Two Families of Monogenic $S_4$ Quartic Number Fields
Hanson Smith


















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Hanson Smith]Hanson Smith

Department of Mathematics, University of Colorado,
Campus Box 395, Boulder, Colorado 80309-0395
hanson.smith@colorado.edu



[2010]11R04, 11R09 11R16







  Consider the integral polynomials f_a,b(x)=x^4+ax+b and g_c,d(x)=x^4+cx^3+d. Suppose f_a,b(x) and g_c,d(x) are irreducible, b| a, and the integers b, d, 256d-27c^4, and 256b^3-27a^4(256b^3,27a^4) are all square-free. Using the Montes algorithm, we show that a root of f_a,b(x) or g_c,d(x) defines a monogenic extension of  and serves as a generator for a power basis of the ring of integers. In fact, we show monogeneity for slightly more general families. Further, we obtain lower bounds on the density of polynomials generating monogenic S_4 fields within the families f_b,b(x) and g_1,d(x). 


[
    [
    December 30, 2023
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§ INTRODUCTION AND OVERVIEW OF RESULTS


Let K be a number field and let _K be its ring of integers. If there exists a monic irreducible polynomial f(x)∈[x] with a root θ such that [θ]=_K, then we say K is monogenic. In other words, K is monogenic if _K admits a power integral basis. A quantity related to monogeneity is the field index. The field index is defined to be the pair-wise greatest common divisor _α∈_K[_K:[α]]. Note that K can have field index 1 and still not be monogenic. Define the minimal index to be min_α∈_K[_K:[α]]. Monogeneity is equivalent to having minimal index equal to 1.
 
Many of the number fields we are most familiar with are monogenic. For example, all quadratic extensions and cyclotomic extensions are monogenic. An example of a non-monogenic field, due to Dedekind <cit.>, is the field obtained by adjoining a root of x^3-x^2-2x-8 to . The problem of classifying monogenic number fields is often called Hasse's problem, as it is believed to have been posed to the London Mathematical Society by Helmut Hasse in the 1960's. See the remark on page 193 of <cit.>. 

We can now state concise, less-general versions of our main results. The following are consequences of Theorems <ref> and <ref> and Theorems <ref> and <ref>, respectively.

 Consider f_b,b(x)=x^4+bx+b with b∈ and let θ be a root. Suppose b and 256-27b are square-free and b≠ 3,5. Then, (θ) is a monogenic S_4 quartic field and θ is a generator of a power integral basis. Further, at least 29.18% of b∈ satisfy these conditions.


 Consider g_1,d(x)=x^4+x^3+d with d∈ and let τ be a root. Suppose d and 256d-27 are square-free and d≠ -2. Then (τ) is a monogenic S_4 quartic field and τ is a generator of a power integral basis. Further, at least 41.849% of d∈ satisfy these conditions.


Heuristically, the best possible percentages in Corollaries <ref> and <ref> seem to be 55.3%. See Remark <ref>. 



§ PREVIOUS WORK


Before a more detailed exposition of our work, we list some results pertaining to Hasse's problem. It has been shown that almost all abelian extension of  with degree coprime to 6 are not monogenic; see Gras <cit.>. Gassert <cit.> shows that all fields obtained by adjoining a root of x^n-a, where a is square-free and a^p≢a modulo p^2 for all primes p| n, are monogenic. In <cit.>, Jones and Phillips identify infinitely many monogenic fields coming from polynomials of the shape x^n+a(m,n)x+b(m,n), where a(m,n) and b(m,n) are prescribed forms. They consider two families of forms, one yielding Galois group S_n and the other A_n. Recently, Bhargava, Shankar, and Wang <cit.> have shown that the density of monic, irreducible polynomials f(x)∈[x] such that a root, θ, of f(x) yields a power basis for the ring of integers of (θ) is 6/π^2=ζ(2)≈ 60.79%. In the same paper, they also show that the density of monic integer polynomials with square-free discriminants is 

    ∏_p (1-1p+(p-1)^2p^2(p+1))≈ 35.82%.

Note that these polynomials are a subset of the monic, irreducible polynomials f(x)∈[x] such that a root, θ, yields a power basis for the ring of integers of (θ). 

