Finite field extensions with the line or translate property for $r$-primitive elements

Let $r,n>1$ be integers and $q$ be any prime power $q$ such that $r\mid q^n-1$. We say that the extension $\mathbb{F}_{q^n}/\mathbb{F}_q$ possesses the line property for $r$-primitive elements if, for every $\alpha,\theta\in\mathbb{F}_{q^n}^*$, such that $\mathbb{F}_{q^n}=\mathbb{F}_q(\theta)$, there exists some $x\in\mathbb{F}_q$, such that $\alpha(\theta+x)$ has multiplicative order $(q^n-1)/r$. We prove that, for sufficiently large prime powers $q$, $\mathbb{F}_{q^n}/\mathbb{F}_q$ possesses the line property for $r$-primitive elements. In particular, when $r=n=2$ and (necessarily) $q$ is odd, we show that $\mathbb{F}_{q^2} /\mathbb{F}_q$ posseses the line property for 2-primitive elements unless $q \in \{3,5,7,9,11,13,31,41\}$. We also discuss the (weaker) translate property for extensions.


Introduction
Let q be a prime power and n ≥ 2 an integer. We denote by F q the finite field of q elements and by F q n its extension of degree n. It is well-known that the multiplicative group F * q n is cyclic; its generators are called primitive elements. The theoretical importance of primitive elements is complemented by their numerous applications in practical areas such as cryptography.
In addition to their theoretical interest, elements of F * q n that have high order, without necessarily being primitive, are of great practical interest because in several applications they may replace primitive elements. Accordingly, recently researchers have worked on the the effective construction of such high order elements, [10,15,16], since that of primitive elements themselves remains an open problem.
With that in mind, we call an element of order (q n − 1)/r, where r | q n − 1, rprimitive, i.e., the primitive elements are exactly the 1-primitive elements. In this line of work, the existence of 2-primitive elements that also possess other desirable properties has been recently considered [8,13].
We call some θ ∈ F q n a generator of the extension F q n /F q if F q n = F q (θ) and, if θ is a generator of F q n /F q , we call the set the set of translates of θ over F q and every element of this set a translate of θ over F q . We say that an extension F q n /F q possesses the translate property for r-primitive elements, if every set of translates contains an r-primitive element. In particular, for r = 1 we simply call it the translate property. A classical result in the study of primitive elements is the following. Theorem 1.1 (Carlitz-Davenport). Let n be an integer. There exist some T 1 (n) such that for every prime power q > T 1 (n), the extension F q n /F q possesses the translate property.
The above was first proved by Davenport [9], for prime q, while Carlitz [3] extended it to the stated form. Interest in this problem was renewed by recent applications of the translate property in semifield primitivity, [12,17,18].
Let θ be a generator of the extension F q n /F q and take some α ∈ F * q n . We call the set L α,θ := {α(θ + x) : x ∈ F q } the line of α and θ over F q . An extension F q n /F q is said to possess the line property for r-primitive elements if every line of this extension contains an r-primitive element. When r = 1, we refer to this property as the line property. A natural generalization of Theorem 1.1 is the following, [6, Corollary 2.4].
Theorem 1.2 (Cohen). Let n be an integer. There exist some L 1 (n) such that for every prime power q > L 1 (n), the extension F q n /F q possesses the line property.
It is clear that all the sets of translates are actually lines (where α = 1), i.e., the line property implies the translate property. Thus, L 1 (n) ≥ T 1 (n).
In this work, we extend Theorems 1.1 and 1.2 to r-primitive elements, by proving the following. Theorem 1.3. Let n and r be integers. There exist some L r (n) such that for every prime power q > L r (n), with the property r | q n − 1, the extension F q n /F q possesses the line property for r-primitive elements. If we confine ourselves to the translate property for r-primitive elements, the same is true for some T r (n) ≤ L r (n).
Then, in Section 4, we focus on the case in which r = n = 2 and provide complete results. First, by proving Theorem 4.6, we effectively estimate L 2 (2) and T 2 (2) theoretically by a sieving method, cf. [7]. This leaves a small number (around 100) of extensions unresolved, the largest prime power remaining being q = 3541. Then, we employ computational methods (extensive as regards the line property) to deal with the remaining extensions.

