriminant and leading coefficient of , respectively. Let be representatives of all the conjugacy classes of maximal subgroups of . For , let be the fixed field of and let be a monic irreducible polynomial in such that is generated by a root of . Suppose that satisfies
(1.1) 
Then there is an index such that has a root in .
It follows from Theorem 1.1 that we may take to be the finite set of rational numbers for which (1.1) does not hold, and we may take to be the affine plane curve defined by the equation . Indeed, the theorem implies that – disregarding elements of – the set consists of the first coordinates of all the rational points on the curves .
In practice this result can be used to explicitly describe the set for any given polynomial . All of the algebraic objects appearing in the theorem – in particular the group , the subgroups , and the polynomials – can be computed using currently available methods in computer algebra. Furthermore, depending on the geometry of the curves , one may be able to determine the sets of rational points on all these curves, thus obtaining a complete characterization of the elements of .
A more general version of Theorem 1.1 is proved in §2, and further details regarding the associated algorithmic questions are given in §3. In order to illustrate the process described above, we include two examples in §4. The first example concerns the polynomial
which is one polynomial in a family discussed by Serre [11, §4.5]. The Galois group of is isomorphic to the alternating group , so a typical specialization will have Galois group . We show that there are infinitely many specializations of with Galois group different from , and that these can parametrized. More precisely, we prove:
In the second example we consider the polynomial
The case of Fermat’s Last Theorem implies that the only rational numbers for which has a rational root are 0 and . We will prove the stronger result that in fact 0 and are the only rational numbers for which is reducible.
Acknowledgements
I thank Pierre Dèbes for several helpful discussions related to the material of §2.
2. HIT via extensions of Dedekind domains
Let be a field of characteristic 0 and let be a polynomial of degree in the variable . We will henceforth regard as an element of the ring and assume that is separable. We define the factorization type of , denoted , to be the multiset consisting of the degrees of the irreducible factors of .
Let be a splitting field of and let be the Galois group of . We assume that is nontrivial. For every element , let denote the specialized polynomial . The Galois group and factorization type of will be denoted by and , respectively.
It follows from HIT that there is a thin^{2}^{2}2See [11, §3.1] for a definition and properties of thin sets. subset of outside of which we have and . We define the exceptional set of , denoted , to be the set of all elements for which either one of these conditions fails to hold:
Our aim in this section is to prove a version of HIT from which one can deduce a method for explicitly describing the set . Our main result in this direction is Theorem 2.6 below.
Let and be the discriminant and leading coefficient of , respectively. Let be the ring
For every intermediate field between and , let denote the integral closure of in . Note that is an extension of Dedekind domains with being a PID. By a prime of (or of we mean a maximal ideal of . If is a prime of and is a prime of , we denote by and the residue fields of and , respectively. Thus,
If divides , we denote the ramification index and residual degree of over by and , respectively.
For every prime of , let be the decomposition group of over and let be the decomposition field of , i.e., the fixed field of . We refer the reader to [10, Chap. I, §§89] for the standard material on decomposition groups and ramification used in this section.
If is any element satisfying , then the evaluation homomorphism given by extends uniquely to a homomorphism . Let be the kernel of this map. We will henceforth identify the residue field with via the map . Note that with this identification, if is an arbitrary polynomial, then upon reducing the coefficients of modulo we obtain the specialized polynomial .
It will be necessary for our purposes in this section to be able to determine how the prime factors in any intermediate field between and . Recall that by a theorem of DedekindKummer, for all but finitely many primes of , the factorization of in can be determined by choosing an integral primitive element of and factoring its minimal polynomial modulo . The finite set of primes that need to be excluded are those that are not relatively prime to the conductor of the ring ; see [10, p. 47, Prop. 8.3] for details. The following lemma provides sufficient conditions on so that will be relatively prime to this conductor, and therefore the DedekindKummer criterion can be applied to .
Lemma 2.1.
Let be an intermediate field between and with primitive element having minimal polynomial . Let
be the conductor of the ring . Suppose that satisfies
Then is relatively prime to . Furthermore, is unramified in .
Proof.
Let be the discriminant of . By a linear algebra argument (see Lemma 2.9 in [10, p. 12]) we have and therefore . Suppose that is a prime of dividing both and . Since we have , so . By definition of this implies that , which is a contradiction. Therefore must be relatively prime to .
The DedekindKummer theorem now allows us to relate the factorization of in to the factorization of in . In particular, the theorem implies that if is ramified in , then has a repeated irreducible factor, which contradicts our assumption that . Therefore must be unramified in . ∎
Lemma 2.2.
Suppose that satisfies . Then the prime is unramified in .
Proof.
Since is the compositum of the fields as ranges over the roots of in , it suffices to show that is unramified in every such field. (See [9, p. 119, Cor. 8.7].) Thus, let be any root of and let . Let be an irreducible factor of having as a root. Dividing by its leading coefficient we obtain a monic irreducible polynomial having as a root; it follows that is the minimal polynomial of over . Let be the discriminant of . Since divides in , divides in . Hence, the hypothesis that implies that . The result now follows from Lemma 2.1. ∎
Proposition 2.3.
Suppose that satisfies , and let be a prime of dividing . Then is isomorphic to .
Proof.
For every element let denote the image of under the quotient map . Recall that the extension is Galois and that there is a surjective homomorphism given by , where for every . Furthermore, since is unramified in by Lemma 2.2, this map is an isomorphism. Hence, in order to prove the proposition it suffices to show that is a splitting field for .
Note that if is a root of , then is a root of . Moreover, if and are distinct roots of , then ; indeed, this follows from the fact that . Thus, reduction modulo is an injective map from the set of roots of to the set of roots of .
Let be the roots of in , and let . Clearly is a splitting field for , and . We will prove that by showing that the group is trivial. Let and let be the element such that . Since is the identity map on , we have for every index , and hence for all . Since and are roots of , this implies that . Thus, fixes every root of , so is the identity element of . Hence is the identity element of . This proves that is trivial and therefore is a splitting field for . ∎
Lemma 2.4.
Let be a prime of and let be a prime of dividing . Then the following hold:

Setting , we have .

Let be an intermediate field between and , and let . If , then .
Proposition 2.5.
Let be an intermediate field between and . Let be a primitive element for and let be the minimal polynomial of . Suppose that satisfies
Then the following are equivalent:

The polynomial has a root in .

There exists a prime of dividing such that .

There exists a prime of dividing such that .
Proof.
By Lemma 2.1, is relatively prime to the conductor of . The DedekindKummer theorem then implies that the degrees of the irreducible factors of in correspond to the residual degrees for primes of dividing . The equivalence of (1) and (2) follows immediately.
We now show that (2) and (3) are equivalent. Suppose that (2) holds, and let be a prime of dividing . By Lemma 2.2, is unramified in and therefore unramified in . Hence, . By Lemma 2.4, . Thus, (3) holds.
Finally, suppose that (3) holds. Let be a prime of dividing such that . Let and . Since and divides , we have . Thus, (2) holds. ∎
Theorem 2.6.
Let be representatives of all the conjugacy classes of maximal subgroups of . For let be the fixed field of , and let be a monic irreducible polynomial in such that is generated by a root of . Suppose that satisfies
Then the following hold:

If , then .

there is an index such that has a root in .
Proof.
We begin by proving (1). Thus, suppose that . Let be monic irreducible polynomials such that
Since , there exists such that is reducible. Let be a root of and let . Since divides and , then . Lemma 2.1 implies that is relatively prime to the conductor of ; we may therefore apply the DedekindKummer theorem to relate the factorization of to the factorization of .
Since is separable and reducible, it must have more than one irreducible factor (up to associates). Hence, there is more than one prime of dividing , and therefore more than one prime of dividing . It follows that if is any prime of dividing , the group is a proper subgroup of . (Indeed, the index is the number of primes of dividing .) Proposition 2.3 now implies that , which proves (1).
We now prove (2). Suppose that and let be a prime of dividing . By Proposition 2.3, the group is a proper subgroup of . Replacing by a conjugate ideal if necessary, we may therefore assume that for some index . The decomposition field then contains , and by Proposition 2.5 applied to the field , this implies that has a root in . This proves one direction of (2). The converse follows by a similar argument. ∎
It follows from the above theorem that the problem of determining the exceptional set of can be reduced to a problem of finding all the rational points on a finite list of curves. More precisely, we have the following.
Corollary 2.7.
With notation as in Theorem 2.6, let be the finite set of all elements such that
For let be the affine plane curve defined by the equation . Let . Then if and only if is the first coordinate of a rational point on one of the curves .
3. Algorithmic aspects
Theorem 2.6 suggests the following algorithm which can be used to study the exceptional set of the polynomial . We state the algorithm first and then explain its precise relation to this problem.
Algorithm 3.1.
Input: A separable polynomial .
Output: A finite set and a finite set .

Create empty sets and .

Include in all the roots of the leading coefficient of .

Include in all the roots of the discriminant of .

Compute the group . More precisely, find a permutation representation of induced by a labeling of the roots of .

Find subgroups representing all the conjugacy classes of maximal subgroups of .

For :

Find a monic irreducible polynomial such that the fixed field of is generated by a root of .

Include in the set .

Include in all the roots of the discriminant of .


Return the sets and .
Theorem 3.2.
Let be a separable polynomial, and let and form the output of Algorithm 3.1 with input . Then the following hold for all :

If , then .

there exists such that has a root in .
Proof.
This is an immediate consequence of Theorem 2.6. ∎
In the case , all of the computational methods needed to carry out the steps of Algorithm 3.1 are known, and most have been implemented in computer algebra systems. Indeed:

A permutation representation of can be computed by using an algorithm of FiekerKlüners [5].

A set of representatives for the conjugacy classes of maximal subgroups of can be obtained using an algorithm of CannonHolt [3].

Given a subgroup , the minimal polynomial of a primitive element of the fixed field of can be found using a method discussed in [6, §3.3].
Most of the above algorithms have been implemented and are included in Magma [1]; the only exception is the computation of Galois groups of reducible polynomials over . Hence, there is at present an obstacle to carrying out Algorithm 3.1 with a reducible polynomial as input. However, this problem is being addressed in current work of Nicole Sutherland, and an implementation of the algorithm of FiekerKlüners for reducible polynomials will be included in a future release of Magma.
In view of the above discussion, it is currently possible to translate the problem of determining the exceptional set of an irreducible polynomial to a problem of determining the sets of rational points on a finite list of algebraic curves. The difficulty of the problem is therefore largely dependent on the genera of these curves; if the genera are not too large, it may be possible to obtain an explicit characterization of the set . For a survey of the presently available methods for computing rational points on curves, we refer the reader to Stoll’s article [13].
4. Examples
Having developed the theoretical and algorithmic material that form the core of this article, we proceed to apply our results to study the exceptional sets of two sample polynomials, one with an infinite exceptional set and one with a finite exceptional set. In order to carry out the necessary computations, an implementation of Algorithm 3.1 in Magma will be used. The source code of our implementation is available in [7].
We include a cautionary remark for the reader who may be interested in reproducing our calculations. The method used by Magma to find primitive elements of fixed fields (which is needed in step 6(a) of Algorithm 3.1) does not always produce the same primitive element for a given field extension. Hence, the output of Algorithm 3.1 that the reader obtains may be different from what is given here. However, in that case the arguments made below can be easily adapted to prove the same results.
4.1. An infinite exceptional set
In [11, §4.5] Serre shows that for even values of , the polynomial
has the alternating group as its Galois group. By HIT, most specializations will have Galois group as well. In the case we obtain the polynomial
with Galois group . We will now determine precisely which specializations of have Galois group different from .
Lemma 4.1.
Let and let . Then the polynomial has no rational root.
Proof.
Suppose that there exists such that . Since , we must have . Defining , the equation implies that
However, a simple argument^{3}^{3}3The solvability of an equation of the form over any given adic field can be tested using a method of Bruin [2, §5.4] which is implemented in the Magma function HasPoint. shows that the above equation has no solution in and therefore no solution in . This contradiction proves the lemma. ∎
Lemma 4.2.
Let and let . Then the polynomial has a rational root if and only if has the form
(4.1) 
for some rational number .
Proof.
Let be the plane curve defined by the equation . The curve is parametrizable; indeed, the rational maps and given by
are easily seen to be inverses.
Suppose that is of the form (4.1). We may then define
so that is a rational point on . Hence, the polynomial has a rational root (namely ).
Conversely, suppose that has a rational root, say . Since , the map is defined at the point . Thus, we may define . We claim that . A straightforward calculation shows that the rational points on the pullback of under are and . Since , the point is different from these two points. Hence , as claimed. The map is therefore defined at , so . In particular, is of the form (4.1). ∎
Proposition 4.3.
Let and let be the Galois group of . Then
Proof.
For the proposition holds because both statements in the above equivalence are true. Indeed, we have
so has order 2. Suppose now that .
Applying Algorithm 3.1 to the polynomial we obtain the set and the polynomials
4.2. A finite exceptional set
In our second example we consider the polynomial
As follows from the case of Fermat’s Last Theorem, the specialized polynomial has a rational root if and only if . We will now prove the following stronger result.
Proposition 4.4.
For , the polynomial is reducible if and only if .
Proof.
Suppose that is reducible. We will show by contradiction that . Thus, suppose that .
Applying Algorithm 3.1 to the polynomial we obtain the set and the polynomials
By Theorem 3.2, one of the polynomials must have a rational root; we accordingly divide the proof into four cases.
Case 1: There exists such that . Defining
the equation implies that
The above equation defines a hyperelliptic curve of genus 2. By a descent argument one can show that the Jacobian variety of has a MordellWeil group of rank 0; it is therefore a straightforward calculation^{4}^{4}4Stoll’s algorithm of 2descent [12] is implemented in Magma and can be accessed via the RankBound function. Once the rank of the Jacobian is known to be 0, the Chabauty0 function carries out the calculation of finding all the rational points on . to determine the set of rational points on . We find that the only rational points are the Weierstrass points, namely . It follows that , which is a contradiction.
Case 2: There exists such that . Letting
we have and , which is clearly impossible. Thus we have a contradiction.
Case 3: There exists such that . Defining
the equation implies that
The above equation defines the elliptic curve with Cremona label 36a1. This curve has rank 0 and a torsion subgroup of order 6; its only affine rational points are
It follows from this that , 2, or . This implies, respectively, that , , or , all of which lead to a contradiction.
Case 4: There exists such that . Letting , the equation implies that
The above equation defines the elliptic curve with Cremona label 36a1, the same curve that appeared in the previous case. Using the above model of the curve, the affine rational points are
It follows that , or 0, which implies that , , or , all of which yield a contradiction.
Since every case has led to a contradiction, we conclude that . This completes the proof of the proposition. ∎
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