The product of a quartic and a sextic number cannot be octic

: In this article, we prove that the product of two algebraic numbers of degrees 4 and 6 over (cid:2) cannot be of degree 8. This completes the classi ﬁ cation of so-called product-feasible triplets (cid:3) ( ) ∈ a b c , , 3 with ≤ ≤ a b c and ≤ b 7 . The triplet ( ) a b c , , is called product-feasible if there are algebraic numbers α β , , and γ of degrees a b , , and c over (cid:2) , respectively, such that = αβ γ . In the proof, we use a proposition that describes all monic quartic irreducible polynomials in (cid:2) [ ] x with four roots of equal moduli and is of independent interest. We also prove a more general statement, which asserts that for any integers ≥ n 2 and ≥ k 1 , the triplet ( ) ( ( ) ) = − a b c n n k nk , , , 1 , is product-feasible if and only if n is a prime number. The choice ( ) ( ) = n k , 4, 2 recovers the case ( ) ( ) = a b c , , 4, 6, 8 as well.

Abstract: In this article, we prove that the product of two algebraic numbers of degrees 4 and 6 over cannot be of degree 8.This completes the classification of so-called product-feasible triplets ( )∈ a b c , , 3 with ≤ ≤ a b c and ≤ b 7. The triplet ( ) a b c , , is called product-feasible if there are algebraic numbers α β , , and γ of degrees a b , , and c over , respectively, such that = αβ γ.In the proof, we use a proposition that describes all monic quartic irreducible polynomials in [ ] x with four roots of equal moduli and is of independent interest.We also prove a more general statement, which asserts that for any integers ≥ n 2 and ≥ k 1, the triplet 1 Introduction In the study by Drungilas et al. [1] a curious version of an abc-problem for sums and products of algebraic numbers has been introduced.(It has nothing to do with the abc-conjecture of Masser and Oesterlé.)According to the definitions in [1], a triplet ( )∈ a b c , , 3 is called • sum-feasible if there exist algebraic numbers α and β of degrees a and b (over ), respectively, such that the degree of + α β is c, • product-feasible if there exist algebraic numbers α and β of degrees a and b (over ), respectively, such that the degree of ⋅ α β is c, • compositum-feasible if there exist number fields K and L of degrees a and b (over ), respectively, such that the degree of their compositum KL is c.
The first named author has been interested in these questions before the publication of the article [1]; independently, the sum question is one of the questions at MathOverflow (http://mathoverflow.net/questions/30151/posed in 2010).
It is clear that the first two definitions are symmetric with respect to a b , , and c in the sense that the triplet ( ) a b c , , is sum-feasible (resp.product-feasible) if and only if so are six possible permutations of the triplet ( ) a b c , , .So, without restriction of generality, we may assume that ≤ ≤ a b c.The case of compositum-feasible triplets is less symmetric.Then, only the degrees a and b (but not c) are involved in a symmetric way.However, since the field KL contains both K and L, we can also assume that ≤ ≤ , , 3 with ≤ ≤ a b c is a sum-feasible, a product-feasible or a compositum-feasible triplet, then (in any of the three cases) we must have ≤ c ab.In the study by Drungilas et al. [1], it was shown that Proposition 1.Each compositum-feasible triplet is also sum-feasible.
Later, in the study by Drungilas and Dubickas [2], it was proved that Proposition 2. Each sum-feasible triplet is also product-feasible.
Consequently, if a triplet is not product-feasible, then it is neither sum-feasible nor compositum-feasible.The simple example ( ) 2, 3, 3 shows that a triplet can be product-feasible but not sum-feasible, so these two questions are not equivalent.Indeed, the product γ of the quadratic number = α e πi 2 3 and the cubic number is conjugate to β.On the other hand, since the integers 2 and 3 are coprime, the sum γ of any quadratic number α and any cubic number β must be of degree 6 (and so = + γ α β cannot be cubic); see, e.g., [3].
In the study by Drungilas et al. [1], all sum-feasible and also all compositum-feasible triplets ( )∈ a b c , , satisfying ≤ ≤ a b c, with ≤ b 6, have been described except for one special case ( ) 6, 6, 8 .the study by Drungilas et al. [4], the missing case ( ) 6, 6, 8 from that classification has been treated by showing that the triplet ( ) 6, 6, 8 is not sum-feasible, and the classification has been extended to ≤ b 7.Then, in the study by Drungilas and Maciulevičius [5], all possible compositum-feasible triplets ( ) a b c , , satisfying ≤ ≤ a b c, with ≤ b 9, have been determined.In the special case when = = a b p is a prime number, a complete characterization of all compositum-feasible triplets ( ) p p ps , , has been given in [6].Recently, in the study by Maciulevičius [7], the second named author described all possible productfeasible triplets ( ) a b c , , satisfying ≤ ≤ a b c, with ≤ b 7, except for the following five cases: 6, 8 , 4, 7, 7 , 4, 7, 14 , 5, 6, 10 , 5, 6, 15 .