Preliminaries
We begin by introducing the notion of freeness. Let m | q n − 1, an element ξ ∈ F * q n is m-free if ξ = ζ d for some d | m and ζ ∈ F * q n implies d = 1. It is clear that primitive elements are exactly those that are q 0 -free, where q 0 is the square-free part of q n − 1. It is also evident that there is some relation between m-freeness and multiplicative order. Throughout this work, a character is a multiplicative character of F * q n , while we denote by χ 0 the trivial multiplicative character. Vinogradov's formula yields an expression for the characteristic function of m-free elements in terms of multiplicative characters, namely: where µ stands for the Möbius function, φ for the Euler function, θ(m) := φ(m)/m and the inner sum suns through multiplicative characters of order d. Furthermore, a direct consequence of the orthogonality relations is that the characteristic function for the elements of F * q n that are k-th powers, where k | q n − 1, can be written as We will use character sums to establish our results. The following estimate is a direct consequence of the main result of [14] and, notably, it is one of the few non-trivial character sum estimates not relying on Weil's results [19].
Proposition 2.2 (Katz). Let θ ∈ F q n be such that F q n = F q (θ) and χ a non-trivial character. Then Let d(R) be the number of divisors of R. The result below provides an asymptotic estimate for this function.  From now on, fix the positive integers r and n and let q be some prime power such that r | q n −1. In addition, suppose that P stands for the set of distinct primes dividing q n − 1. It follows that q n − 1 = p∈P p ap , where a p ≥ 1 for all p ∈ P. Additionally, we have that r = p∈P p bp , where for every p ∈ P, 0 ≤ b p ≤ a p .
We partition P as follows: It is clear that the above sets are pairwise disjoint and that P s ∪ P t ∪ P u = P. It is straightforward to check that r = st and that u is the radical of the part of q n − 1 that is co-prime with r. Lemma 2.1, implies that the set of u-free elements, contains all the σ-primitive elements, where for some 0 ≤ σ p ≤ a p . In addition, the u-free elements that are r-th powers are the σ-primitive elements with σ as in . Note that b pi + 1 ≤ a pi . Now from the set of u-free elements, that are also r-th powers, exclude those that are not f i -th powers for every i = 1, . . . , k. We are left with exactly the σ-primitive elements, where σ is as in ( In particular, in any case In other words, with the notation of Section 2, the characteristic function for r-primitive elements of F * q n can be expressed as where x ∈ F * q n . Moreover, for every i = 1, . . . , k, notice that since e i | f i , we have that an f i -th power is also an e i -th power, i.e., for Next, recall that, for i = 1, . . . , k, e i = p Finally, we insert the above and the expressions (2.1) and (2.2) into (3.3), and obtain (3.4) where x ∈ F * q n and (χ 1 χ 2 ψ 1 · · · ψ λ ) stands for the product of the corresponding characters, itself a character.
Next, fix some α, θ ∈ F * q n , such that F q n = F q (θ). Let N (θ, α) be the number of r-primitive elements of the form α(θ + x), where x ∈ F q . It suffices to show that In addition, notice that the orders of all the factors of the product (χ 1 χ 2 ψ 1 · · · ψ k ) are co-prime. Hence the product itself is trivial if and only if all its factors are trivial. With this in mind, Proposition 2.2 implies that, unless all the characters χ 1 , χ 2 , ψ 1 ,. . . ,ψ k are trivial, In (3.5), we separate the term that corresponds to d 1 = d 2 = δ 1 = . . . = δ k = 1 and, with the above in mind, we obtain Notice that for all Furthermore, it is well-known that, for every d | q n − 1, there exist exactly φ(d) characters of order d. Hence the latter condition can be also written as where we recall that d(m) stands for the number of divisors of m ∈ Z. Now, observe that usf 1 · · · f k | q n − 1, wherefore Proposition 2.3 implies that Further, observe that where the left side of the above inequality depends solely on r. It follows that, for q large enough, (3.7) holds. Hence N (θ, α) = 0. The proof of the first statement of Theorem 1.3 is now complete, while the second statement (about T r (n)) follows immediately from the fact that all the sets of translates in an extension are also lines of this extension.

The translate and line properties for 2-primitive elements in quadratic extensions
We progress to an effective version of Theorem 1.3, when r = n = 2. We begin with providing special cases of some of the preliminary results, presented in Section 2. For example, for n = 2, the following improvement of Proposition 2.2, see [6, Lemma 3.3], holds.
Furthermore, let W (R) be the number of the square-free divisors of R. The following special case of Proposition 2.3, provides an efficient bound for this function.
