Acta Mathematica Academiae Scientiarum Hungaricae 69. (1995)

1995 / 1-2. szám - Győry K.: On a problem of A. M. Odlyzko on algebraic units of bounded degree

ON A PROBLEM OF A. M. ODLYZKO ON ALGEBRAIC UNITS OF BOUNDED DEGREE 3 Proof of the Theorem. Let £i,...,£m be algebraic units of degree ^ n with m = M(n) such that £,- — £j is a unit for all г ф j. If m ^ 2n + 1 then we are done. Hence assume that to > 2n + 1. Consider the number field К = Q(f'i,..., £2n+i )• Its degree, denoted by k, is at most n2n+1. Each £; with i > 2n + 1 is of degree at most n over К. For a given positive integer q with 1 ^ q ^ n, consider those i with 2n -f 1 < i ^ m for which £г- is of degree q over K, and denote by Mq the number of i under consideration. For each unit £,• of degree q over K, denote by £t-a\..., the conjugates of £i over К. Then for each j witji 1 ^ j 5Í 2n + 1, £j — £^ is an algebraic unit for p — 1 There exist algebraic integers хц,... xqi in К such that Further, this product is a unit in К. Putting F is a decomposable form of degree 2n + 1 with coefficients in Ok- For each £i under consideration, the corresponding tuple (1, хц,..., xqi) is a solution of the equation (6) F(l,xi,...,xq) G Ok in (1 ,xu...,xq) eO^1. We apply now our Lemma to equation (6). Denote by d the degree of the normal closure over Q of the splitting field of F over K. We have d ^ k\. We notice that if the units £,• and £,/ are of degree q over К with 2n + 1 ^ ^ г, i' ^ to, i ф ithen they lead to the same solution (1, x\,..., xq) of (6) if and only if £i and £,/ are conjugates to each other over К. Hence, by our Lemma we infer that Mq úqibkdf 4 'k ^ expexp{38</(rc2n+1)!}. This implies that M(n) 5í Mq ^ exp exp{ 39n(n2"+1)!} 9=1 which completes the proof of the Theorem. □ Remark 2. It is clear from the proof that using our Lemma with the bound (5) in place of (4), we obtain estimate (2) instead of (1). П (£i — £t'P*) = £) + xu£4j 1 +■•• + %• p=1 2n+l F(x 0, • • • j %q) — j j ( x0£ j T 2-1 £j + ■ • • + "Pqi) j j=1

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