Many of the approaches to Hasse's problem have focused on fields with a given Galois group. We summarize the state of the art for degree 4 number fields. For a nice treatise on approaches to monogeneity using index form equations, see Gaál's book <cit.>, “Diophantine Equations and Power Integral Bases." In particular, Chapter 6 deals with the quartic case. A general algorithm for solving the quartic index form equations is presented, with the author expanding upon specific cases. 

A biquadratic field is an extension having /2×/2 as the Galois group. A biquadratic extension can be written (√(m),√(n)). Gaál, Pethő, and Pohst <cit.> parametrize the field indices that can occur based on congruence conditions on m and n. Gras and Tanoé's article <cit.> gives necessary and sufficient conditions for a biquadratic field to be monogenic. Jadrijević <cit.> describes the minimal and field indices of the two families (√((c-2)c),√((c+2)c)) and (√((c-2)c),√((c+4)c)). This investigation is continued for the family (√((c-2)c),√((c+4)c)) in <cit.>. When c and c+4 are square-free, Gaál and Jadrijević <cit.> show (√(2c),√(2(c+4))) is not monogenic, compute an integral basis, and determine the elements of minimal index. 

Dihedral quartic fields have received a significant amount of attention. In <cit.>, Huard, Spearman, and Williams compute the discriminant and an integral basis of quartic fields with quadratic subfields. Further, they find infinitely many monogenic D_8 fields. Specifically, they show that for each square-free c there are infinitely many fields of the form (√(c),√(a+b√(c))) that are monogenic. Gaál and Szabó <cit.> solve index form equations to show that the power integral bases found in <cit.> are the only possible power integral bases. In <cit.>, Kable resolves the question of monogeneity when the D_8 field in question has an imaginary quadratic subfield and establishes some bounds in all cases. Using their algorithm from <cit.>, Gaál, Pethő, and Pohst <cit.> compute “small" indices of totally real quartic fields with Galois group either /4 or D_8 and discriminant of absolute value less than 10^6. The indices may not be minimal since the algorithm they implemented checks only for solutions to the index form equation with absolute value less than 10^6.

A pure quartic field is a field obtained by adjoining a root of a polynomial of the form x^4-a to . In <cit.>, Funakura gives necessary and sufficient conditions for pure monogenic quartic fields. Gaál and Remete <cit.>, characterize the only power integral bases of a number of infinite families of pure quartic fields using binomial Thue equations and extensive calculations on a supercomputer. 

The simplest quartic fields are given by a root of x^4-tx^3-6x^2+tx+1, where t≠±3,0. They are totally real with Galois group /4. If t^2+16 is not divisible by an odd square, Olajos <cit.> has shown that the only two simplest quartics that are monogenic occur when t=2 and t=4. In <cit.>, Gras shows there are only two monogenic imaginary cyclic quartic fields. These are (ζ_5) and (ζ_16-ζ_16).

For A_4 fields, Spearman <cit.> shows x^4+18x^2-4tx+t^2+81 defines an infinite family of monogenic fields when t(t^2+108) is square-free. 

With <cit.>, Gaál considers five families of totally complex quartic polynomials. The polynomials are shown to be irreducible and the Galois groups are classified; A_4, D_8, /4, and /2×/2 all occur. Further, Gaál computes all power integral bases of the orders generated by the roots. 

As for S_4 quartics, work by Bérczes, Evertse, and Győry <cit.> restricts multiply monogenic orders. A recent paper <cit.> by Gassert, Smith, and Stange shows x^4-6x^2-tx-3 with t+8 and t-8 square-free defines an infinite family of monogenic S_4 quartic fields. It is worth noting that the methods of <cit.> are distinct from much of the other literature in that arithmetic properties of elliptic curves are central to proving monogeneity.



§ RESULTS

In this paper we identify two families of monogenic quartic fields:


Let a and b be integers such that 256b^3-27a^4(256b^3,27a^4) is square-free. Suppose that f_a,b(x)=x^4+ax+b is irreducible and let θ be a root. Further, suppose every prime, p, dividing (256b^3,27a^4) satisfies one of the following conditions:



  * p divides a and b, but p^2 does not divide b.


  * p=2, p b, and (a,b) is congruent to one of the following pairs in /4×/4:  (0,1), (2,3).