Lemma 4.2. Let R, a be positive integers and let p 1 , . . . , p j be the distinct prime divisors of R such that p i ≤ 2 a . Then W (R) ≤ c R,a R 1/a , where c R,a = 2 j (p 1 · · · p j ) 1/a .
In particular, d R := c R,8 < 4514.7 for every R.
Proof. The statement is an immediate generalization of [7, Lemma 3.3] and can be proved using multiplicativity. The bound for d R can be easily computed.
Next, observe that for r = n = 2, since r | q n − 1, we must further assume that q is odd, in which case, evidently, 4 | q 2 − 1. Now, fix some θ ∈ F q 2 such that F q 2 = F q (θ) and some α ∈ F * q 2 . Further, let R ′ be the square-free part of the odd part of q 2 − 1 and take some R | R ′ . Also, let N R (θ, α) stand for the number of R-free elements, that are squares, but not fourth powers, in the set {α(θ + x) : x ∈ F q }. For our purposes, it suffices to show that N R ′ (θ, α) = 0.
Our next aim is to relax the conditions of Proposition 4.3. For this purpose, we adapt the sieving techniques of Cohen-Huczynska, [7]. Proposition 4.4 (Sieving inequality). Let m | R ′ and θ, α ∈ F * q 2 such that F q 2 = F q (θ). In addition, let {r 1 , . . . , r s } be a set of divisors of m such that gcd(r i , r j ) = r 0 for every i = j and lcm(r 1 , . . . , r s ) = m. Then Proof. For any l | R ′ , let S l be the set of l-free elements of the form α(θ + x), where x ∈ F q , that are squares, but not fourth powers. In other words, |S l | = N l (θ, α). Accordingly, we may work with |S l | instead of N l (θ, α).
We will use induction on s. The result is trivial for s = 1. For s = 2 notice that S r1 ∪ S r2 ⊆ S r0 and that S r1 ∩ S r2 = S m . The result follows after considering the cardinalities of those sets.
Next, assume that our hypothesis holds for some s ≥ 2. We shall prove our result for s + 1. Set r := lcm(r 1 , . . . , r s ) and apply the s = 2 case on {r, r s+1 }. The result follows from the induction hypothesis.
Proposition 4.5. Let q, α, θ and R ′ be as above. Additionally, let ε and ε ′ be as above and assume that ε > 0.
The case when q ≡ 3 (mod 4) follows in the same fashion, but with (4.4) in mind.
We are now ready to proceed with the numerical aspects.
4.1. Numerical aspects. All the mentioned computations and algorithms were implemented with the SageMath software. Since, in some cases, finding a computationally efficient or viable way to perform our calculations was non-trivial, the basic steps of our calculations are described in detail. Furthermore, we note that a modern mid-range laptop can perform the computations of this subsection in less than two minutes.
The above, with the help of Lemma 4.2, implies that the case q ≥ q 0 = (2 · 4514.7) 4 ≃ 6.65 · 10 15 is settled. Next, let t(q) stand for the number of prime factors of q 2 − 1. A quick computation reveals that, if t(q) ≥ 14, then q ≥ q 0 , i.e., the case t(q) ≥ 14 is settled. Let p(i) stand for the i-th prime (for example p(2) = 3). Based on Proposition 4.5, we employ the following algorithm that takes t 1 ≤ t 2 as input and goes through the following steps: If Algorithm 4.1 returns true, then the case Let us now explain the validity of Algorithm 4.1. Assume that the returned value is true for some t 1 ≤ t 2 . Take some q, such that t 1 ≤ t(q) ≤ t 2 and write t(q) , where the p i 's are the (distinct) prime factors of q 2 − 1 in ascending order. It is clear that W (q 2 − 1) = 2 t(q) . Thus a condition for our purposes, implied by Proposition 4.5, is Of course, p i ≤ p(i), which impliess that ε 1 ≤ ε = 1 − s−1 i=0 1/p t1−i , and that t(q) ≤ t 2 , that is, the quantity q 1 computed in Step 2 is in fact larger than the right side of (4.7); hence, if q ≥ q 1 , then (4.7) holds. The number c in Step 3 stands for the maximum number of prime divisors a number not larger than q 2 1 − 1 can admit.
We successfully apply Algorithm 4.1 for the pairs (t 1 , t 2 ) = (11, 13) and (10, 10); consequently, the case t(q) ≥ 10 is settled. Thus, we may now assume that t(q) ≤ 9 and focus on the case q ≤ (2 · 2 9 ) 2 = 1,048,576. The interval 3 ≤ q ≤ 1,048,576 contains precisely 82,247 odd prime powers. We first exploit Proposition 4.3. A quick computation reveals that, in the interval in question, there are exactly 2,425 odd prime powers, where (4.3) or (4.4), respectively, do not hold when all the relevant quantities are explicitly computed.Among these, q = 1,044,889 is the largest.
