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A polynomial analogue of Jacobsthal function A. B. Kalmyninab, S. V. Konyagina a Steklov Mathematical Institute of Russian Academy of Sciences, Moscow, Russia b National Research University Higher School of Economics, Moscow, Russia
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     § 1. Introduction and main resultsLet N be a natural number. The Jacobsthal function j(N) is defined as the length of the largest gap between numbers which are coprime to N. In other words,
\begin{equation*}
j(N)=\max_{m}\{\text{for some } x\in \mathbb N \text{ the inequality } (x+i,N)>1 \text{ holds for all }i\leqslant m\}.
\end{equation*}
\notag
Estimates for j(N) are used to establish results on large gaps between consecutive primes. For instance, the current best lower bound for p_{n+1}-p_n is Theorem A (see [1]). Let y\geqslant 19 and Then
\begin{equation*}
j(P(y))\gg y\frac{\ln y\ln\ln\ln y}{\ln\ln y}=(1+o_{y\to \infty}(1))\ln P(y)\frac{\ln\ln P(y)\ln\ln\ln\ln P(y)}{\ln\ln\ln P(y)}.
\end{equation*}
\notag
In particular, for X\geqslant e^{16}, As for the upper bounds for j(N), Iwaniec showed [2] that the following holds. In this paper, we study a polynomial analogue of the Jacobsthal function. Namely, let f(x) be a non-constant polynomial with integer coefficients. Define the function j_f(x) by
\begin{equation*}
j_f(N)=\max_{m}\{\text{for some } x\in \mathbb N \text{ the inequality } (x+f(i),N)>1 \text{ holds for all }i\leqslant m\}.
\end{equation*}
\notag
So, instead of the intervals not containing numbers coprime to N we study shifts of polynomial sequences. Here, we are going to present a lower bound for j_f(P(y)) similar to the classical Rankin’s bound [3]. To formulate the resulting statement, let us first define a number M(f), which one can call an average size of the maximal preimage of a non-zero point under a map f\colon \mathbb F_p\to \mathbb F_p. Definition 1. Suppose that f(x)\in \mathbb Z[x]. If p is a prime number, denote by M_p(f) the maximal number of solutions x\in \mathbb F_p to the equation f(x)=y for fixed y\in \mathbb F_p\setminus\{0\}. The quantity M(f) is defined as a unique M with for all real X\geqslant 2 if such a number M exists.It is easy to see that if f_d(x)=x^d, then M(f_1)=1 and M(f_2)=M(f_3)=2. In the third section, we will express M(f) in terms of certain Galois groups associated with f(x). We compute M(f_d) explicitly for all d and also M(f) for a typical polynomial of degree d. The second section is devoted to the proof of the following bound for j_f(P(y)). Theorem 1. Suppose that, for a polynomial f\in \mathbb Z[x], the factorization of f contains \ell_f distinct linear and h_f distinct non-linear factors. Then, for y\geqslant 19,
\begin{equation*}
\begin{aligned} \, j_f(P(y)) &\gg y(\ln y)^{\ell_f-1}\biggl(\frac{(\ln\ln y)^2}{\ln\ln\ln y}\biggr)^{h_f} \biggl(\frac{\ln y\ln\ln\ln y}{(\ln\ln y)^2}\biggr)^{M(f)} \\ &=(1+o_{y\to\infty}(1))\ln P(y)(\ln\ln P(y))^{\ell_f-1} \\ &\qquad\times\biggl(\frac{(\ln\ln\ln P(y))^2}{\ln\ln\ln\ln P(y)}\biggr)^{h_f} \biggl(\frac{\ln\ln P(y)\ln\ln\ln\ln P(y)}{(\ln\ln\ln P(y))^2}\biggr)^{M(f)}. \end{aligned}
\end{equation*}
\notag
Remark 1. Let us observe that the value of j_f is invariant under shifts by a constant, that is, for all n\in \mathbb Z we have j_f=j_{f+n}. Choosing n=-f(0), we can always force our polynomial to be divisible by x. In particular, for a polynomial f of degree at least 2 we always have
\begin{equation*}
j_f(P(y))\gg y\frac{(\ln\ln y)^2}{\ln\ln\ln y}\frac{\ln y\ln\ln\ln y}{(\ln\ln y)^2}=y\ln y.