  * p=3, p a, and (a,b) is congruent to one of the following pairs in /9×/9:  (1,3), (1,6), (2,0), (2,3), (4,0), (4,6), (5,0), (5,6), (7,0), (7,3), (8,3), (8,6). 
 

Then, (θ) is monogenic and θ is a generator of the ring of integers.

 

Let c and d be integers such that d is square-free and 256d-27c^4 is not divisible by the square of an odd prime. If 4|(256d-27c^4), we require that (c,d) is congruent to either (0,1) or (2,3) in /4×/4. Suppose that g_c,d(x)=x^4+cx^3+d is irreducible and let τ be a root. Then, (τ) is monogenic and τ is a generator of the ring of integers.
 

If we restrict the above families we can classify the Galois groups and analyze densities. Note the infinitude of the restricted families below shows the more general families described above are infinite.
 

With the notation as in Theorem <ref>, consider f_b,b(x)=x^4+bx+b. Suppose the coefficients of f_b,b(x) satisfy the conditions given in Theorem <ref>. If b≠ 3,5, then (θ) has Galois group S_4. Moreover, the density of polynomials satisfying Theorem <ref> among polynomials of the form x^4+bx+b with b∈ arbitrary is at least 51-4π^24π^2≈ 29.18%.



With the notation as in Theorem <ref>, consider g_1,d(x)=x^4+x^3+d. Suppose the coefficients of g_1,d(x) satisfy the conditions given in Theorem <ref>. If d≠ -2, then (τ) has Galois group S_4. Moreover, the density of polynomials satisfying Theorem <ref> among polynomials of the form x^4+x^3+d with d∈ arbitrary is at least 14-π^2π^2≈ 41.85%.


The primary reason for choosing the restricted families in the above theorems was so that we could easily analyze their densities. Within the larger class of polynomials which we prove yield monogenic fields, one can find other restrictions on the coefficients that yield families with a specific Galois group. However, in these cases studying density becomes more difficult, as one is concerned with square-free values of higher degree polynomials. Our methods could achieve similar results for polynomials of the form x^4+ax^2+b or x^4+c. However, these families have already been well-studied.

The outline of our paper is as follows:  To prove Theorems <ref> and <ref> our main tool is the Montes algorithm, which we will briefly describe in Section <ref>. In Section <ref>, we show that the restricted families are irreducible and have Galois group S_4. Section <ref> is concerned with applying the Montes algorithm to prove monogeneity. Lastly, in Section <ref>, we analyze the densities of our restricted families. 



 §.§ Acknowledgements
  The author would like to thank Katherine Stange and Alden Gassert for their help and encouragement. The author would also like to thank Sebastian Bozlee for the careful proofreading. 



§ THE MONTES ALGORITHM

We prove monogeneity with a simple application of the Montes algorithm. We follow <cit.> for our exposition of the algorithm. Those interested in more general situations are advised to consult <cit.>. For the purposes of our work, the goal of the Montes algorithm is to compute the p-adic valuation v_p([_K:[θ]]). 

We begin by fixing notation. Let f(x)∈[x] be monic and irreducible, θ a root of f(x), K=(θ), _K the ring of integers of K, and p a prime in . We extend the p-adic valuation on  to [x] in the following manner. If g(x)=b_0+b_1 x+⋯ +b_k x^k, define v_p(g(x))=min_0≤ j≤ k(v_p(b_j)).

Now we describe a version of the Montes algorithm. Consider the reduction of f(x) modulo p. Let ϕ(x) be an irreducible factor of f(x) modulo p and let ϕ(x) be a lift of ϕ(x) to [x]. We may write

    f(x)=a_0(x)+a_1(x)ϕ(x)+⋯ +a_r(x)ϕ(x)^r

where a_i(x)∈[x] has degree strictly less than (ϕ(x)). We call this the ϕ-adic development of f. To any coefficient, a_i(x), of the ϕ-adic development of f we attach the point (i, v_p(a_i(x))) in the plane. The lower convex envelope of these points is called the ϕ-Newton polygon of f. The polygon determined by the sides of the ϕ-Newton polygon with negative slope is called the principal ϕ-polygon of f. We denote this polygon by N. The integer lattice points on or below N contain the arithmetic information we are interested in. Specifically, the ϕ-index of f is (ϕ) times the number of points in the plane with integral coordinates that lie on or below N, strictly above the x-axis, and strictly to the right of the y-axis. We denote this number, the number of points in the integer lattice satisfying the above conditions, by _ϕ(f). 


