On the degree of compositum of two number fields
Abstract: We prove that the sum of two algebraic numbers both of degree 6 cannot be of degree 8. The triplet (6, 6, 8) was the only undecided case in the previous characterization of all positive integer triplets (a, b, c), with a ≤ b ≤ c and b ≤ 6, for which there exist algebraic numbers α, β and γ of degrees a, b and c, respectively, such that α + β + γ = 0. Now, this characterization is extended up to b ≤ 7. We also solve a similar problem for \documentclass{article}\usepackage{amssymb}\begin{document}\pagestyle{empt… Show more
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Cited by 5 publications
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For all but one positive integer triplet (a, b, c) with a b c and b 6, we decide whether there are algebraic numbers α, β and γ of degrees a, b and c, respectively, such that α + β + γ = 0. The undecided case (6,6,8) will be included in another paper. These results imply, for example, that the sum of two algebraic numbers of degree 6 can be of degree 15 but cannot be of degree 10. We also show that if a positive integer triplet (a, b, c) satisfies a certain triangle-like inequality with respect to every prime number then there exist algebraic numbers α, β, γ of degrees a, b, c such that α + β + γ = 0. We also solve a similar problem for all (a, b, c) with a b c and b 6 by finding for which a, b, c there exist number fields of degrees a and b such that their compositum has degree c. Further, we have some results on the multiplicative version of the first problem, asking for which triplets (a, b, c) there are algebraic numbers α, β and γ of degrees a, b and c, respectively, such that αβγ = 1.Proof: Suppose that K and L are number fields. Then, by the primitive element theorem, K = Q(α) and L = Q(β) for some α ∈ K and β ∈ L. Furthermore, the compositum KL = Q(α, β) can be expressed as KL = Q(α+tβ) and also as KL = Q(α(t+β)) for all but finitely many rational numbers t. See the proof of Theorem 4.6 in [16] for the case α + tβ. The proof for α(t+β) is the same. Indeed, consider the field K t = Q(α(t+β)). Since Q ⊆ K t ⊆ Q(α, β), there are two distinct rational numbers t and t for which K t = K t . Assume without loss of generality that α(t +β) = 0. Then, as the quotient of α(t + β) and α(t + β) belongs to K t , we obtain (t − t )/(t + β) = (t + β)/(t + β) − 1 ∈ K t . Thus β ∈ K t . This implies α ∈ K t , so that K t = Q(α, β).Since [K : Q] = a, [L : Q] = b, [KL : Q] = c, choosing an appropriate t ∈ Q, in the additive case we see that the degrees of α, tβ and −α − tβ are a, b, c, respectively. In the multiplicative case, the degrees of α, t + β and α −1 (t + β) −1 are a, b, c.
For all but one positive integer triplet (a, b, c) with a b c and b 6, we decide whether there are algebraic numbers α, β and γ of degrees a, b and c, respectively, such that α + β + γ = 0. The undecided case (6,6,8) will be included in another paper. These results imply, for example, that the sum of two algebraic numbers of degree 6 can be of degree 15 but cannot be of degree 10. We also show that if a positive integer triplet (a, b, c) satisfies a certain triangle-like inequality with respect to every prime number then there exist algebraic numbers α, β, γ of degrees a, b, c such that α + β + γ = 0. We also solve a similar problem for all (a, b, c) with a b c and b 6 by finding for which a, b, c there exist number fields of degrees a and b such that their compositum has degree c. Further, we have some results on the multiplicative version of the first problem, asking for which triplets (a, b, c) there are algebraic numbers α, β and γ of degrees a, b and c, respectively, such that αβγ = 1.Proof: Suppose that K and L are number fields. Then, by the primitive element theorem, K = Q(α) and L = Q(β) for some α ∈ K and β ∈ L. Furthermore, the compositum KL = Q(α, β) can be expressed as KL = Q(α+tβ) and also as KL = Q(α(t+β)) for all but finitely many rational numbers t. See the proof of Theorem 4.6 in [16] for the case α + tβ. The proof for α(t+β) is the same. Indeed, consider the field K t = Q(α(t+β)). Since Q ⊆ K t ⊆ Q(α, β), there are two distinct rational numbers t and t for which K t = K t . Assume without loss of generality that α(t +β) = 0. Then, as the quotient of α(t + β) and α(t + β) belongs to K t , we obtain (t − t )/(t + β) = (t + β)/(t + β) − 1 ∈ K t . Thus β ∈ K t . This implies α ∈ K t , so that K t = Q(α, β).Since [K : Q] = a, [L : Q] = b, [KL : Q] = c, choosing an appropriate t ∈ Q, in the additive case we see that the degrees of α, tβ and −α − tβ are a, b, c, respectively. In the multiplicative case, the degrees of α, t + β and α −1 (t + β) −1 are a, b, c.
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 Q {\mathbb{Q}} cannot be of degree 8. This completes the classification of so-called product-feasible triplets ( a , b , c ) ∈ N 3 \left(a,b,c)\in {{\mathbb{N}}}^{3} with a ≤ b ≤ c a\le b\le c and b ≤ 7 b\le 7 . The triplet ( a , b , c ) \left(a,b,c) is called product-feasible if there are algebraic numbers α , β \alpha ,\beta , and γ \gamma of degrees a , b a,b , and c c over Q {\mathbb{Q}} , respectively, such that α β = γ \alpha \beta =\gamma . In the proof, we use a proposition that describes all monic quartic irreducible polynomials in Q [ x ] {\mathbb{Q}}\left[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 n\ge 2 and k ≥ 1 k\ge 1 , the triplet ( a , b , c ) = ( n , ( n − 1 ) k , n k ) \left(a,b,c)=\left(n,\left(n-1)k,nk) is product-feasible if and only if n n is a prime number. The choice ( n , k ) = ( 4 , 2 ) \left(n,k)=\left(4,2) recovers the case ( a , b , c ) = ( 4 , 6 , 8 ) \left(a,b,c)=\left(4,6,8) as well.
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