\end{equation*}
\notag
Indeed, since M(f)\geqslant 1, for \ell_f\geqslant 2 or \ell_f=1 and h_f\geqslant 1, we get the desired inequality. The last remaining case is \ell_f=1, h_f=0, that is, f=c(ax+b)^d for some a,b,c\in \mathbb Z, ac\neq 0. In the last section we show that M(f_d)=\tau(d)\geqslant 2 for f_d(x)=x^d, d\geqslant 2, hence, in our case, M(f)=M(f_d)\geqslant 2, and the inequality follows. It turns out that the value M(f) always exists and is rational. To formulate our result on M(f), let us first introduce the quantity m(G,H;X). Definition 2. Let G be a finite group acting on a set X and H be a subgroup of G. For g\in G, denote by X^g the set of fixed points of g. Then Theorem 2. For a field L, let L_f denote the splitting field of the polynomial f(x)- y over the field L(y) of rational functions over L. Denote by K the intersection of \mathbb Q_f and \overline{\mathbb Q}, that is, the maximal constant subfield of \mathbb Q_f. Let G_f and G_f^+ be the Galois groups of \mathbb Q_f=K_f over \mathbb Q(y) and K(y), respectively. These groups have a natural action on the set X of all solutions to f(x)=y and G_f^+ may be viewed as a subgroup of G_f fixing K. We have Our study of j_f is motivated by Theorem 5 in [4]. This result gives an upper bound for the least integer \gamma_k>0 such that all the numbers \gamma_k+j^d for 1\leqslant j\leqslant k are not sums of two squares. The problem of estimating \gamma_k was raised by the first two authors. § 2. Proof of the main theoremTo prove our main result, we are going to choose x such that x+f(i) is not coprime to P(y) for many first values of i by choosing residues x_p \mod p for all p\leqslant y. We will choose x_p \mod p for primes p\leqslant y so that, for all i=1,\dots,m there is some p\leqslant y for which f(i)+x_p \equiv 0\pmod{p}. Next, we take x to be a solution of the simultaneous congruences x\equiv x_p\pmod{p} for p\leqslant y. It now follows that j_f(P(y)) \geqslant m. In the proof, we will use the following well-known result on sifted sets. Lemma 1. Suppose that \kappa>0, z\geqslant 2. Assume that \{a_n\} is a sequence of non- negative real numbers such that, for all d \,|\, P(z), where g(d) is a multiplicative function with g(p)\leqslant \kappa, g(p)<p for all primes p, |r_d|\leqslant g(d) and z\ll X. Then whereThis result is a version of the fundamental lemma of sieve theory (see, for example, Theorem 2.2 in [5]). In particular, we have the following corollary. Corollary 1. Let \kappa, z, g(d), V(z), and X be as above. Suppose that, for any p\leqslant z, the set \Omega_p\subset \mathbb Z/p\mathbb Z contains g(p) elements. Let S(X,\Omega) be the number of n\leqslant X such that n\, \operatorname{mod} p \notin \Omega_p for all p\leqslant z. Then Proof. To see this, we set, for any p\leqslant z, and let Q(z;p) be the least positive integer solution to the congruences
\begin{equation*}
Q(z;p)\equiv P(z;p) \pmod p\quad \text{and}\quad Q(z;p)\equiv -1 \pmod{P(z;p)}.