This implies that
is product-feasible if and only if n is a prime number.
In the proof of Theorem 4, assuming that there are α and β of degree 4 and 6 over , respectively, whose product αβ is of degree 8, we will first show that the conjugates of α must all be of the same modulus.In 1969, Robinson [9] described algebraic integers whose conjugates including α itself are all of the same moduli; see also [10] for the description of algebraic numbers with conjugates of two distinct moduli.Here, we need a more specific result for quartic algebraic numbers α whose all four conjugates have equal moduli.The proposition below is a new ingredient that turns out to be a useful tool in the proof of Theorem 4.
Proposition 7. Assume that ( ) p x is a monic quartic polynomial in [ ] x , which is irreducible over and whose four roots have equal moduli.Then, ( ) p x must be of one of the following forms: , where ∈ s r , and < s r 4 2 , where ∈ r and ≠ ′ u u are the conjugates of a real quadratic algebraic number satisfying This approach apparently cannot be used in the proof of a more general Theorem 6, although some parts of the proof of Theorem 4 will be used in the more general setting of Theorem 6.
Of course, Theorem 6 immediately implies Theorem 4 by selecting ( ) ( ) = n k , 4,2.However, we present both proofs of the case ( , , 4, 6, 8 , since a separate proof of Theorem 4 contains some ideas that may be useful in treating similar problems for algebraic numbers of small degrees, but not only.As observed by one of the referees, in the first proof, the case ( ) 4, 6, 8 is obtained via descent, using the fact that the triplet ( ) 2, 4, 6 is not product-feasible.The main auxiliary result in the proof of Theorem 6 is Lemma 11.
In the next section, we will prove Proposition 7 and also give several auxiliary lemmas, which will be useful in the proofs of Theorems 4 and 6.Then, in Section 3, we will prove Theorem 4. The proof of Theorem 6 is a bit more involved.First, in Section 4, we will give some preparation under the assumption that the triplet is product feasible.Second, in Section 5, using some previous results, we will show that the triplet ( , where n is a prime number and ≥ k 1, is product-feasible.Finally, we will complete the proof of Theorem 6 by getting a contradiction for composite n. Assume first that p has a real root.By the assumptions of the proposition, not all four roots of p can be real.Thus, p must have a pair of complex conjugate roots and another real root.So, without loss of generality, we may assume that the four roots of p in equation ( 1) are { for some The product of a quartic and a sextic number  3 4 is a negative rational number, say, −r, where ∈ > r 0 .Furthermore, since p has four roots, at least one real, on the circle | | = z α, by Ferguson's result [11] (or even by an earlier result of Boyd [12]), we must have ( ) x .The polynomials p and g have the same constant coefficient, so g is a monic linear polynomial − x r, which implies that p is as in ( ) i .Next, we consider the case when p has no real roots.Then, there are > ϱ 0 and < < < ϕ ξ π 0 such that the roots of . This time, by equation ( 1), we obtain 0 .Furthermore, by Vieta's formulas, and Likewise, by Vieta's formula, we deduce (3) . Therefore, setting , which is the case . Here, the inequality must be strict, i.e., < s r 4

2
, since otherwise ( ) p x has a real root.It remains to consider the case ≠ a 0

3
. Then, = ≠ a a ϱ 0 and by equation ( 1), we obtain Here, , and = a r 0 2 .Since ∈ > r 0 , these four numbers are all rational if and only if + ′ u u and ′ uu are both rational.Furthermore, u and ′ u are both real by equations ( 4) and (5).They are both irrational, since otherwise ( ) p x would be reducible.Since u and ′ u are the roots of the quadratic polynomial .Consequently, p is a polynomial of the form described in case ( ) iii .This completes the proof of the proposition.
In the proofs of Theorems 4 and 6, we will also use the following two lemmas:  The next lemma will be used in the proof of Theorem 6.
By Lemma 8 with ( ) ( ) = a b , 4,6, the full list of conjugates of αβ of degree 8 is, for instance, for some In equation (10) we can choose any distinct products α β i t and α β j l , which do not belong to Γ 1 .In particular, for indices i and j, we must have ≠ ≠ Γ Γ and Γ Γ.
Adding and multiplying all eight conjugates in equation (10), we obtain where ≔ + +⋯+ ∈ r β β β . This yields and hence, Squaring equation ( 12) and multiplying it by α β i t 2 2 , we find that Thus, β t is a root of a degree 4 polynomial over the field ( ) α α , i 2 .As we have shown earlier, all the conjugates of α have equal moduli.Hence, there are three possible cases for the minimal polynomials of α that are listed in Proposition 7.
In case ( ) i , we have = α αε i for some But then the degree of β t over ( ) α is at most 4. Since the degree of β over ( ) , we obtain , so that equation ( 13) leads to the same contradiction again.In the alternative case, when = Γ Γ 1 2 , we obtain where 1, 2, …,6 .This time, by equation ( 11), we cannot take = i 2, so that { } ∈ i 3, 4 .Multiplying all equalities in equation ( 14) we obtain = α α 12 .Consequently, α 6 is a rational number . Adding all equalities in equation ( 14), derive , since ≠ α α 2 .Thus, by equation ( 12), we must have This means that = δ αβ i t is a rational number or a quadratic number.Evidently, δ cannot be rational, since α i and β t are of distinct degrees 4 and 6, respectively.On the other hand, if δ were quadratic, then the product of the quadratic number δ and the quartic number ∕α 1 i would be the sextic number β t .However, the triplet ( ) 2, 4, 6 is not product-feasible by [7,Theorem 1].This completes the proof of Theorem 4.