To illustrate how the ϕ-Newton polygon is obtained, consider f(x)=x^6+3x^5+x^4+15x^3+9x^2+18x+27. We reduce modulo 3 and obtain x^4(x^2+1). Working with the irreducible factor x, we take the lift x and the x-adic development is again our original polynomial f(x)=x^6+3x^5+x^4+15x^3+9x^2+18x+27. Now the x-Newton polygon is:


[scale=1]
    [black, very thick] (4,0) – (6,0);  
    
    [<->,thick] (0,4)
        |- (7,0);
    [black, thick] (0,3) – (1,2);
    [black, thick] (1,2) – (4,0);
    [black, thick] (4,0) – (6,0);    
    [black] (0,3) circle (2pt) node[anchor=east] (0,3);
    [black] (1,2) circle (2pt) node[anchor=east] (1,2);
    [black] (2,2) circle (2pt) node[anchor=west] (2,2);
    [black] (3,1) circle (2pt) node[anchor=south] (3,1);
    [black] (4,0) circle (2pt) node[anchor=north] (4,0);
    [black] (5,1) circle (2pt) node[anchor=south] (5,1);
    [black] (6,0) circle (2pt) node[anchor=north] (6,0);
    [red] (2,1) circle (2pt) node[anchor=north] (2,1);
    [red] (1,1) circle (2pt) node[anchor=north] (1,1);
    [] at (3.5,-1) The x-Newton polygon for f(x);



The principal x-polygon merely excludes the side between (4,0) and (6,0). Further, accounting for (1,1), (2,1), and (1,2), we see _x(f)=3.


Continuing with our description of the Montes algorithm, to any integral x-coordinate 0≤ i≤ r of the principal ϕ-polygon N, we attach the residual coefficient c_i∈_p[x]/ϕ(x), defined to be

    c_i={[                                         0, if  (i,v_p(a_i(x)))  lies strictly above N;                                                                     or v_p(a_i(x))=∞.;                                           ;            a_i(x)p^v_p(a_i(x))∈_p[x]/ϕ(x),              if (i,v_p(a_i(x))) lies on N. ]}

Note we have covered all cases since (i,v_p(a_i(x))) cannot lie below N, as N is the lower convex hull of the (i,v_p(a_i(x))).

Let S be one of the sides of N. Suppose S has slope λ=-he where h,e are positive, coprime integers. Define the length of S, denoted l, to be the length of the projection onto the x-axis. The ramification index of S is e, the denominator of λ. The degree of S, denoted d, is le.

Let t be the x-coordinate of the initial vertex of S. We define the residual polynomial attached to S to be 

    R_λ(f)(y)=c_t+c_t+ey+⋯+c_t+(d-1)ey^d-1+c_t+dey^d∈_p[x]/ϕ(x)[y].

 

Now we state the Theorem of the index, our key tool in proving monogeneity. This is Theorem 1.9 of <cit.>.

Choose monic polynomials ϕ_1,…, ϕ_k whose reduction modulo p are the different irreducible factors of f(x). Then, 
    v_p([_K:[θ]])≥_ϕ_1(f)+⋯ + _ϕ_k(f).

Further, equality holds if and only if, for every ϕ_i, each side of the principal ϕ_i-polygon has a separable residual polynomial.


 The Montes algorithm is concerned with separability. With the notation as above, suppose f(x)≡γ(x)ψ(x) modulo p where γ(x) is separable and (γ(x),ψ(x))=1. Then, γ(x) contributes nothing to v_p([_K:[θ]]). To see this, let η(x) be an irreducible factor of γ(x) and consider the η(x)-adic development of f(x):

    f(x)=a_0(x)+a_1(x)η(x)+⋯ +a_r(x)η(x)^r.

Because f(x) has only one factor of η(x) modulo p, we note p a_1(x). Hence the principal η-polygon has only one side and that side terminates at (1,0). Thus _η(f)=0. Furthermore, the residual polynomial will be separable since linear polynomials are always separable. 