\end{equation*}
\notag
Setting
\begin{equation*}
a_m=\begin{cases} 1 &\text{if }m={\displaystyle\prod_{p\leqslant z}\prod_{r \in \Omega_p}\bigl(P(z;p)n+rQ(z;p)\bigr)} \text{ for some }n\leqslant X, \\ 0 &\text{otherwise}, \end{cases}
\end{equation*}
\notag
it is easily seen that (m,P(z))=1 and a_m=1 if and only if the corresponding n does not lie in any \Omega_p for p\leqslant z. The conditions of Lemma 1 also clearly hold. Corollary is proved. We also need the following well-known consequence of Chebotarev’s density theorem. Lemma 2. Let f be an irreducible polynomial with integer coefficients. If r_p(f) is the number of roots of f in \mathbb F_p, then there are two constants c_1,c_2>0 such that, for x\geqslant 3, Proof. Let K be the splitting field of f over \mathbb Q. Galois group G of K over \mathbb Q acts on the set of roots of f, this action is transitive. It is easy to see that, for large enough p, the number r_p(f) is equal to the number of fixed points of this action for the associated Frobenius conjugacy class. Therefore, by Chebotarev’s theorem, the proportion of the primes p for which r_p(f)=k is equal to that of the elements of G with exactly k fixed points. By Burnside’s lemma, the average number of fixed points for a transitive action is equal to 1, therefore, the average of r_p(f) over primes is also 1. Asymptotic formulas with required remainder terms follow from the effective version of Chebotarev’s density theorem for K/\mathbb Q, see Theorem 1.3 in [6]. Lemma 2 is proved. Let us now prove Theorem 1. Proof of Theorem 1. Let A be a large constant and B be an even larger constant. We set
\begin{equation*}
\begin{gathered} \, m=\frac{y}{B}(\ln y)^{\ell_f-1}\biggl(\frac{(\ln\ln y)^2}{\ln\ln\ln y}\biggr)^{h_f} \biggl(\frac{\ln y\ln\ln\ln y}{(\ln\ln y)^2}\biggr)^{M(f)}, \\ z_0=(\ln y)^A,\qquad z_1=\exp\biggl(\frac{\ln\ln \ln y\ln y}{A\ln\ln y}\biggr). \end{gathered}
\end{equation*}
\notag
We are going to choose x_p in three steps. After the first two steps, we compute the number of unsifted numbers i\leqslant m. To be more precise, if we have an already chosen x_p for some values of p, then the number i is called sifted if, among the used primes, there is a p such that For the first step, let us choose
\begin{equation*}
x_p=0 \ \text{ for }\ p\leqslant z_0\ \text{ and for }\ z_1<p<\frac{y}2.
\end{equation*}
\notag
Now, suppose that the number i\leqslant m is not sifted after the first step. Let k(x) be a factor of f(x) of degree 1. Then k(i) has no prime factors p\in [2,z_0]\cup (z_1,y/2). Since k(i)\ll m\ll y(\ln y)^{\ell_f+M(f)}, for large enough A we conclude that |k(i)| is either a prime or a z_1-smooth number. For large enough A, the number of z_1-smooth numbers that are \ll m can be estimated as
\begin{equation*}
\Psi(O(m),z_1)\ll \frac{m}{(\ln y)^{\ell_f+M(f)+2}}=o\biggl(\frac{y}{\ln y}\biggr).
\end{equation*}
\notag
For the second step, let us choose x_p for all z_0<p\leqslant z_1 as follows: by the definition of M(f), for any such p there is a non-zero y_p\in \mathbb F_p\setminus \{0\} such that the congruence f(i)\equiv y_p \pmod p has M_p(f) solutions. We set x_p\equiv -y_p \ (\operatorname{mod} p).