The case ( (
) , 1 are integers.Throughout this section, we assume that there exist algebraic numbers α and β satisfying The beginning of the argument is essentially the same as that in Section 3. Since ( ) β and ( ) αβ are subfields of ( ) α β , , we find that [ ( ) ] α β , : is divisible by both b and c and therefore divisible by On the other hand, and Thus, we have the following diagram: The product of a quartic and a sextic number  7 Let ≔ α α 1 , α 2 , …, α a be the distinct conjugates of α over , and let Figure 2 and Lemma 8 imply that the numbers in the set are all conjugate over .Since the set Γ contains the number αβ, it must have exactly c distinct elements.This means that [ ( ) ] = α β c : i j for any indices ∈ i A and ∈ j B. Therefore, the same degree diagram as that in Figure 2 holds for any pair of indices ∈ i A, ∈ j B (Figure 3).We conclude this section with the following lemma and its corollary: of α over , the degree of α i over ( ) α j is equal to − a 1.
Proof.Assume that ≠ α α i j are two distinct conjugates of α, so { } ∈ ⧹ i A j .By the diagram in Figure 3, we can assume that = j 1 and so = α α j .Since β is of degree b over ( ) α , all the numbers are conjugate to ≔ γ αβ over .Let the remaining conjugates of γ over be Take the polynomials where ( ) In particular, this implies that ≠ γ αβ c l for ∈ l B, since otherwise, by Figure 3, the degree of γ c over the field ( ) α would be , which is a contradiction.Hence, This implies [ ( ) 1. Thus, ℓ α is of degree − a 1 over ( ) = K α , and hence so must be the degree of α i over K for each { } ∈ ⧹ i A j .This completes the proof of the lemma.□ Corollary 12.Under the conditions of Lemma 11, we assume that there exists ∈ m such that ( ) ∈ α α i m j for two distinct conjugates α i and α j of α over .If = ≥ a n 3, then there exists a rational number r 6 such that Thus, there is ∈ l B for which α β i l and α β j l both lie in the intersection ∩ Γ Γ for every ∈ t A. So α is of degree at most ∕ n 2 over , which is not the case since α is of degree > ∕ n n 2 over .
The product of a quartic and a sextic number  9 Likewise, if = a n is odd, then α has a real conjugate α i , and so the product r 7 of all the conjugates of α can be written as α i n (in both cases > α 0  Consequently, the degree of ζ over its subfield must be at least − n 1.However, the degree of ζ over equals ( ) φ n , where φ stands for Euler's totient function [14,Theorem 3.1].For each composite n, we have ( ) < − φ n n 1, so the degree of ζ over is less than − n 1, a contradiction.This completes the proof of Theorem 6.

Lemma 8 . [ 1 ,
Proposition 21] Suppose that α and β are algebraic numbers of degrees a and b over , respectively.conjugates of β.If β is of degree b over ( )

Lemma 11 . 1
Let ≥ n 3 and ≥ k 2 be integers.Assume that α and β are algebraic numbers of degrees = over , respectively, whose product αβ is of degree = c nk.Then, for any pair of conjugates ≠ α α i j

l c c 1 1 Figure 3 :
Figure 3: Degree diagram for α i , β j and α β i j .

αα n 7 2 7 2 2 .
≠ i j.By Corollary 12, we conclude that the conjugates of α over all have the same moduli.We now consider two cases depending on whether the composite number = a n is even or odd.If n is even and α has no real conjugates, then the product r 7 of all conjugates can be written in the form ( ) = ∕ r .If α has a real conjugate, say ′ α , then − ′ α is also its conjugate (as the number of nonreal conjugates is even), so ( ) for any nonreal α i .Therefore, in both cases, there are two distinct indices ∈ i j A , By Corollary 12, we conclude that = so that the minimal polynomial of α over is − -feasible if and only if n is a prime number.The choice ( Keywords: algebraic numbers, product-feasible triplets, field extensions MSC 2020: 11R04, 11R21, 11R32, 12F10
α , then all the numbers α β i j , ≤ ≤ The case = k 1 of Theorem 6 follows from the previous result of the second named author: