Enumerative Galois theory for cubics and quartics

Article history: Received 8 May 2019 Received in revised form 28 April 2020 Accepted 17 June 2020 Available online xxxx Communicated by Kartik Prasanna MSC: primary 11R32 secondary 11C08, 11G35


Introduction
Consider monic polynomials f (X) = X n + a 1 X n−1 + · · · + a n−1 X + a n (1.1) of a given degree n 3, with integer coefficients. Recall that the Galois group G f of f is the automorphism group of its splitting field. As G f acts on the roots of f , it can be embedded into the symmetric group S n ; the only information that we will need about inseparable polynomials is that their Galois group is not isomorphic to S n . The enumeration of polynomials with prescribed Galois group is an enduring topic.

Van der Waerden's conjecture
Van der Waerden [27] showed that a generic polynomial has full Galois group, and a popular objective has been to sharpen his bound on the size  We have paraphrased slightly: van der Waerden suggested that monic, irreducible, non-S n polynomials of degree n are rarer than monic reducibles, counted in this way. It follows from the proof in [5] that if n 3 then R n (H) = c n H n−1 + O n (H n−2 (log H) 2 ), (1.2) for some constant c n > 0. Chela [5] stated this without an explicit error term, and in Appendix B we explain how the error term in (1.2) comes about. Van der Waerden's conjecture may therefore be equivalently stated as follows. Hitherto, no case of this conjecture was known. In the cubic case n = 3, Lefton [21] showed that E 3 (H) ε H 2+ε , a record that has stood unbeaten for over four decades. We establish the following asymptotic formula for E 3 (H), thereby resolving the cubic case of van der Waerden's conjecture. Note from [5] that c 3 = 8( π 2 6 + 1 4 ), so we draw the following equivalent conclusion.
Theorem 1.4. The number of monic, irreducible, non-S 3 cubic polynomials It was thought that the second author [12] had come close to settling the quartic case n = 4 over a decade ago, asserting the estimate (1.4) However, we have discovered an error in Eq. (7) therein, which appears to damage the argument beyond repair-see [16, p. 613] for the correct expressions. To our knowledge, the strongest unconditional bound to date is E 4 (H) ε H 2+ √ 2+ε , obtained in [14]. The inequality (1.4) is known conditionally [30,Theorem 1.4]. We establish the following asymptotic formula for E 4 (H), thereby settling the quartic case of van der Waerden's conjecture. 2.91, and note from [5] that c 4 = 16(ζ(3) + 1 6 ), so if only irreducible polynomials are considered then the exponent is lower than 3. Theorem 1.6 shows that irreducible non-S 4 quartics are less numerous than reducible quartics, and is equivalent to Theorem 1.5.

Specific groups
We now address the general problem of counting polynomials with prescribed Galois group. For G S n , let us write N G,n = N G,n (H) for the number of monic, irreducible, integer polynomials, with coefficients bounded by H in absolute value, whose Galois group is isomorphic to G. The second author showed in [13] that N G,n n,ε H n−1+ 1 [S n :G] +ε , (1.6) and in [14] that The latter article established that if n 3 then breaking a record previously held by van der Waerden [27], Knobloch [20], Gallagher [17] and Zywina [31].
Recall that if f is irreducible then G f acts transitively on the roots of f . Thus, in the cubic case n = 3, the only possibilities for the Galois group of an irreducible cubic polynomial are S 3 and A 3 . The polynomials counted in Theorem 1.4 are the A 3 cubics, and the others are either reducible or have full Galois group. Our bound N A 3 ,3 ε H 1.5+ε dramatically improves upon Lefton's longstanding record of N A 3 ,3 ε H 2+ε .
Using the C programming language, we found that (for the code, see Appendix A). From the additional data point N A 3 ,3 (500) = 52420, one might empirically estimate the exponent as log(355334/52420)/ log 4 ≈ 1.38. The best lower bound that we know of is coming from the one-parameter family X 3 + tX 2 + (t − 3)X − 1 given for example in Smith's tables [25, §12]. So the correct exponent, if well-defined, lies between 1 and 1.5. Now consider the quartic case n = 4. In this case there are five possibilities for G f , namely S 4 , A 4 , D 4 , V 4 and C 4 , see [19]. Here D 4 is the dihedral group of order 8, and A 4 , V 4 are respectively the alternating and Klein four groups. As usual C 4 is the cyclic group of order 4. We write S H for the set of monic, irreducible quartics with coefficients in Z ∩ [−H, H], and for G ∈ {S 4 , A 4 , D 4 , V 4 , C 4 } we define We ascertain the order of magnitude for the number of D 4 quartics. To our knowledge, this is the first time that the order of magnitude of N G,n has been obtained, for G S n . Theorem 1.7. We have In addition, we show that V 4 and C 4 quartics are less numerous.
Finally, to complete the proof of Theorem 1.6, we establish the following upper bound for A 4 quartics. H; the latter cited result is based on a quantitative version of Hilbert's irreducibility theorem. We can construct a family of quartics that implies a sharper lower bound for N V 4 than what we were able to find in the literature: the construction given in §7.1 shows that N V 4 H 3/2 . We summarise our state of knowledge concerning the quartic case as follows: The story is still far from complete. We expect that in time asymptotic formulas will emerge for every N G,4 (H). Below we provide the values of N G,4 (150), evaluated using the C programming language (for the code, see Appendix A).
This suggests that the upper bounds for A 4 , V 4 and C 4 quartics may be far from the truth. We remark that our counting problem differs substantially from the corresponding problem for quartic fields, for which Bhargava [1] showed that in some sense a positive proportion of quartic fields have Galois group D 4 . For an explanation of why the results are consistent, see [30,Remark 5.1].

Parametrisation, concentration, and root separation
This leads us to the diophantine equation and we can parametrise the solutions using algebraic number theory. This equation is discussed in [7, §14.2.3] and elsewhere [11], but here we also need to deal with common divisors between the variables, and these can be enormous. Accounting for the common divisors gives rise to parametrised families of (I, J, Y ) encompassing all solutions to the diophantine equation (1.8). The broad idea is to count those pairs (I, J) with the parameters lying in given dyadic ranges, and then to count possibilities for the corresponding a, b, c subject to those ranges. To illustrate the concentration method, consider the discriminant. On one hand, this is O(H 4 ), being quartic in a, b, c. On the other hand, based on (1.7), we would expect it to have size roughly H 6 . For concreteness, one of the parametrised families of solutions to (1.8) is imposing a constraint on s, t. Writing λ = t/s, one interpretation is that if s is not small then λ(λ − 1)(λ + 1) is small, so the ratio t/s must be close to a root of the polynomial X(X − 1)(X + 1). In other words, either s ≈ 0 or t ≈ 0 or s ≈ t or s ≈ −t. This restriction on the pair (s, t) delivers a saving.
Four instances of concentration arise in our proof. In the first, the concentrating polynomials are linear, and the rewards are easily harvested. In the second, the concentrating polynomials are cubic, and the roots are well-separated, owing to (i) Mahler's work [22] involving what is now known as the Mahler measure [26], and (ii) the discriminant always being bounded well away from zero. In the third, the concentrating polynomials are quadratic, and we can consider a difference of perfect squares. In the final instance, the concentrating polynomials are cubic, but are "close" to being quadratic, and we can again consider a difference of perfect squares.