We proceed to the sieving part, i.e., Proposition 4.5. Namely, we attempt to satisfy the conditions of Proposition 4.5 as follows. Until we run out of prime divisors of k, or until ε ≤ 0, we add to the set of sieving primes (that is, the primes p 1 , . . . , p s in Proposition 4.5) the largest prime divisor not already in the set. If, for one such set of sieving primes, the condition of Proposition 4.5 is valid, then the desired result holds for the prime power in question.
This procedure was successful, for all the 2,425 prime powers mentioned earlier, with the 101 exceptions of Table 4.1. So, to sum up, we have proved the following.  Table 4.1, α ∈ F * q n and θ ∈ F q 2 \ F q , there exists some x ∈ F q such that α(θ + x) is 2-primitive. In particular, L 2 (2) ≤ 3541.
It follows that |A| represents the number of set of translates of F q 2 /F q that include a 2-primitive element. It is clear that the set F q 2 \ F q is, in fact, partitioned into the distinct sets of translates of F q 2 /F q . Additionally, |F q 2 \ F q | = q(q − 1) and, since every set of translates has cardinality q, it follows that there are exactly q − 1 distinct sets of translates. Thus, F q 2 /F q has the translate property if and only if, at some point, |A| reaches q − 1. This is checked in line 16. On the contrary, if this number never reaches q − 1, this extension does not have the translate property, see line 21.
We ran Algorithm 4.2 for all the 101 prime powers of Table 4.1 and it returned true, with the exception of q = 5, 7, 11, 13, 31 and 41. We note that for all these computations, a modern mid-range laptop spent about 2.5 hours of computer time. Summing up, we have the following: Theorem 4.7. For every odd prime power q = 5, 7, 11, 13, 31 or 41 the extension F q 2 /F q possesses the translate property for 2-primitive elements. In particular, T 2 (2) = 41.

4.3.
Direct verification of the line property. We turn our attention to the line property. Fix some α ∈ F * q 2 and note that the lines of α and the various θ's over F q define yet another partition of F q 2 \ F q . For example, if α = 1 this partitioning coincides with the one that the sets of translates define. This partitioning, however, is not unique to every α ∈ F * q 2 , as we shall now demonstrate. Let α 1 , α 2 ∈ F * q 2 be such that α 1 /α 2 = b 0 ∈ F q . It follows that an arbitrary line that α 1 defines, along with some generator θ of the extension F q 2 /F q , is of the form is, one of the lines that α 2 defines. Consequently, α 1 and α 2 are associated with the same partitioning.
From this observation, we use Algorithm 4.3 which is based on Algorithm 4.2. Let us now explain its validity. The NotInLine procedure is merely a generalization of the NotInTranslate procedure of Algorithm 4.2, wherein the element γ ∈ A ′ ∪ ζA ′ is now considered. The procedure CheckLines follows the same steps as the main procedure of Algorithm 4.2, with the difference that, instead of the sets of translates, the partition is now dictated by the lines that γ defines. Note that for γ = 1 the check that is performed in this step is identical to the one performed in Algorithm 4.2. Finally, the main procedure of Algorithm 4.3, begins by building the set G = A ′ ∪ ζA ′ . Since a is primitive, a j+(q 2 −1)/2 = −a j ; so, in order to find a suitable A ′ , only the exponents 1, . . . , (q 2 − 1)/2 need to be considered. After the set G is built, the algorithm checks the output of CheckLines for all γ ∈ G.
14 days, all on our mid-range modern laptop. By way of comparison, the computer time consumed for smaller prime powers varied from 3-5 days, when q ≃ 1000 to a few seconds for q ≃ 100. The considerable cost for the larger numbers highlights the significance of strong theoretical methods that could minimize or, ideally, eliminate the computer dependency of our methods. For instance, a theoretical elimination of the two largest prime powers of Table 4.1 would reduce the computer time spent by more than two months.
Remark. As the line property implies the translate property, the exceptional extensions appearing in Theorem 4.7 also appear in Theorem 4.8. Unsurprisingly, the opposite is not true as the extensions F q 2 /F q , for q = 3 and 9, possess the translate property but not the line property for 2-primitive elements.