Galois groups over rational function fields and Explicit Hilbert Irreducibility
Abstract.
Let be a polynomial in two variables with rational coefficients, and let be the Galois group of over the field . It follows from Hilbert’s Irreducibility Theorem that for most rational numbers the specialized polynomial has Galois group isomorphic to and factors in the same way as . In this paper we discuss methods for computing the group and obtaining an explicit description of the exceptional numbers , i.e., those for which has Galois group different from or factors differently from . To illustrate the methods we determine the exceptional specializations of three sample polynomials. In addition, we apply our techniques to prove a new result in arithmetic dynamics.
1. Introduction
Let have positive degree in the variable , and let be the Galois group of over . For any rational number we may consider the polynomial and its Galois group, which we denote by . Hilbert’s Irreducibility Theorem (henceforth abbreviated HIT) implies that as varies in , most specializations have Galois group isomorphic to and factor in the same way as . However, there may exist rational numbers such that is not isomorphic to or factors differently from ; 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 the set .
A standard step in the proof of HIT is to show that there exist a finite set and algebraic curves having the following property: for , belongs to the set if and only if is a coordinate of a rational point on one of the curves . Our method for explicitly describing the exceptional set of is based on a constructive proof of this result, which we summarize in the following theorem.
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 the theorem that we may take to be the set of all for which (1.1) does not hold, and we may take to be the plane curve defined by the equation . The problem of explicitly describing the set can therefore be reduced to the following:

Compute the polynomials , and

Determine all the rational points on the curves .
The second step of course does not have an algorithmic solution at present, though there are several techniques available for approaching the problem; see [21] for a survey. For the first step, however, all of the necessary computational tools are available.
To achieve step (1), one must begin by computing a permutation representation of the Galois group ; this can be done using methods of Fieker and Klüners [7]. Though these authors mainly discuss the case of irreducible polynomials over , their methods can be extended to work more generally. For instance, Fieker [6] adapted the algorithm to compute Galois groups of irreducible polynomials over . In the present paper we discuss the modifications needed for Fieker’s implementation and we further extend the method so that it applies to reducible polynomials over . This generalized algorithm for computing Galois groups over has been implemented by the second author and is included in Magma V2.23 [3].
Once the group has been computed, its maximal subgroups can be obtained using an algorithm of Cannon and Holt [4]. Finally, the fixed field of any subgroup of can be computed using known methods; see [10, §3.3] and our discussion in §3.3. Hence, given the polynomial it is possible to compute defining equations for the curves . Functionality for this computation will be available in Magma V2.24 via the intrinsic HilbertIrreducibilityCurves.
In summary, by using currently available methods in computational group theory and Galois theory, and by applying techniques for determining rational points on curves, it is possible in many cases to obtain a complete characterization of the exceptional set of a polynomial .
This article is organized as follows. We devote §2 to the proof of Theorem 1.1, and §3 to a discussion of the algorithms for computing Galois groups and fixed fields over . In order to illustrate the process described above, we include three examples in §4.
The first example concerns the polynomial , which has a finite exceptional set. The case of Fermat’s Last Theorem implies that the only rational numbers for which has a rational root are 0 and . We prove that in fact 0 and are the only rational numbers for which is reducible.
In the second example we consider the polynomial , which is irreducible and has Galois group isomorphic to the symmetric group . Our analysis will show that, in addition to the obvious reducible specialization , there is an infinite family of reducible specializations. More precisely, we prove that for ,
Moreover, when has the above form we show that factors as a product of two irreducible cubic polynomials.
The third example relates to the polynomial , which is one polynomial in a family discussed by Serre [18, §4.5]. The Galois group of is isomorphic to the alternating group , so a typical specialization will have Galois group . However, there are infinitely many exceptions to this: we prove that
Furthermore, for numbers of the above form, we determine precisely which groups arise as varies. We show in particular that the groups and arise for infinitely many such numbers .
In §5 we apply our methods to prove a new result in arithmetic dynamics. Let be a rational function, and for let denote the fold composition of with itself. We say that a rational number is periodic under if there exists such that ; in that case, the least such is called the period of . An important open problem in arithmetic dynamics is a uniform boundedness conjecture of Morton and Silverman [15] which in particular would imply the following: there exists a constant such that for every rational function of degree 2 and every period , has no rational point of period . This conjecture has been refined in various special cases. For example, Poonen [17] studied the family of maps of the form and Manes [14] studied maps of the form . Manes conjectures that no such map can have a rational point of period , and shows that there exist at most finitely many such maps having a rational point of period 5. We prove the following stronger statement: for all but finitely many maps of the form , there exist a positive proportion of prime numbers such that does not have a point of period 5 in the adic field .
2. An explicit form of HIT
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^{1}^{1}1See [18, §3.1] for a definition 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 2.7 below.
It should be noted that the expert will be familiar with several of the results proved in this section. However, we have included complete proofs of most statements due to the lack of a reference treating this subject at the desired level of generality, in particular allowing the polynomial to be reducible.
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 [16, Chap. I, §§89] for the standard material on decomposition groups and ramification used in this section.
If is any element satisfying , 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 well known 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 [16, 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 [16, 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 all the roots of in , it suffices to show that is unramified in every such field. (See [13, 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 . By Lemma 2.1, is unramified in . ∎
We recall the notion of an isomorphism of group actions. If and are groups acting on sets and , respectively, then we say that there is an isomorphism of group actions between and if there exist an isomorphism and a bijection such that for all and all .
Proposition 2.3.
Suppose that satisfies , and let be a prime of dividing . Then there is an isomorphism of group actions , where acts on the roots of and acts on the roots of .
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. We claim 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 .
The map is thus an isomorphism . Moreover, the fact that implies that the actions of and are isomorphic. ∎
Remark 2.4.
Lemma 2.5.
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 2.6.
Let be an intermediate field between and . Let be a primitive element for and let be its minimal polynomial. 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 .

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 2.5, . Thus, (3) holds.
Conversely, suppose that (3) holds. Let be a prime of dividing such that . Let and . Since and divides , we have . Thus, (2) holds.
The equivalence of (3) and (4) is clear. ∎
We can now prove the main result for this section.
Theorem 2.7.
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
(2.1) 
Then the following hold:

If , then .

there is an index such that has a root in .
Proof.
We prove (1) by contradiction. Thus, suppose that and . By Proposition 2.3, the latter condition implies that the group acts on the roots of in the same way that acts on the corresponding roots of . Since , there must be an irreducible factor of such that is reducible. Note that acts transitively on the roots of , but since is reducible and separable, does not act transitively on its roots. Thus we have a contradiction, proving (1).
For the proof of (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 . By Proposition 2.6 applied to the field , this implies that has a root in . This proves one direction of (2). The converse follows by a similar argument. ∎
3. Computation of Galois groups over
We restrict now to the case . It is clear from Theorem 2.7 that in order to better understand the exceptional set of a given polynomial it is necessary to compute the Galois group , the maximal subgroups of , and their corresponding fixed fields. In this section we discuss the Galois group and fixed field algorithms.
3.1. Galois groups of irreducible polynomials over
The article [7] describes an algorithm to compute Galois groups of irreducible polynomials over . As noted in [7] Section 7.7, this algorithm can be adjusted to compute Galois groups of polynomials over fields other than the rational field. For example, [23, 22] discusses this for polynomials over global rational and algebraic function fields. An algorithm for computing Galois groups of polynomials over has been implemented in [6] and included in Magma V2.15. We describe here some of the adjustments to the algorithm in [7] that are necessary for these computations. We address these adjustments using the same headings as [23, 22] after providing a brief summary of the algorithm used. For a full exposition of the algorithm see [23] and [22], Algorithms 1 and 11, respectively.
We describe here the algorithm used in [7] with no degree restrictions. Let be a polynomial of degree over with splitting field over . The algorithm of Stauduhar [19] starts with a group which is known to contain the Galois group , and then traverses the maximal subgroups of until it either finds one which contains , or finds that no maximal subgroup contains , in which case has been determined.
Algorithm 3.1 (Compute the Galois group of a polynomial).
Input: A polynomial of degree over .
Output: The Galois group of .