New and old identities
Our investigation of the quartic case begins with classical criteria [19] involving the discriminant and cubic resolvent of a monic, irreducible quartic polynomial (1.5). When the Galois group is D 4 , V 4 or C 4 , the cubic resolvent has an integer root, which we introduce as an extra variable x. Changing variables to use e = b − x instead of b, we obtain the astonishing symmetry (3.3), which we believe is new. For emphasis, the identity is Using ideas from the geometry of numbers and diophantine approximation leads to the upper bound The proof then motivates a construction that implies the matching lower bound (1.10) The analysis described above roughly speaking provides an approximate parametrisation of the D 4 , V 4 and C 4 quartics, by certain variables u, v, w, x, a, where a is as in (1.5). To show that N V 4 and N C 4 satisfy the stronger upper bound O(H 2 log H), we use an additional piece of information in each case; this takes the form of an equation y 2 = P u,v,w,a (x), where P u,v,w,a is a polynomial and y is an additional variable. We require upper bounds for the number of integer solutions to this diophantine equation in (x, y), and these bounds need to be uniform in the coefficients. We are able to ascertain that the curve defined is absolutely irreducible, which enables us to apply a Bombieri-Pila [3] style of result by Vaughan [28,Theorem 1.1].
Our study of A 4 quartics starts with the standard fact that the discriminant is in this case a square. Deviating from previous work on this topic, we employ the invariant theory of GL 2 actions on binary quartic forms (or, equivalently, unary quartic polynomials), see [2]. The discriminant can then be written as (4I 3 − J 2 )/27, where Our strategy is first to count integer solutions (I, J, y) to and then to count integer solutions (a, b, c, d) to the system (1.11). In the latter step, we require upper bounds that are uniform in the coefficients. Further manipulations lead us to an affine surface Y I,J , which we show to be absolutely irreducible. A result stated by Browning [4, Lemma 1], which he attributes to Heath-Brown and Salberger, then enables us to cover the integer points on the surface by a reasonably small family of curves. By showing that Y I,J contains no lines, and using this fact nontrivially, we can then decompose each curve in the family into irreducible curves of degree greater than or equal to 2, and finally apply Bombieri-Pila [3].
For convenient reference, we record below a version of the Kappe-Warren criterion [19], as given in an expository note of Keith Conrad's [8,Corollary 4.3]. The distinction between D 4 and C 4 is done slightly differently between those two documents; Conrad's description of this is readily deduced from [9, Theorem 13.1.1] and the identity (3.2). We will see in §3 that the cubic resolvent of a monic, quartic polynomial with integer coefficients is a monic, cubic polynomial with integer coefficients. Also note that if f (X) ∈ Z[X] is irreducible then its discriminant Δ is a non-zero integer.

Theorem 1.10 (Kappe-Warren criterion). For a monic, irreducible quartic f (X) ∈ Z[X]
, whose cubic resolvent is r(X), the isomorphism class of the Galois group G f is as follows.

Organisation
The cubic case is handled in §2. In §3 we establish (1.9), and in §4 we prove the complementary lower bound (1.10). In §5, we establish Theorem 1.8, thereby also completing the proof of Theorem 1.7. In §6 we prove Theorem 1.9, thereby also completing the proof of Theorem 1.6. Finally, in §7, we show that and also the code used to compute N A 3 ,3 (2000). In Appendix B, we verify the error term in (1.2). In Appendix C, we show that if the discriminant 4I 3 − J 2 is non-zero then the set of binary forms with given invariants I and J contains no rational lines; this is related to Lemma 6.1 and is of independent interest.

Notation
We adopt the convention that ε denotes an arbitrarily small positive constant, whose value is allowed to change between occurrences. We use Vinogradov and Bachmann-Landau notation throughout, with the implicit constants being allowed to depend on ε. We write #S for the cardinality of a set S. If g and h are positive-valued, we write g h if g h g. Throughout H denotes a positive real number, sufficiently large in terms of ε. Let μ(·) be the Möbius function. Both authors thank the Mathematisches Forschungsinstitut Oberwolfach and the Fields Institute for excellent working conditions, and the second author would like to thank the Mathematical Institute at the University of Oxford for hosting him during a sabbatical. We thank Victor Beresnevich, Manjul Bhargava, Tim Browning, John Cremona, James Maynard, Samir Siksek, Damiano Testa, Frank Thorne, and Stanley Xiao for helpful discussions. Finally, we are grateful to the anonymous referee for a careful reading and for particularly helpful comments.

The cubic case
In this section, we establish Theorem 1.4. As discussed in the introduction, this is counting monic, A 3 cubic polynomials with integer coefficients bounded by H in absolute value, and we will show that be an irreducible cubic polynomial with G f A 3 and a, b, c ∈ [−H, H]. Then its discriminant Δ is a non-zero square. A short calculation reveals that where I and J are as defined in (1.7). In particular, there exists Y = 3 √ Δ ∈ 3N satisfying (1.8).