§ GALOIS GROUPS AND IRREDUCIBILITY

Consider the two families f_a,b(x)=x^4+ax+b and g_c,d(x)=x^4+cx^3+d. These polynomials have discriminants Δ_f=b^3(256-27a^4b) and Δ_g=d^2(256d-27c^4). To prove monogeneity, we require the conditions outlined in Theorems <ref> and <ref>. However, to obtain families with Galois group S_4, we impose further restrictions. Namely, we require a=b≠ 3,5 for f_a,b(x) and c=1, d≠ -2 for g_c,d(x). There are less restrictive S_4 families, but we have chosen these parameters so that we can analyze the densities of these families.

In this section we are concerned with proving the first claims of Theorems <ref> and <ref>, which we restate.


The polynomials f_b,b(x)=x^4+bx+b and g_1,d(x)=x^4+x^3+d where b, d, 256-27b, and 256d-27  are square-free, b≠ 3,5, and d≠ -2 are irreducible and have Galois group S_4.


Before proving Theorem <ref>, we state two results we will need. We begin with some definitions. Given a quartic polynomial h(x)=x^4+a_3x^3+a_2x^2+a_1x+a_0 with roots α_1,α_2,α_3,α_4, we define the resolvent cubic to be

    R_h(y)=y^3-a_2y^2+(a_3a_1-4a_0)y-a_3^2a_0-a_1^2+4a_2a_0.


R_h has roots α_1α_2+α_3α_4, α_1α_3+α_2α_4, and α_1α_4+α_2α_3. Given h(x), a depressed quartic is obtained by the substitution x=X-a_34 and has the form

    h_dep(X)=X^4+(-3a_3^28+a_2)X^2+(a_3^38-a_3a_22+a_1)X+(-3a_3^4256+a_3^2a_216-a_3a_14+a_0).

If we have a depressed quartic h_dep(x)=x^4+b_2x^2+b_1x+b_0, we define the resolvent cubic to be 

    R_h,dep(z)=z^3+2b_2z^2+(b_2^2-4b_0)z-b_1^2.

Though R_h(y) and R_h,dep(z) are both called the resolvent cubic, they are actually different polynomials even if h(x) is depressed to begin with. More specifically, the substitution y=z-a_3^24+a_2 sends R_h(y) to R_h,dep(z). Thus R_h has a root in  if and only if R_h,dep has a root in .

Now we recall a classical theorem. One can see <cit.> for a clear, elementary exposition.

With the notation as above, h(x) factors into quadratic polynomials in [x] if and only if at least one of the following hold:


  *  R_h,dep has a nonzero root in ^2. That is, R_h,dep has a root that is the square of a nonzero rational number.

  *  b_1=0 and b_2^2-4b_0∈^2.



We will also use the following result of Kappe and Warren <cit.> to determine the Galois groups. One can also consult <cit.> for a nice exposition with ample examples.

 Let h(x) be a quartic polynomial that is irreducible over  and let Δ_h be the discriminant. Further, let G_h be the Galois group of h. Then, with the notation as above, the first two columns of the following table imply the third column.


 
 Δ_h     R_h     G_h

 not a square     irreducible     S_4 
 
 a square     irreducible     A_4 
 
 not a square     reducible     D_8 or /4 

 a square     reducible     /2×/2 
 
 

 


We proceed with the proof of Theorem <ref>.
 
We begin with the irreducibility f_b,b(x)=x^4+bx+b. If b is not ± 1, then f_b,b is Eisenstein at any prime dividing b. If b=± 1, then the rational root test shows that there is not a root in . To show f_± 1,± 1 does not split into quadratic factors we consider R_f_±1,±1,dep(z)=z^3∓ 4z-1. The rational root test shows R_f_±1,±1,dep does not have a root in , let alone ^2. Since b_2=±1, Theorem <ref> shows f_±1,±1 is irreducible. Note that since R_f_±1,±1,dep is irreducible, R_f_±1,±1 is irreducible.

It remains to consider R_f_b,b=y^3-4by-b^2 for b≠± 1. Suppose we have a root k. The rational root test shows k∈ divides b^2. If p is a prime divisor of b, then reduction modulo p shows p divides k. Since b is square-free, b divides k. Write k =bk_0. We have b^2(bk_0^3-4k_0-1)=0. Thus, bk_0^3-4k_0-1=0. If any prime divides k_0, reduction modulo that prime yields a contradiction. Hence, k_0=± 1. If k_0=1, then b=5, and if k_0=-1, then b=3. Therefore, if b≠ 3,5, then R_f_b,b is irreducible.  

For g_1,d(x)=x^4+x^3+d, suppose we have a root l. By Gauss's lemma, l∈. We have l^3(l+1)=-d. Since d is square-free, we must have l=1 and d=-2. Thus for d≠ -2, we conclude g_1,d does not have a root in . To see that g_1,d does not factor into quadratics, we make the change of variables x=X-14 to obtain the depressed quartic 

    X^4-38X^2+18X-3256+d.

Here b_1=18 so condition (<ref>) of Theorem <ref> does not hold. Consider R_g_1,d(y)=y^3-4dy-d. If d≠± 1, then R_g_1,d is Eisenstein at any prime dividing d and hence irreducible. If d=± 1, the rational root test shows R_g_1,d is irreducible. Thus condition (<ref>) of Theorem <ref> does not hold since R_g_1,d has a root in  if and only if R_g_1,d,dep has a root in . We conclude g_1,d is irreducible.

We have demonstrated that, with the conditions given in Theorem <ref>, f_b,b,g_1,d,R_f_b,b, and R_g_1,d are all irreducible. A quick computation shows that Δ_f_b,b and Δ_g_1,d are not squares. Thus, Theorem <ref> shows that f_b,b and g_1,d have Galois group S_4.  



 Let β be a root of R_g_1,d. Note that g_1,d and R_g_1,d both have discriminant d^2(256d-27). The methods of Section <ref> show β generates a power basis for the ring of integers of the cubic field (β) exactly when τ, a root of g_1,d, generates a power basis for the ring of integers of the quartic field (τ). 




§ MONOGENEITY



For the following we recall a classical formula from algebraic number theory. Let K be a number field obtained by adjoining a root, α, of some monic irreducible polynomial h(x)∈[x]. Write 𝒪_K for the ring of integers, (K) for the discriminant of K, and Δ_h for the discriminant of h(x). Let p be a prime. We have

    v_p((K))+2v_p([𝒪_K:[α]])=v_p(Δ_h).

Note this implies any prime dividing [𝒪_K:[α]] also divides Δ_h. Before we proceed with the proof, we recall Theorem <ref>:
  

Let a and b be integers such that 256b^3-27a^4(256b^3,27a^4) is square-free. Suppose that f_a,b(x)=x^4+ax+b is irreducible and let θ be a root. Further, suppose every prime, p, dividing (256b^3,27a^4) satisfies one of the following conditions:



  *  p divides a and b, but p^2 does not divide b.


  *  p=2, p b, and (a,b) is congruent to one of the following pairs in /4×/4:  (0,1), (2,3).


  *  p=3, p a, and (a,b) is congruent to one of the following pairs in /9×/9:  (1,3), (1,6), (2,0), (2,3), (4,0), (4,6), (5,0), (5,6), (7,0), (7,3), (8,3), (8,6). 
 

Then, (θ) is monogenic and θ is a generator of the ring of integers.



Recall Δ_f=256b^3-27a^4. Let p be a prime dividing Δ_f. We will show that v_p([𝒪_(θ):[θ]])=0. First, suppose p|Δ_f, but p(256b^3,27a^4). Since 256b^3-27a^4(256b^3,27a^4) is square-free, we see 

    1=v_p(Δ_f)=v_p(((θ)))+2v_p([𝒪_(θ):[θ]]).

Thus v_p([𝒪_(θ):[θ]])=0. 

So we consider primes p dividing (256b^3,27a^4). Suppose p satisfies condition (<ref>). We apply the Montes algorithm. Considering f_a,b(x) modulo p we obtain x^4. Thus the only irreducible factor we must consider is x. Taking the lift ϕ(x)=x, the principal x-polygon of f_a,b(x) has one side, originating at (0,1) and terminating at (4,0). Thus _x(f_a,b)=0. The residual polynomial attached to this side is y-b/p, which is clearly separable. By Theorem <ref>, v_p([𝒪_(θ):[θ]])=0. 

Now suppose p=2 satisfies condition (<ref>). We apply the Montes algorithm. Note that 2 necessarily divides a, so modulo 2 we have

    f_a,b(x)≡ x^4+b≡ (x+1)^4.

The (x+1)-adic development of f_a,b(x) is 

    f_a,b(x)=(x+1)^4-4(x+1)^3+6(x+1)^2+(a-4)(x+1)+b-a+1.

To show monogeneity, we need _x+1(f_a,b)=0. Thus we want v_2(b-a+1)=1. One checks this is equivalent to the criteria given in condition (<ref>). The residual polynomial is linear and hence separable. Thus, Theorem <ref> tells us v_2([𝒪_(θ):[θ]])=0.

Finally, suppose p=3 satisfies condition (<ref>). We begin applying the Montes algorithm. Note that 3 necessarily divides b, so that modulo 3 we have

    f_a,b(x)≡ x^4+ax≡ x(x^3+a).

From Remark <ref>, the separable factor x contributes nothing to the index.

For the factor (x^3+a), we have two cases:

Case 1: Suppose a≡ 1 modulo 3. Thus f_a,b(x)≡ x(x+1)^3 modulo 3, we take the (x+1)-adic development

    f_a,b(x)=(x+1)^4-4(x+1)^3+6(x+1)^2+(a-4)(x+1)+b-a+1.

In order to have _x+1(f_a,b)=0, we need v_3(b-a+1)=1. This is satisfied by the following pairs (a,b) in /9×/9:  (1,3), (1,6), (4,0), (4,6), (7,0), (7,3). The residual polynomial is linear and hence separable. Applying Theorem <ref>, we conclude v_3([𝒪_(θ):[θ]])=0.

Case 2: Suppose a≡ -1 modulo 3. Thus f_a,b≡ x(x-1)^3 modulo 3, we take the (x-1)-adic development

    f_a,b(x)=(x-1)^4+4(x-1)^3+6(x-1)^2+(a+4)(x-1)+b+a+1.

In order to have _x-1(f_a,b)=0, we need v_3(b+a+1)=1. This is satisfied by the following pairs (a,b) in /9×/9:  (2,0), (2,3), (5,0), (5,6), (8,3), (8,6). The residual polynomials is linear and hence separable. Applying Theorem <ref>, we conclude v_3([𝒪_(θ):[θ]])=0.

Since we have covered all primes dividing the discriminant of f_a,b, we see [𝒪_(θ):[θ]]=1. We conclude that (θ) is monogenic and θ generates the ring of integers. 


Before proving Theorem <ref>, we remind ourselves of the statement:


Let c and d be integers such that d is square-free and 256d-27c^4 is not divisible by the square of an odd prime. If 4|(256d-27c^4), we require that (c,d) is congruent to either (0,1) or (2,3) in /4×/4. Suppose that g_c,d(x)=x^4+cx^3+d is irreducible and let τ be a root. Then, (τ) is monogenic and τ is a generator of the ring of integers.



Recall Δ_g=d^2(256d-27c^4). Let p be a prime dividing Δ_g. We will show v_p([𝒪_(τ):[τ]])=0. First, suppose p|(256d-27c^4), but p d and p≠ 2. By assumption, v_p(256d-27c^4)=1. Hence

    1=v_p(Δ_g)=v_p(((τ)))+2v_p([𝒪_(τ):[τ]]).

Thus v_p([𝒪_(τ):[τ]])=0.

Now suppose p| d. Applying the Montes algorithm, we consider g_c,d(x) modulo p. We have two cases:


Case 1:  Suppose p| c. The reduction of g_c,d(x) is simply x^4, so we only consider the irreducible factor x. Taking the lift ϕ(x)=x, the principal x-polygon of g_c,d(x) has one side, originating at (0,1) and terminating at (4,0). Thus _x(g_c,d)=0. The residual polynomial attached to this side is y-d/p, which is clearly separable. Thus, by Theorem <ref>, v_p([𝒪_(τ):[τ]])=0.

Case 2:  Suppose p c. Modulo p we have

    g_c,d(x)≡ x^4+cx^3≡ x^3(x+c).
 
We treat the irreducible factor x exactly as above. Again, the principal x-polygon is one-sided and the residual polynomial is separable. We conclude _x(g_c,d)=0.