Choose a finite place of . Compute a splitting field for over the completion of at .

Find a group which the Galois group of is contained in.

While has maximal subgroups which could contain :

For each conjugacy class of maximal subgroups of , compute a relative invariant polynomial for a representative maximal subgroup .

Apply a Corollary to Theorem 5 of [19] : For a conjugacy class of maximal subgroups of not yet decided on do

Compute the precision needed in the roots of for transformation by then evaluation in . Compute the roots of to precision in the splitting field .

For the representatives of the right cosets of in :

Decide whether is the image of an element of . If not, then continue with the next coset. If so then, if for other representatives of right cosets, so set and restart the loop (3) with the new .
Otherwise a descent into this conjugacy class may be reattempted after applying another Tschirnhausen transformation.



Return .
We now discuss how each of the steps of the above algorithm can be carried out in the case where is irreducible.
 Choosing a good prime:

(Step 1). A good prime is necessary for computing a completion of and a splitting field over this completion. The image of must be squarefree over the residue field at . Instead of the completion being a adic field (completion of the rationals) or a series field over the field of constants (completion of a global rational function field) we complete in two directions and compute a completion as a series field over a adic field. For this we need two primes, an integer prime for computing a adic field and a polynomial prime to compute a series field over this adic field.
The choice of a good polynomial prime can be undertaken in the same way as for the global function fields; see [23] Section 3.1 or [22] Section 8.1. In contrast to the global case we consider only primes. Let be the degree of , the LCM of the degrees of the factors of the image of mapped over , and let be the number of factors of the image of mapped over . Similar to the case of global function fields we choose a prime with the smallest if such occurs for a prime we have considered; otherwise a prime we have considered with largest .
To choose a good integer prime for the adic part of the completion we construct the number field , where is the prime polynomial chosen. Then we map to a polynomial over and compute a prime which is good for the computation of the Galois group of , a polynomial over a number field. Lemma 2.16 of [8] contains some necessary conditions such primes must satisfy. In addition to this we choose the prime so that the extension of the adic field is not of too large degree to be expensive to work in nor of too small degree that computations will require excessive precision. This trade off is discussed in [23, 22] Section 3.1 and 8.1, respectively.
 Computing roots:

(Step 1) We construct the field which will contain all the roots of the image of . An extension of a adic field is used as a splitting field when computing Galois groups of polynomials over or a number field. A splitting field is used when computing Galois groups of polynomials over or an extension thereof. The splitting field is a combination of these. The local field contains the roots of mapped over the completion of at .
We take as the image of in and use the map given by the completion mapping at into , and then combine with the mappings . To compute the roots of we first compute the roots of in to the required adic precision and Hensel lift to the required adic precision.
 A starting group:

(Step 2) Section 3.3 of [23, 22] applies also for polynomials over . While all Galois groups are contained in , it can be more efficient if a smaller group containing the Galois group can be computed.
Subfields can provide us with the information to compute a smaller starting group.

Compute the subfields [9] of the field extension and the Galois groups of the normal closures of these subfields.

Compute the intersection of the wreath products, corresponding to the block system in [8] Theorem 3.1, of with the Galois groups of the normal closures of subfields of degree for all subfields of .

 Invariants:
 Mapping back to the function field:

(Step 3iiiA) Given a series we check whether the coefficients of map back to elements of . To the resulting series now in we apply the homomorphism which maps to and the coefficients to polynomials over using a homomorphism mapping the generator of to a root of in , where is the adic precision of . Lastly we take the remainder of this resulting polynomial mod .
 Tschirnhausen transformations:
 Determining a descent:

(Step 3iiiA) Most of the discussion in [23] Section 3.8 and [22] Section 8.8 applies here, including bounding the degree of the evaluation of an invariant at the roots of . However, just as we required two primes to define a splitting field, we also require two bounds – one on the polynomial degree of an evaluation, and one on the size of the coefficients of that polynomial. The minimum infinite valuation can be computed in the same way as for the global function fields. This can be used to compute a precision for a series over the integers (computed using complex roots) which is a bound for the complex size of the integral coefficients. This is used to compute a bound on the norm as times the square of a bound on the evaluation of the invariant at a transformed root of size , where is a Tschirnhausen transformation. The absolute precision of times is used to bound the degree of the evaluation of an invariant mapped back to . The maximum coefficient of is then used to bound the coefficients of this mapped evaluation.
 Precision:

Since we have two completions in our splitting field construction, we require a precision for each completion, a adic precision and a series precision. These are computed from the bound computed above. The series precision is taken to be the absolute precision of the series bound and the adic precision is computed from the largest coefficient of the series bound using Proposition 3.12 in [1].
3.2. Galois groups of reducible polynomials over
Section 7.6 of [7] mentions that their algorithm can be used to compute Galois groups of reducible polynomials. Adjustments of the algorithm necessary to compute Galois groups of reducible polynomials over global rational and algebraic function fields are discussed in [23, 22] and included in Magma V2.18. Here we describe the necessary adjustments, which are included in Magma V2.23, to use the algorithm of [7] to compute Galois groups of reducible polynomials over . We use Algorithm 2 of [22] (Algorithm 12 in [23]) and address these adjustments using the same headings used there. This algorithm uses the product of the Galois groups of the factors of to gain a starting group in Step 2 of Algorithm 3.1, and also does some post processing.
 Choosing a good prime:
 Computing roots in the splitting field over the completion:

The local field must be computed such that it contains the roots of over for all factors of . The field can then be used as a splitting field.
 Check disjointedness of splitting fields:
 Invariants:
 Determination:

The computation of the precision necessary is as in Section 3.1, using the minimum infinite valuation (negative of the maximum degree) of all scaled roots of . Here also we can substitute with a smaller value and compute an unproven group which we can later prove is the Galois group of .
 Multiple and linear factors:
3.3. Computing a fixed field of a subgroup of a Galois group
The procedure needed for this computation, which was implemented in Magma by Fieker and Klüners, is independent of the coefficient ring of the polynomial. We summarize an algorithm in a similar way to [23] Algorithm 1 ([22] Algorithm 11). Though the details differ between coefficient rings, the necessary adjustments are already addressed in the various descriptions of the Galois group algorithm given in [7, 23, 22] and above. This algorithm applies to both reducible and irreducible polynomials.
Algorithm 3.2 (Compute a fixed field of a subgroup of a Galois group).
Given a subgroup of a Galois group of a polynomial of degree , and given the data used to compute from , compute a defining polynomial for the fixed field of .

Compute a relative invariant polynomial and the right transversal .

Compute a Tschirnhausen transformation such that
using roots of to some low precision in the splitting field used for the Galois group computation.

Compute a bound on the evaluation of the invariant at the roots of and the roots of to a precision that allows the bound to be used.

Compute the polynomial with roots

Map the coefficients of back to the coefficient ring of , and return the resulting polynomial. Note that this is a defining polynomial for the fixed field of .
Remark 3.3.
The polynomial returned by Algorithm 3.2 will be of degree ; this can cause difficulties in practice when is large. In a sample computation we carried out for a polynomial of degree 30, the group had a maximal subgroup with . While we were able to compute the fixed fields for all subgroups of index at most 15, the fixed field of could not be determined.
3.4. Proof of Galois groups of polynomials over
Galois groups computed using lower precision than necessary can be proved to be correct or incorrect using absolute resolvents as in [8] Algorithm 5.1 and [7] Section 7.4. We consider here the adjustments to these algorithms that are needed for polynomials over .
Suppose we know that . Algorithm 5.1 of [8] will determine whether or . It does this by computing a resultant and two factors, and , of to precision 1 based on an orbit which is not a orbit. This factorization is lifted to a factorization with precision , where is computed from a bound on the coefficients of the factors of . If corresponds to a true factor of , then ; otherwise . We need to determine the appropriate bound from which the precision can be computed as in the computation of the Galois group. Currently we can bound the degree of the evaluation of an invariant and the size of those coefficients (when the coefficient ring of is ). Letting be the roots of , we can use these bounds in order to bound the coefficients of the polynomial
and its factors by the quantity
where is a bound on obtained as in the step “Determining a descent” of Section 3.1.
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 three sample polynomials. The following algorithm will be our main tool.
Algorithm 4.1.
Input: A separable polynomial .
Output: A finite set and a finite set .

Create empty sets and .

Include in all the rational roots of the discriminant of .

Include in all the rational roots of the leading coefficient 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 rational roots of the discriminant of .


Return the sets and .
For steps 4 and 6(a) we use the methods discussed in §3, and for step 5 we use an algorithm of Cannon and Holt [4]. All of our computations were done using Magma V2.23, which includes implementations of these algorithms. The intrinsic function HilbertIrreducibilityCurves in Magma V2.24 will be an implementation of Algorithm 4.1.
For later reference we record the following consequence of Theorem 2.7.
Proposition 4.2.
Let be a separable polynomial with Galois group , and let and form the output of Algorithm 4.1 with input . Then for all we have:

If , then .

there exists such that has a rational root.
4.1. A finite exceptional set
In our first 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 prove the following stronger result.
Proposition 4.3.
Let . Then is reducible if and only if .
Proof.
Suppose that is reducible and that . Applying Algorithm 4.1 to the polynomial we obtain the set and the polynomials
By Proposition 4.2, at least one of the polynomials must have a rational root; we accordingly divide the proof into four cases.
Case 1: There exists such that . Defining and
the equation implies that This equation defines the elliptic curve with Cremona label 36a3, which has has rank 0, and its only affine rational point is . It follows that and thus , which is a contradiction. Hence this case cannot occur.
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 . Letting and , the equation implies that . This equation defines the elliptic curve with Cremona label 36a1, which has rank 0 and a torsion subgroup of order 6; its only affine rational points are , , and . Since , we must have and therefore , which is a contradiction.
Case 4: There exists such that . Letting , the equation implies that . This 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 , , and . 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 . ∎
4.2. An infinite family of exceptional factorizations
Let , which is an irreducible polynomial with Galois group isomorphic to the symmetric group . In this example we will determine precisely for which rational numbers the specialization is reducible, and how factors in that case.
Proposition 4.4.
Let be a nonzero rational number such that is reducible. Then has the form for some .
Proof.
Applying Algorithm 4.1 to we obtain the set and the polynomials
By Proposition 4.2, one of the polynomials must have a rational root. We will show that cannot be 2 or 3, from which the proposition follows easily.
Suppose that for some . Letting
we obtain . This equation defines an elliptic curve with exactly two rational points, namely the point at infinity and the point . Hence we must have , which implies that . However, the equation implies that , which is a contradiction.
By a similar argument one can show that the only rational solutions to the equation are , , and ; hence is impossible for since . ∎
Lemma 4.5.
Let and be the curves in defined by the equations and , respectively. Then .
Proof.
The curves and are both nonhyperelliptic curves of genus 3, but they admit a map to an elliptic curve of rank 0; this allows us to determine their rational points. We give the proof only for , since the argument is very similar for .
There is a map from to the elliptic curve given by . The curve has rank 0, and its only rational points are and . Any rational point on must necessarily have , and will therefore map to the point on . However, this is impossible since . Hence has no rational point. ∎
Proposition 4.6.
Suppose that is of the form for some rational number . Then factors as
Moreover, both cubic factors of are irreducible.
Proof.
Substituting in the polynomial and factoring, we obtain the above factorization. Lemma 4.5 implies that neither factor of can have a rational root, and therefore both factors are irreducible. ∎
4.3. An infinite family of exceptional Galois groups
In [18, §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 . In this example we will determine precisely for which rational numbers the Galois group is different from , and which groups arise for such numbers . Our main results are Propositions 4.9 and 4.12.
Lemma 4.7.
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
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). ∎
Lemma 4.8.
Let and let . Then the polynomial has no rational root.
Proof.
Suppose that is such that . Since , we must have . Defining , the equation implies that
which is clearly impossible for . This contradiction proves the lemma. ∎
Proposition 4.9.
Let and let be the Galois group of . Then