PUBLICATIONS DE L’INSTITUT MATHÉMATIQUE Nouvelle série, tome 77(91) (2005), 67–70

TWO EXERCISES CONCERNING THE DEGREE OF THE PRODUCT OF ALGEBRAIC NUMBERS

Artūras Dubickas

Communicated by Aleksandar Ivić

Abstract. Let k be a field, and let α and β be two algebraic numbers over k of degree d and ℓ, respectively. We find necessary and sufficient conditions under which deg(αβ) = dℓ and deg(α + β) = dℓ. Since these conditions are quite difficult to check, we also state a simple sufficient condition for such equalities to occur.

Let k be a field, and let k^{a} be an algebraic closure of k. Suppose that α ∈ k^{a} is of degree d over k. If β ∈ k^{a} has degree ℓ over k then [k(α, β) : k] ⩽ dℓ, so any γ ∈ k(α, β) has degree at most dℓ over k. In particular, αβ and α + β both have degree at most dℓ over k. Furthermore, for a ‘generic’ β of degree ℓ we have equality, namely, αβ and α+β are both of degree dℓ. For some problems concerning linear forms in conjugate algebraic numbers and the Mahler measure of an algebraic number (over ℚ) we have α ∈ k^{a} satisfying certain conditions (see, e.g., [1], [3]) and need to enlarge the set of such numbers by either multiplying or by adding a ‘generic’ β (of degree ℓ) in the sense that αβ (or α + β) has ‘generic’ degree dℓ. How one can be sure that a particular β have the required properties?

In this note we state some simple sufficient, necessary and necessary and sufficient conditions on β in order that αβ (or α+β) is of maximal possible degree. We begin with the following necessary and sufficient condition.

Theorem 1. Suppose that α ∈ k^{a} is of degree d over k and β ∈ k^{a} is of degree ℓ over k. Then αβ is of degree dℓ over k if and only if β is of degree ℓ over k(α) and α ∈ k(αβ). Similarly, α + β is of degree dℓ over k if and only if β is of degree ℓ over k(α) and α ∈ k(α + β).

Proof. The proof follows easily from the following standard diagram:

2000 Mathematics Subject Classification: 11R04, 11R32, 12E99.

Key words and phrases: field, algebraic number, degree, root of unity.

k  E = k(α) ∩ k(β)  k(β)
\  \
k(α)  k(α, β)

Indeed, since

[k(αβ) : k] ⩽ [k(α, β) : k] = [k(α, β) : k(α)][k(α) : k] = [k(α, β) : k(α)]d ⩽ dℓ,

we have [k(αβ) : k] = dℓ if and only if k(α, β) = k(αβ) and [k(α, β) : k(α)] = ℓ. Of course, k(α, β) = k(αβ) implies that α ∈ k(αβ). But then also β ∈ k(αβ) and so α ∈ k(αβ) implies that k(α, β) = k(αβ) too. Consequently, the conditions k(α, β) = k(αβ) and α ∈ k(αβ) are equivalent. On the other hand, [k(α, β) : k(α)] = ℓ = [k(β) : k] if and only if the minimal polynomial of β over k is irreducible over the field k(α), that is β has degree ℓ over k(α). This proves the theorem for αβ. The proof of the theorem for the sum α + β is precisely the same. □

Set E = k(α) ∩ k(β) (see the diagram). The degree of β over E is equal to the degree of β over k(α) (see, for instance, [2]). So if E is a proper extension of k then the degree of β over k(α) is smaller than ℓ. Consequently, Theorem 1 implies that E = k(α) ∩ k(β) = k is a necessary condition for deg(αβ) = dℓ (and for deg(α + β) = dℓ) to occur.

Unfortunately, the condition α ∈ k(αβ) of Theorem 1 is quite difficult to check. This raises the question on whether there is a simple method of finding many different β satisfying deg(αβ) = d deg β. The next theorem gives a sufficient condition for this equality to occur.

Theorem 2. Suppose that α is an algebraic number of degree d over a field k of characteristic zero, and let K be a normal closure of k(α) over k. If L = k(β) is a normal extension of k of degree ℓ and L ∩ K = k then deg(α + β) = dℓ. If, in addition, β is torsion-free then deg(αβ) = dℓ.

Recall that (as in [3]) β is called torsion-free if β′/β is not a root of unity for any β′ ≠ β, where β′ and β are conjugate over k. The condition on β to be torsion-free is necessary in the multiplicative part of Theorem 2. Indeed, the example k = ℚ, α = \sqrt{2}, β = \sqrt{3} with d = ℓ = 2 and ℚ(\sqrt{2}) ∩ ℚ(\sqrt{3}) = ℚ shows that αβ = \sqrt{6} is of degree 2 over ℚ, although ℚ(\sqrt{2})/ℚ and ℚ(\sqrt{3})/ℚ are normal extensions and dℓ = 4. Of course, if β is not torsion-free, we can add to it an element k_{0} ∈ k and consider β_{0} = β + k_{0} instead. Since L = k(β) = k(β_{0}) for any k_{0} ∈ k, it is sufficient to take k_{0} for which β_{0} = β + k_{0} is torsion-free. (Below, we will show that such k_{0} exists: see Theorem 3.) In the above example we can take k_{0} = 1. Then β_{0} = 1 + \sqrt{3} and αβ_{0} = \sqrt{6} + \sqrt{2} is of degree 4 over ℚ.

Proof of Theorem 2. The conditions of the theorem imply that LK is a Galois extension of k (see [4] for all standard facts about Galois extensions which are used here). Therefore α′ + β′ is conjugate to α + β for arbitrary pair α′, β′, where α′ and α are conjugate over k and β′ is conjugate to β over k. Hence deg(α + β) ⩽ dℓ with inequality being strict if and only if α + β = α′ + β′ with certain α′ ≠ α and β′ ≠ β. Assume that α + β = α′ + β′. Then L ∩ K = k implies that γ := α − α′ = β′ − β ∈ k, because α′ ∈ K, β′ ∈ L. Let σ be an automorphism of K taking α to α′. Suppose that σ is of order t > 1, so that σ^{t}(α) = α. Then by adding t equalities γ = σ^{j}(α) − σ^{j+1}(α) corresponding to j = 0, 1, …, t − 1 we obtain that tγ = 0. Since char k = 0, this can only occur if γ = 0, giving α′ = α and β′ = β, a contradiction.

Similarly, deg(αβ) ⩽ dℓ, where deg(αβ) < dℓ if and only if αβ = α′β′ with certain α′ ≠ α and β′ ≠ β. Now, a similar argument shows that γ := β′/β = α/α′ can lie in k, but only if γ is a root of unity (see also [5]). More precisely, if σ : β → β′ is of order t then

(\frac{β′}{β} )^{t} = γ^{t} = \frac{σ(β)}{β} \frac{σ^{2}(β)}{σ(β)} … \frac{β}{σ^{t−1}(β)} = 1

so β is not torsion-free, a contradiction. This proves Theorem 2. □

We will conclude by showing the following.

Theorem 3. For each β ∈ k^{a}, where k is a field of characteristic zero, there is a k_{0} ∈ k such that β + k_{0} is torsion-free.

Proof. Suppose that there is a β ∈ k^{a} such that β + k_{0} is not torsion-free for each k_{0} ∈ ℤ, where ℤ is a prime subfield of k. Then, for some fixed β′ (which is conjugate to β over k and β′ ≠ β), ω := (β′ + k_{0})/(β + k_{0}) is a root of unity for infinitely many k_{0} ∈ ℤ. By Corollary 1.3 of [2], the degree of ω over k is bounded, so there is an absolute constant n_{0} ∈ ℕ, n_{0} > 1, and infinitely many k_{0} ∈ ℤ for which (β′ + k_{0})^{n_{0}} = (β + k_{0})^{n_{0}}. Subtracting the left-hand side of this equality from its right-hand side and dividing by β − β′ we obtain that

ξ_{0} + ξ_{1}k_{0} + ⋯ + ξ_{n_{0}−1}k_{0}^{n_{0}−1} = 0,

where the coefficients ξ_{j} = \binom{n_{0}}{j}(β^{n_{0}−j} − β^{′n_{0}−j})/(β − β′) ∈ k(β, β′), j = 0, 1, …, n_{0} − 1, do not depend on k_{0}. Now, by taking any n_{0} distinct elements k_{0} (among infinitely many) in order that a respective determinant would be non-zero, we deduce that ξ_{0} = ξ_{1} = ⋯ = ξ_{n_{0}−1} = 0. However, ξ_{n_{0}−1} = n_{0} ≠ 0, a contradiction.

□

This research was partially supported by the Lithuanian State Science and Studies Foundation and by INTAS grant no. 03-51-5070.

References

[1] J. D. Dixon and A. Dubickas, The values of Mahler measures, Mathematika, (to appear).

[2] P. Drungilas and A. Dubickas, On subfields of a field generated by two conjugate algebraic numbers, Proc. Edinburgh Math. Soc. 47 (2002), 119–123.

[3] A. Dubickas, On the degree of a linear form in conjugates of an algebraic number, Illinois J. Math. 46 (2002), 571–585.

[4] S. Lang, Algebra, 3rd ed., Graduate texts in mathematics 211, Springer-Verlag, New York, Berlin, 2002.

[5] C. J. Smyth, Conjugate algebraic numbers on conics, Acta Arith. 40 (1982), 333–346.

Department of Mathematics and Informatics (Received 01 08 2004) Vilnius University Naugarduko 24 Vilnius 03225, Lithuania

arturas.dubickas@maf.vu.lt