Parametrisation
Let where u, v ∈ N with u cubefree, and letg = uv 2 .
As u is cubefree, observe that g | 2I. Write We factorise the left hand side of (2.2) in the ring R : Note that R is a principal ideal domain, and is therefore a unique factorisation domain. The greatest common divisor of x + y √ −3 and x − y √ −3 divides both 2x and 2y √ −3, and so it divides 2 √ −3. Write Note that R has discriminant −3, so 3 is the only rational prime that ramifies in R. Thus, either u is cubefree in R, or else u = 9u for some cubefree u ∈ R not divisible by √ −3. The cubefree component of an element ρ of R is well defined up to multiplication by the cube of a unit, that is, up to sign: one prime factorises ρ and divides by a maximal cubic divisor. Now u is the cubefree component of 2de, up to multiplication by ±1 or Consider the norm which in particular is multiplicative, and note that R ⊂ Q( u; the other case N (e) u is similar. As any element of R is uniquely represented as a 1 2 Z-linear combination of 1 and √ −3, we may write with q, r, s, t ∈ Z, and so As (x, y) = 1, we must have (s, t) 2, and our bound q 2 +3r 2 In fact we can say more. From (2.2) and (2.3), we compute-using N (·) or otherwisethat Recall that either u is cubefree in R, or else u = 9u for some cubefree u ∈ R not divisible by √ −3. Therefore u is the cubefree component of 4(q 2 + 3r 2 ), up to multiplication by ±1 or ±( √ −3) 3 , and in particular u 4(q 2 + 3r 2 ). We already saw that q 2 + 3r 2 u, so we conclude that

Scales, and Lefton's approach
We consider solutions for which A |a| < 2A, where A ∈ [1, H] is a power of two. In the main part of the proof we only wish to choose the coefficient a at the end, however it is convenient to fix the scale A from the outset. There are O(log H) such scales.
Lefton's approach [21] is to choose a A and b H, and then to observe [21, Lemma 2] that the equation As Y = 3 √ Δ, we may write this as Y HA.
We also choose scales G, V, T ∈ N, powers of 2, in O((log H) 3 ) ways; these constrain our parameters tog Note from (2.1) and (2.4) that The plan is to count pairs (I, J) of integers subject to the above ranges and satisfying (1.8) for some Y ∈ N with Y HA, and then to count (a, b, c) ∈ Z 3 with |a| A and |b|, |c| H corresponding to our choice of the pair (I, J). We need a method that is efficient when T is reasonably small, and another method that is efficient when G is reasonably small. Note that In the previous subsection, we saw that given I, J with (4I 3 −J 2 )/3 a square there exist parameters v, q, r, s, t with certain properties. The pair (I, J) is determined in O(H ε ) ways by v, q, r, s, t, uniformly in the relevant ranges. Indeed, the variables x and y are as in (2.3), and uz 3 is then determined via (2.2). Next, the variable u is a divisor of uz 3 , of which there are O(H ε ), and finally we know g, g, I, J. The upshot is that we have reduced our task of counting pairs (I, J) to that of upper bounding the number of quintuples (v, q, r, s, t) that can possibly arise in this way.

A linear instance of the concentration method
We begin by considering the case We now assume that The contribution from this case is therefore bounded above by By (2.5) we conclude that there are O(H 1+ε T ) possibilities for (I, J) in total.

Root separation
The approach in the previous subsection is effective when T is reasonably small. Here we develop an approach that works well when G is reasonably small. We assume that |t| |s|, so that |t| T ; the other scenario is similar. We begin by choosing v V and q √ G/V . We begin with the case r = 0. Choose r = 0 with r √ G/V , define a polynomial F by F(X) = rX 3 + 3qX 2 − 9rX − 3q, and write κ = s/t. From (2.6) we obtain Using what is now known as the Mahler measure [26], Mahler analysed the separation of roots of polynomials. It is this that enables us to capitalise efficiently on the concentration inherent in the cubic inequality (2.7). Mahler established, in particular, a lower bound for the minimum distance between two roots, in terms of the degree, discriminant, and the sum of the absolute values of the coefficients of the polynomial [22,Corollary 2]. Applying this to the polynomial F with roots κ 1 , κ 2 , κ 3 yields One might not immediately realise that the discriminant of F should necessarily be positive and fairly large. However, this is indeed the case, and it happens to be a constant multiple of N (d) 2 . From the formula for the discriminant of a cubic polynomial, we compute that = (18(q 2 + 3r 2 )) 2 (|q| + |r|) 4 .
We now have The upshot is that the other parameters determine O HA rGV T 2 + 1 possibilities for s. Bearing in mind (2.5), this case contributes at most solutions. If instead r = 0, then (2.4) implies that q √ G/V , and with κ = s/t we obtain and so This case permits at most We conclude that there are O(H ε (H + A √ G)) possibilities for the pair (I, J).