Next, we are going to estimate the number R of i\leqslant m that remain unsifted after these two steps. Our goal will be to establish the inequality for all large y if A and B are taken to be large enough. For p\leqslant \sqrt{y}, let \Omega_p^{\mathrm{I}} consist of all residues t \ \operatorname{mod} p such that k(t)\equiv 0 \ (\operatorname{mod} p) for some linear factor k(x) of f. Next, if p\in [2,z_0]\cup (z_1,\sqrt{y}), we set
\begin{equation*}
\Omega_p^{\mathrm{II}}\,{=}\,\{t \ \operatorname{mod} p\colon \exists\,\text{non-linear irreducible factor }q(x)\text{ of }f(x)\text{ with }q(t)\equiv 0 \ (\operatorname{mod} p)\}.
\end{equation*}
\notag
For p\notin[2,z_0]\cup(z_1,\sqrt{y}), we define \Omega_p^{\mathrm{II}}=\varnothing. Finally, for z_0<p\leqslant z_1 we set
\begin{equation*}
\Omega_p^{\mathrm{III}}=\{t\ \operatorname{mod} p\colon f(t)\equiv y_p\ (\operatorname{mod} p)\}
\end{equation*}
\notag
and \Omega_p^{\mathrm{III}}=\varnothing for other values of p. Set \Omega_p=\Omega_p^{\mathrm{I}}\cup \Omega_p^{\mathrm{II}}\cup \Omega_p^{\mathrm{III}}. We claim that if i\leqslant m is unsifted after two steps, then either i\ll \sqrt{y} or k(i) is z_1-smooth for some linear factor k(x) of f (the number of such i is o(y/\ln y) by the choice of A) or i\mod p\notin \Omega_p for any p\leqslant \sqrt{y}. Indeed, suppose that i\ \operatorname{mod} p\in \Omega_p for some p. There are three cases to consider.
Case 1: k(i) is divisible by p for some linear factor k(x) of f. In this case, either k(i)=\pm p and i\ll \sqrt{y}, or |k(i)| is not a prime number, hence k(i) must be z_1-smooth or else i must be sifted after the first step. Case 2: p\in [2,z_0]\cup (z_1,\sqrt{y}) and for some irreducible non-linear factor q(x) of f(x), we have q(i)\equiv 0\ (\operatorname{mod} p). Then f(i) is also divisible by p, hence p must be sifted after the first step. Case 3: z_0<p\leqslant z_1 and f(i)\equiv y_p. Then i is sifted after the second step. From this we conclude that
\begin{equation*}
R\leqslant S(m,\Omega)+O\bigl(\sqrt{y}+\Psi(O(m),z_1)\bigr)=S(m,\Omega)+o\biggl(\frac{y}{\ln y}\biggr)
\end{equation*}
\notag
by our choice of A. We will now apply Corollary 1 to estimate S(m,\Omega). Let g(p)=|\Omega_p|. If, for some p\leqslant \sqrt{y}, we have g(p)=p, then, clearly, S(m,\Omega)=0. Otherwise notice that all three parts of \Omega_p contain at most d=\deg f elements, hence Corollary 1 is applicable for \kappa=3d. Hence
\begin{equation*}
S(m,\Omega)\ll m\prod_{p\leqslant \sqrt{y}}\biggl(1-\frac{g(p)}{p}\biggr)\ll m\exp\biggl(-\sum_{p\leqslant \sqrt{y}}\frac{g(p)}{p}\biggr).
\end{equation*}
\notag
Now, let k_1(x),\dots,k_{\ell_f}(x) and q_1(x),\dots,q_{h_f}(x) be all the linear and non-linear irreducible factors of f(x), respectively. Each two fixed irreducible polynomials have no common roots modulo large enough primes. Therefore, there is a constant p_0 such that, for p>p_0, the sets \Omega_p^{\mathrm{I}}, \Omega_p^{\mathrm{II}} and \Omega_p^{\mathrm{III}} are pairwise disjoint. Moreover, taking p_0 large enough, we can assure that |\Omega_p^{\mathrm{I}}|=\ell_f for all p>p_0 and |\Omega_p^{\mathrm{II}}|=r_p(q_1)+\dots+r_p(q_{h_f}) for p\in (p_0,z_0]\cup (z_1,\sqrt{y}). Also, by definition, |\Omega_p^{\mathrm{III}}|=M_p(f) for all z_0<p\leqslant z_1. Consequently,
\begin{equation*}
\sum_{p\leqslant \sqrt{y}}\frac{g(p)}{p}=\sum_{p_0<p\leqslant \sqrt{y}}\frac{\ell_f}{p}+\sum_{h\leqslant h_f}\, \sum_{p\in (p_0,z_0]\cup (z_1,\sqrt{y})}\frac{r_p(q_h)}{p}+\sum_{z_0<p\leqslant z_1}\frac{M_p(f)}{p}.
\end{equation*}
\notag
By the Mertens theorem, the first sum is
\begin{equation*}
\sum_{p_0<p\leqslant \sqrt{y}}\frac{\ell_f}{p}=\ell_f \ln\ln y+O(1).
\end{equation*}
\notag
Lemma 2 gives, for every h\leqslant h_f,
\begin{equation*}
\sum_{p\in (p_0,z_0]\cup (z_1,\sqrt{y})}\frac{r_p(q_h)}{p}=\ln\ln y-\ln\ln z_1+\ln\ln z_0+O(1).
\end{equation*}
\notag
Finally, by the definition of M(f), the third sum can be evaluated as
\begin{equation*}
\sum_{z_0<p\leqslant z_1}\frac{M_p(f)}{p}=M(f)(\ln\ln z_1-\ln\ln z_0)+O(1).
\end{equation*}
\notag
Plugging all of this into our estimate, we get
\begin{equation*}
\begin{aligned} \, S(m,\Omega) &\ll m(\ln y)^{-\ell_f}\biggl(\frac{\ln z_1}{\ln y\ln z_0}\biggr)^{h_f} \biggl(\frac{\ln z_0}{\ln z_1}\biggr)^{M(f)} \\ &=m(\ln y)^{-\ell_f}\biggl(\frac{\ln\ln\ln y}{A^2(\ln\ln y)^2}\biggr)^{h_f} \biggl(\frac{A^2(\ln\ln y)^2}{\ln y\ln\ln\ln y}\biggr)^{M(f)}=\frac{A^{2M(f)-2h_f}y}{B\ln y}. \end{aligned}
\end{equation*}
\notag
Since the implied constant does not depend on A and B, we can take B large enough to obtain so that
\begin{equation*}
R\leqslant S(m,\Omega)+o\biggl(\frac{y}{\ln y}\biggr)\leqslant \frac{y}{3\ln y}.
\end{equation*}
\notag
Since for the third step we have \pi(y)-\pi(y/2)=y/((2+o(1))\ln y) primes p\leqslant y left, one can sift one of remaining numbers at a time to ensure that every number i\leqslant m is sifted out in these three steps. This concludes the proof of Theorem 1. § 3. Computation of M(f)In this section, we prove Theorem 2 on the value of M(f). The quantity M(f) is the logarithmic average of M_p(f) over prime values of p. The number M_p(f) is defined as the maximal number of solutions for f(x)=y with x\in \mathbb F_p for a fixed non-zero y\in \mathbb F_p. We will show that M(f) always exists and is a rational number, and we also reduce its computation to group actions. In the introduction, we associated f(x) with several Galois groups, which have a natural action on the set X of all roots of the equation f(x)-y, where y is a formal variable. Namely, G_f is the Galois group of f(x)-y over \mathbb Q(y) and G_f^+ is the Galois group of the same polynomial over K(y). It is easy to see that G_f^+ is a normal subgroup of G_f. Indeed, it is clear that K is a finite normal extension of \mathbb Q. If G_f^0 is its Galois group, then there is a natural surjective homomorphism \pi\colon G_f\to G_f^0. More precisely, every automorphism of the splitting field of f(x)-y over \mathbb Q(y) sends constants to constants. Such an automorphism is an automorphism over K(y) if and only if all the constant elements are fixed, hence G_f^+=\ker \pi is a normal subgroup. Birch and Swinnerton-Dyer [7] were able to reduce the question on the number of solution of f(x)=y to certain other Galois groups. Lemma 3. Let G_p^+ be the Galois group of f(x)-y over \overline{\mathbb F}_p(y) and G_p be the Galois group of the same polynomial over \mathbb F_p(y). For n\leqslant d=\deg f, let m_n be the number of G_p-invariant orbits of action of G_p^+ on ordered