Considering the factor x+c, we note it is separable. From Remark <ref> we see _x+c(g_c,d)=0 and the residual polynomial is separable. We apply Theorem <ref> to see v_p([𝒪_(τ):[τ]])=0.

For the final scenario, suppose 4|(256d-27c^4) and 2 d. Modulo 2 we have

    g_c,d(x)≡ x^4+d≡ (x-1)^4.

Beginning the Montes algorithm, the (x-1)-adic development is 

    (x-1)^4+(c+4)(x-1)^3+(3c+6)(x-1)^2+(3c+4)(x-1)+c+d+1.

To ensure _x-1(g_c,d)=0, we need v_2(c+d+1)=1. One checks this is equivalent to the conditions given in the theorem statement. Finally, if v_2(c+d+1)=1 the residual polynomial is linear and hence separable. Thus, by Theorem <ref>, v_2([𝒪_(τ):[τ]])=0.

Since we have covered all primes dividing the discriminant of g_c,d, we conclude [𝒪_(τ):[τ]]=1. Thus (τ) is monogenic and τ generates the ring of integers. 





§ DENSITY

In this section, we will show the families of monogenic S_4 fields defined by the polynomials f_b,b(x)=x^4+bx+b and g_1,d(x)=x^4+x^3+d with the conditions imposed in Theorem <ref> are infinite. In fact, we will give a lower bound on the density of each family. The families f_b,b and g_1,d are parametrized by b and d respectively. So, by density, we mean the natural density of b∈ or d∈ yielding monogenic fields.

To begin with, it is well-known that the natural density of square-free integers is 

    1ζ(2)=6π^2≈ 60.79%.

See <cit.> for example. Now let S(x;m,k) denote the number of square-free integers that do not exceed x and are congruent to m modulo k. We will also need a result of Prachar from <cit.>:



    S(x;m,k)∼6xπ^2k∏_p| k(1-1p^2)^-1     (x→∞)

for (m,k)=1 and k≤ x^2/3-ϵ. 


Before the proof, we recall the second claim of Theorem <ref>:

The density of monogenic S_4 fields within the number fields defined by f_b,b(x)=x^4+bx+b is at least 51-4π^24π^2≈ 29.18%. 



From Theorem <ref>, to show that there are infinitely many monogenic fields defined by a root of f_b,b, it suffices to show there are infinitely many square-free b such that 256-27b is square-free. The density of square-free b is 6π^2. By Theorem <ref>, the density of square-free numbers congruent to 256 modulo 27 among numbers congruent to 256 modulo 27 is 

    6π^2(1-19)^-1=274π^2.

Thus, at worst, the density of monogenic fields in this family is 

    6π^2-(1-274π^2)=51-4π^24π^2≈ 29.18%.



As above, before beginning the proof, we recall the second claim of Theorem <ref>:

The density of monogenic S_4 fields within the number fields defined by g_1,d(x)=x^4+x^3+d is at least 14-π^2π^2≈ 41.85%.


 
From Theorem <ref>, to show that there are infinitely many monogenic fields defined by a root of g_1,d, it suffices to show there are infinitely many square-free d such that 256d-27 is square-free. The density of square-free d is 6π^2. By Theorem <ref>, the density of square-free numbers congruent to 27 modulo 256 among numbers congruent to 27 modulo 256 is 

    6π^2(1-14)^-1=243π^2.

Thus, at worst, the density of monogenic fields in this family is 

    6π^2-(1-243π^2)=14-π^2π^2≈ 41.85%.




As above, let θ be a root of f_b,b(x)=x^4+bx+b and τ a root of g_1,d(x)=x^4+x^3+d. Computationally, it appears that 55.3% of fields of the form (θ) have θ as a generator of 𝒪_(θ). Likewise, it appears that 55.3% of fields of the form (τ) have τ as a generator of 𝒪_(τ). If (τ) is monogenic, it seems that τ is almost always a generator of the ring of integers, since (τ) appears to be monogenic about 55.3% of the time. However, (θ) seems to be monogenic about 58.7% of the time. Thus there are some cases where (θ) is monogenic, but θ does not generate the ring of integers. We obtained these heuristics using SageMath <cit.> and testing b and d between -2,500,000 and 2,500,000. 






abbrvnat

 