An approximately quadratic inequality
From the previous two subsections, we glean that the number of allowed pairs (I, J) is at most We also have J − 3aI + a 3 = 27c H.
As A > 999 √ H, we know that J = 27c − 9ab + 2a 3 and a have the same sign, so The left hand side above is cubic in x, but x is fairly small, so we can approximate the cubic by a quadratic in order to exploit concentration. The triangle inequality gives Observe that x 0 is a positive real number, since Recall that x ∈ Z − √ I is a discrete variable. The number of possibilities for x is therefore bounded above by a constant times min X, Once we know x, the triple (a, b, c) is determined in at most two ways. The total number of monic, A 3 cubics with |a| A is therefore bounded above by

A remarkable symmetry
In this section, we establish (1.9). Theorem 1.10 tells us that if f is irreducible and G f is isomorphic to D 4 , V 4 or C 4 if and only if the cubic resolvent has an integer root. Moreover, it follows from the triangle inequality that if H 150, f ∈ S H and r(x) = 0 then |x| 2H. The proposition below therefore implies (1.9). Then We set about proving this. Multiplying (3.1) by 4, we obtain We begin with the case in which both sides of (3.3) are 0. For each c there are at most τ (2c) choices of (x, a). Therefore, by an average divisor function estimate, the number of choices of (x, a, c) is O(H log H). Having chosen x, a, c with xa = 2c, there are then O(H) possible (d, e). We conclude that the number of solutions for which xa = 2c is O (H 2 log H). It remains to treat solutions for which xa = 2c. Write This forces a 2 − 4e = uw 2 and xa − 2c = ±uvw for some w ∈ N. Our strategy will be to upper bound the number of lattice points (u, v, w, x, a) with u = 0 in the region defined by |x|, |a| 2H and This is more than adequate, so in the sequel we assume that uv 2 > 40H. Now (3.4) implies that x v √ u. There are v choices of w, and since As w v, we now have In particular, there are possibilities for a. We obtain the upper bound completing the proof.

A construction
In this section, we establish (1.10). Our construction is motivated by the previous section. Let δ be a small positive constant. We shall choose positive integers x, a, u, w ≡ 12 mod 18, v≡ 4 mod 6 with u squarefree, in the ranges Let us now bound from below the number of choices (u, v, w, x, a). If we choose u, v ∈ N with u H 2−2δ , v 99 and We compute that Observe that the conditions for some constant c 0 > 0. Partial summation now gives We claim that the polynomial f defined by (1.5) lies in S H , and that G f is isomorphic to D 4 , V 4 or C 4 . We now confirm this claim.
Plainly |a| H. Moreover, since and similarly 0 < 4(b − x) < H. Now the triangle inequality gives |b| x + H/4 < H. Finally, we check that  v, w, x, a). Suppose the quadruple (a, b, c, d) is obtained via this construction. Then x is a root of the cubic resolvent of f , so there are at most three possibilities for x. Since u, v, w ∈ N with u squarefree, the equations now determine the triple (u, v, w). Thus, a quadruple (a, b, c, d) can be obtained from (u, v, w, x, a) in at most three ways via our construction, and so we've constructed at least a constant times H 2 (log H) 2 polynomials in this way. This completes the proof of (1.10).

V 4 and C 4 quartics
In this section we prove Theorem 1.8, and thereby also establish Theorem 1.7. From §3, we know that if f ∈ S H and G f is isomorphic to

V 4 quartics
By Theorem 1.10, the discriminant Δ of f is a square. We have the standard formula [16, §14.6] We make the substitutions (5.1) using the software Mathematica [29], obtaining the factorisation Note that the denominator of the left hand side is non-zero, for the irreducibility of f implies that Δ = 0. We now equate the right hand side with y 2 , for some y ∈ Z. Given u, v, w, a, the integer point (x, y) must lie on one of the two curves C ± u,v,w,a defined by Therefore N V 4 is bounded above, up to a multiplicative constant, by H 2 log H plus the number of sextuples (u, v, w, x, a, y) ∈ N 3 × Z 3 satisfying |x|, |a| 8H, (3.4), (3.5), (3.6) and (x, y) ∈ C + u,v,w,a ∪ C − u,v,w,a . We first consider the contribution from (u, v, w, a) for which C ± u,v,w,a is reducible over Q. In this case is a square in Q[x], so 4a 2 uw 2 + 64uv 2 ± 32auvw = 0.
The contribution from this case is therefore bounded above by a constant times Start by choosing u, w for which 40H < uw 2 H 2 . There are then and then v is determined by (5.4) in at most two ways. Now so the number of possibilities for x is bounded above by a constant times Thus, the contribution from this case is bounded above by a constant times We have shown that there are O(H 2 log H) sextuples satisfying |x|, |a| 8H, (3.4), (3.5), (3.6) and (5.3) such that C ± u,v,w,a is reducible over Q.
It remains to address the situation in which C ± u,v,w,a is absolutely irreducible. We will ultimately apply Vaughan's uniform count for integer points on curves of this shape [28, Suppose w v and uv 2 40H. Then x, a √ H, so the number of solutions is bounded above by a constant times Similarly, if v w and uw 2 40H then there are O(H 2 log H) solutions. Next, we consider the scenario in which w v and uv 2 > 40H. Using (3.4), this implies As |x| > 1 2 v √ u and w v, we now have Since we arrive at the inequality In particular, given u, v, w there are Choose u, v, w ∈ N and a ∈ Z such that C ± u,v,w,a is absolutely irreducible. Note (5.5), and put L = 12H v √ u + 1. Now [28, Theorem 1.1] reveals that (5.3) has O(L 1/2 ) solutions (x, y), with an absolute implied constant. As w v, the number of solutions is therefore bounded by a constant multiple of The final case, wherein v w and uw 2 > 40H, is very similar to the previous one. We have considered all cases, and conclude that

C 4 quartics
We follow a similar strategy to the one that we used for V 4 . The root of the cubic resolvent is x, so from Theorem 1.10 we find that (x 2 − 4d)Δ is a perfect square. Observe from (5.1) that x 2 − 4d = uv 2 . Factorising the right hand side of (5.2), we thus obtain for some y ∈ Z. Given u, v, w, a, this defines a pair of curves Z ± u,v,w,a . As u = 0, the curve Z ± u,v,w,a is absolutely irreducible if and only if the curve C ± u,v,w,a defined in (5.3) is absolutely irreducible. The remainder of the proof can be taken almost verbatim from §5.1. We conclude that and this completes the proof of Theorem 1.8. In light of (1.9) and (1.10), we have also completed the proof of Theorem 1.7.

A 4 quartics
In this section, we establish Theorem 1.9. We again use Theorem 1.10, which in particular asserts that A 4 quartics have square discriminant. It remains to show that the diophantine equation disc(X 4 + aX 3 + bX 2 + cX + d) = y 2 has O(H 5 2 + 1 √ 6 +ε ) integer solutions for which |a|, |b|, |c|, |d| H and y ∈ Z \{0}. We have the standard formula [2] Δ := disc(X 4 + aX 3 where I and J are as defined in (1.11). The idea now is to count integer triples (I, J, y) solving (1.12) with I H 2 and y = 0, and to then count quadruples of integers (a, b, c, d)  Fix I, J for which 4I 3 = J 2 . From (1.11), we have Therefore and so In each case, we substituted the form of the line into g(a, c, d) = 0 and expanded it as a polynomial in t. Equating coefficients then provided seven equations.
In Case I, we used the software Mathematica [29] to obtain 4I 3 −J 2 = 0 by elimination of variables. The proof reveals, in fact, that there are no complex lines, but all we need is for there to be no rational lines. Here is the code. In Case II the t 4 coefficient is −729, and in Case III the t 3 coefficient is −110592, so these cases can never occur. We have deduced 4I 3 − J 2 = 0 from the existence of a rational line, completing the proof.
Observe that Y I,J is the zero locus of the polynomial Lemma 6.2. The affine surface Y I,J is absolutely irreducible.
Proof. Assume for a contradiction that Y I,J is not absolutely irreducible. Then there exist polynomials f 0 (a, c), g 0 (a, c), and h 0 (a, c), defined over Q, for which 0 (a, c)).