n-tuples of distinct roots of f(x)=y. Then, for m_n=0, the system f(x_1)=f(x_2)=\dots=f(x_n) has no solutions in \mathbb F_p with x_i\neq x_j for i\neq j, and, for m_n>0, the number of solutions is For convenience, let us formulate Chebotarev’s density theorem. First, define the Frobenius element. Definition 3. Let K/\mathbb Q be a finite Galois extenstion, \mathcal O_K be its ring of integers, and G_K be its Galois group. Suppose that p is an unramified rational prime number and \mathfrak p\subset \mathcal O_K is a prime ideal lying over p. The Frobenius element of \mathfrak p is an element \sigma_{\mathfrak p}\in G_K with for all x\in \mathcal O_K.For different prime ideals \mathfrak p lying over fixed prime number p Frobenius elements are conjugated, therefore, up to conjugation, we can define the Frobienus element \sigma_p. The following lemma is a consequence of Chebotarev’s density theorem. Lemma 4. Let K/\mathbb Q be a finite Galois extension with Galois group G_K. Suppose that C\subset G_K is a conjugacy class and \pi(x;K,C) is the number of primes p\leqslant x such that the Frobenius element \sigma_p lies in C. Then, for some c_K>0, For a proof, see [6]. Let us now prove Theorem 2. Proof of Theorem 2. In [7], it is proved that, for large enough p, the group G_p^+ can be identified with the Galois group of f(x)-y over \mathbb C(y). It is easy to see that this group coincides with G_f^+. Next, large primes p are unramified in K, so we fix some prime ideal \mathfrak p lying over p in \mathcal O_K and consider the corresponding Frobenius element \sigma_{\mathfrak p}. The largest constant subfield of (\mathbb F_p)_f can be identified with \mathcal O_K/\mathfrak p. Moreover, G_p acts on constants via Frobenius automorphism and its powers. For large p, the group G_p is generated by G_f^+ and the preimage of \sigma_{\mathfrak p}, that is,
\begin{equation*}
G_p=\langle \pi_p^{-1}(\sigma_{\mathfrak p}),G_f^+ \rangle,
\end{equation*}
\notag
where \pi_p is the homomorphism of restriction of G_p-action on constants (its image is the Galois group of \mathcal O_K/\mathfrak p over \mathbb F_p), which allows us to consider G_p as a subgroup of G_f. Let g be any element of \pi_p^{-1}(\sigma_{\mathfrak p}). Let us show that, in this case, the maximal n with m_n>0 is equal to \max_{h\in G_f^+}|X^{hg}|. Indeed, G_f^+ is normal in G_p, hence the latter group is generated by g and G_f^+, so a G_f^+-orbit is G_p-invariant if and only if it is g-invariant. This means that if our orbit takes form G_f^+(x_1,\dots, x_n), then, for any h_1\in G_f^+, there exists h_2\in G_f^+ such that The subgroup G_f^+ is normal, so there exists h_3 with gh_2=h_3g. Therefore, our condition is equivalent to the following: for any h_3 one can find h_1 such that x_1,\dots,x_n are fixed by hg, where h=h_1^{-1}h_3. This proves the desired identity. Thus, for large p the number M_p(f) is equal to \max_{h\in G_f^+}|X^{hg}|, where g\in \pi_p^{-1}(\sigma_{\mathfrak p})\subset \pi_p^{-1}(\sigma_p)=:C and \sigma_p is the conjugacy class of \sigma_{\mathfrak p} in G_f^0. Since G_f^+ is normal, this number does not depend on the choice of g and \mathfrak p. Chebotarev’s density theorem shows that
\begin{equation*}
\begin{aligned} \, \sum_{p\leqslant x}M_p(f) &=\sum_{C, g_0\in C}\frac{|C|}{|G_f^0|}\max_{h\in G_f^+} |X^{h\pi_p^{-1}(g_0)}| \operatorname{li}(x)+O\bigl(x\exp\bigl\{-c_K\sqrt{\ln x}\,\bigr\}\bigr) \\ &=\operatorname{li}(x)\frac{1}{|G_f^0|}\sum_{g_0\in G_f^0}\max_{h\in G_f^+} |X^{h\pi_p^{-1}(g_0)}|+O\bigl(x\exp\bigl\{-c_K\sqrt{\ln x}\,\bigr\}\bigr), \end{aligned}
\end{equation*}
\notag
where C in the first sum runs over all conjugacy classes of G_f^0. Next, G_f^0=G_f/G_f^+ and the value \max_h |X^{hg}| depends only on the class of g\in G_f in the quotient G_f/G_f^+. This means that the last sum is equal to m(G_f,G_f^+;X). Therefore,
\begin{equation*}
\sum_{p\leqslant x}M_p(f)=m(G_f,G_f^+;X)\operatorname{li}(x)+O\bigl(x\exp\bigl\{-c_K\sqrt{\ln x}\,\bigr\}\bigr)
\end{equation*}
\notag
and, by partial summation, we obtain
\begin{equation*}
\sum_{p\leqslant x}\frac{M_p(f)}{p}=m(G_f,G_f^+;X)\ln\ln x+O(1),
\end{equation*}
\notag
which concludes the proof. Theorem 2 is proved. Remark 2. For a few natural examples, the number m(G,H;X) turns out to be integer, at least when H is normal in G and actions of both G and H are faithful and transitive. However, there are examples of non-integrality: the group G=S_4\times \mathbb Z/2\mathbb Z acts transitively on a set X with 6 elements. Taking H=A_4, we obtain m(G,H;X)=7/2. This example was found by P. Kosenko, who also managed to find some examples of non-integrality for |X| equals 8, 9, and 10. Incidentally, S_4\times \mathbb Z/2\mathbb Z is the Galois group of generic even sextic polynomial. Unfortunately, we were not able to produce examples of polynomials f(x) of degree 6 with G_f=S_4\times \mathbb Z/2\mathbb Z and G_f^+=A_4. Since now we have a tool for computation of M(f), let us find M(f) for f_d and for a typical polynomial of degree d. We call a polynomial of degree d generic if G_f^+\cong S_d. Theorem 3. Let d be an integer. (i) M(x^d)=\tau(d), where \tau(d) is the divisor function. (ii) M(f)=d, for any generic polynomial f(x) of degree d. Proof. To prove the first part, we note that, for f(x)=x^d, Indeed, in this case it is easy to see that if y\neq 0 and x^d=y has at least one solution, then the number of solutions is equal to the number or dth roots of unity in \mathbb F_p. Next, using the identity we get, for any x\geqslant 2,
\begin{equation*}
\sum_{p\leqslant x}\frac{M_p(f_d)}{p} =\sum_{p\leqslant x}\frac{(d,p-1)}{p}=\sum_{p\leqslant x} \frac{\sum_{d_1\,|\, (d,p-1)}\varphi(d_1)}{p} =\sum_{d_1\,|\, d}\varphi(d_1)\sum_{\substack{p\leqslant x\\ p\equiv 1\ (\operatorname{mod} {d_1})}}\frac{1}{p}.
\end{equation*}
\notag
Next, from Prime Number Theorem for primes in arithmetic progressions we find
\begin{equation*}
\sum_{\substack{p\leqslant x\\ p\equiv 1\ (\operatorname{mod} {d_1})}}\frac{1}{p}=\frac{\ln\ln x}{\varphi(d_1)}+O(1).
\end{equation*}
\notag
Substituting this identity into the above formula, we get the desired identity: the factors \varphi(d_1) and 1/\varphi(d_1) cancel out and we are left with the sum \sum_{d_1\,|\, d}1=\tau(d). This proof does not rely on Theorem 2, but can be easily reformulated in terms of the natural action of the affine group \mathrm{Aff}(\mathbb Z/ d\mathbb Z) on \mathbb Z/d\mathbb Z. Indeed, in our case K=\mathbb Q(\zeta_d) is the cyclotomic field. Every automorphism of \mathbb Q_f sends \zeta_d to \zeta_d^a for some a\in (\mathbb Z/d\mathbb Z)^* and y to \zeta_d^b y for some b\in \mathbb Z/d\mathbb Z. The map (a,b)\mapsto (x\mapsto ax+b) is an isomorphism G_f\cong \mathrm{Aff}(\mathbb Z/d\mathbb Z). Since G_f^+ fixes \zeta_d, we also see that G_f^+ corresponds to the shift subgroup y\mapsto y+b and is isomorphic to \mathbb Z/d\mathbb Z.
As for generic polynomials, we have G_f^+\cong S_d, hence G_f=G_f^+ and for all g\in G_f as needed. Theorem 3 is proved.The paper [7] contains, among other things, some criteria for f(x) to be generic. In particular, authors prove that if f'(x) has simple roots and f(\alpha)\neq f(\beta) for any two distinct roots, then f(x) is generic. Of course, this means that the coefficients of any non-generic polynomial must satisfy a certain polynomial relation. This, in turn, implies that most polynomials of degree d are generic. It is also clear that our proof of M(f)=d for generic polynomials uses only G_f=G_f^+. Obviously, the latter condition is actually equivalent to M(f)=d. Since G_f^0=G_f/ G_f^+, it is also equivalent to K=\mathbb Q, that is, all the constants in \mathbb Q_f must be rational. Let us call such polynomials ordinary. Currently, we are not aware of any simple way to test whether a given polynomial is ordinary, besides more narrow criteria of Birch and Swinnerton-Dyer. The polynomials that we call generic are called Morse functions by J.-P. Serre in his book on Galois theory (see § 4.4 in [8]). It is also mentioned that Hilbert established an isomorphism G_f\cong S_n for all such polynomials, see also [9]. Similarly to the results on gaps between primes, our results imply the following. Corollary 2. For any f(x)\in \mathbb Z[x], there exists a constant c_f>0 such that, for all X\geqslant e^{16}, there is a positive integer n\leqslant X with \Omega(n+f(m))>A(n) for all 1\leqslant m\leqslant Y, where
\begin{equation*}
Y=c_f\ln X(\ln\ln X)^{\ell_f-1}\biggl(\frac{(\ln\ln\ln X)^2}{\ln\ln\ln\ln X}\biggr)^{h_f} \biggl(\frac{\ln\ln X\ln\ln\ln\ln X}{(\ln\ln\ln X)^2}\biggr)^{M(f)},
\end{equation*}
\notag
\Omega is the number of prime factors counted with multiplicity, and A(n) is the number of irreducible factors of f(x)+n. It is also worth noting that our proofs can be extended to integer-valued polynomials f(x) which are not necessarily in \mathbb Z[x]. AcknowledgementsWe would like to thank Peter Kosenko for providing an abstract example of non-integrality of m(G,H;X) (see Remark 2) and Mathoverflow user Joachim König for providing the main idea for the proof of Theorem 2: https://mathoverflow.net/a/430001/101078. We also thank anonymous referees for correcting several inaccuracies, especially in § 3, and for pointing out the references to Serre and Hilbert. 
Citation in format AMSBIB
\Bibitem{KalKon24} |
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