By Jan-Hendrik Evertse

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**Example text**

N are multiplicatively dependent if there are x1 , . . , xn ∈ Z, not all 0, such that α1x1 · · · αnxn = 1. Otherwise, α1 , . . , αn are called multiplicatively independent. 21. Let n 1. Let α1 , . . , αn , β1 , . . , βn , γ ∈ Q be such that α1 , . . , αn = 0, α1 , . . , αn are multiplicatively independent, (β1 , . . , βn ) ∈ Qn . Then eγ α1β1 · · · αnβn is transcendental. Proof. Suppose that δ := eγ α1β1 · · · αnβn is algebraic. 16) log δ = γ + β1 log α1 + · · · + βn log αn + 2kπi.

D is a permutation of τ1 , . . , τd . The exponents τ1 (αi1 )+· · ·+τd (αid ) need not be distinct. We group together the terms with equal exponents. Let γ1 , . . , γs be the distinct values among τ1 (αi1 ) + · · · + τd (αid ) (0 i1 , . . , in d), and for k = 1, . . , s, denote by Jk the set of tuples (i1 , . . , id ) such that τ1 (αi1 ) + · · · + τd (αid ) = γk . 10) τ1 (βi1 ) · · · τd (βid ). ,ik )∈Jk Notice that each τ ∈ Gal(L/Q) permutes the pairs (γ1 , δ1 ), . . , (γs , δs ). A priori, all coefficients δk might be 0.

Ii) Let β1 , . . , βn ∈ K be algebraic over k and let α ∈ K be a zero of X n +β1 X n−1 + · · · + βn . Then α is algebraic over k. Proof. (i) Consider the field k(α, β). 6, all elements of k(α, β) are algebraic over k. (ii) Let L = K(β1 , . . , βn ), M = L(α). 4, M is finite over L. 5, M is finite over K. 6, α is algebraic over k. 8 implies that the set of α ∈ K that are algebraic over k form a field, the algebraic closure of k in K. 8 is that if α ∈ K is algebraic over the algebraic closure of k in K, then α belongs to this algebraic closure.