Construction for V 4
Consider where b, t ∈ N with b ≡ 0 mod 4, t≡ 1 mod 4 Observe that the cubic resolvent splits into linear factors over the rationals. If we can show that f is irreducible over Q, then it will follow from Theorem 1.10 that G f V 4 . Plainly f (x) > 0 whenever x ∈ R, so f (X) has no rational roots, and therefore no linear factors. Suppose for a contradiction that f (X) is reducible. Then by Gauss's lemma f (X) = (X 2 + pX + q)(X 2 + rX + s), for some p, q, r, s ∈ Z. Considering the X 3 coefficient of f gives r = −p.
We begin with the case p = 0. Then considering the X coefficient of f gives s = q. Now It remains to consider the case p = 0. Now In particular b 2 − 4t 2 is a square, which is impossible because b 2 − 4t 2 ≡ 12 mod 16.
Both cases led to a contradiction. Therefore f is irreducible, and we conclude that G f V 4 . Our construction shows that N V 4 H 3/2 .

Construction for A 4
We use a construction motivated by [23,Theorem 1.1]. Consider the family of quartic polynomials Next, consider the cubic resolvent of f , given by This is also irreducible in Z[X, u, v], as r 1,1 (X) = X 3 − 18X 2 − 4X + 8 is irreducible in Z[X]. Hence, by Hilbert's irreducibility theorem [6, Theorem 2.5], almost all specialisations u, v ∈ N with u, v √ H/5 give rise to an irreducible f (X) ∈ Z[X] whose cubic resolvent is also irreducible. Finally, a short calculation reveals that disc(f (X)) = (16(27uv 4 + u 3 )) 2 , so these polynomials have Galois group G f A 4 . They are distinct, so N A 4 (H) H.
degree n. We may assume, for ease of notation, that H is an integer. It may help the reader to know that where ζ(·) denotes the Riemann zeta function and k n denotes the Euclidean volume of the region R ⊂ R n−1 defined by As Chela explains from the outset, van der Waerden had already shown that the number of f given by (1.1) having a factor of degree k ∈ [2, n/2] with |a i | H for all i is O(H n−2 log H). Thus, we need only to count polynomials with a linear factor X + v, so suppose that there are T (v) of these.
To deal with the issue of over-counting, Chela bounds the number of polynomials with at least two (not necessarily distinct) linear factors. Chela's reasoning is that these polynomials have a quadratic factor, and if n 4 then this reveals that there are O(H n−2 ) such polynomials. In the case n = 3 this reasoning breaks down, but a standard mean value estimate for the arithmetic function To this end, since if X − 1 divides f (X) then 0 = f (1) = 1 + a 1 + · · · + a n , the final task is to count polynomials with a 1 + · · · + a n = −1.
For h ∈ Z and N ∈ Z 2 , write L(N, h) for the number of vectors (a 1 , . . . , a n ) ∈ Z n such that max i |a i | N, a 1 + · · · + a n = h. The quantity L(N, 0) equivalently counts lattice points (a 1 , . . . , a n−1 ) in the region N R, so by standard geometry of numbers [24, Lemma 1] we obtain (B.1).

Appendix C. Binary quartic forms with given invariants
In this appendix, we prove the following result related to Lemma 6.1. In words, it asserts that given I, J ∈ C for which the discriminant 4I 3 − J 2 is non-zero, the space of binary quartic forms with these invariants contains no complex lines. A rational line on the variety induces a complex line on the variety, by equating coefficients, so a consequence is that there are no rational lines. See [2] for further information about the invariants I and J of a binary quartic form aX 4 + bX 3 Y + cX 2 Y 2 + dXY 3 + eY 4 . In each case, we expanded the expressions for I and J as polynomials in t. Equating coefficients then provided seven equations, and we used the software Mathematica [29] to obtain 4I 3 − J 2 = 0 by elimination of variables. For example, in Case I, the code is as follows. Cases II, IV, and V also lead to 4I 3 − J 2 = 0, whilst Case III can never occur. We have deduced 4I 3 − J 2 = 0 from the existence of a complex line, completing the proof of the theorem.