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Sample text

N belong to some interval say [a−1 , a], a > 1. Thus the elementary polynomials in the σi (x) will also belong to some interval of the same form. Now they are the coefficients of the characteristic polynomial of xi , which has integer coefficients since x ∈ Uk . Thus there are only finitely many possible characteristic polynomials of elements x ∈ C , hence only finitely many possible roots of minimal polynomials of elements x ∈ C , which shows that x can belong to C for only finitely many x.

N+1 ) be an ideal of ok . There exists an integral ideal b such that ab = (α) (α)p where α ∈ ok by the exercise. It follows that there is β ∈ b such that βa (α)p. Therefore (β/α)a = βb−1 ⊂ ok , and then we obtain (β/α)αi ∈ ok for 1 ≤ i ≤ n + 1. On the other hand, we have (β/α)a p. Then there exists j such that (β/α)αj ∈ / p. In other words, set γi = (β/α)αi , then γi ∈ ok for 1 ≤ i ≤ n+1 but γj ∈ / p for some j. Hence γj = 0 in the residue field ok /p. Multiplying by β/α, we have γ1 x1 + · · · + γn+1 xn+1 = 0.

Also |am | = |am |c∞ because the am ∈ Z, and hence |am | 1 < < M c, c 2 |Pm |c∞ where the first inequality only holds if |σm α|∞ > |σm+1 α|∞ and the second inequality holds for any m. As follows, we shall show that |σµ α|c∞ > |α| > |σµ+1 α|c∞ , for any µ, |α| > |σ1 α|c∞ and σn α|c∞ > |α| all do not hold. It follows immediately that |α| = |σα|c∞ for some embedding σ, which completes the proof. We need only consider the case of |σµ α|c∞ > |α| > |σµ+1 α|c∞ , for some µ. For two cases |α| > |σ1 α|c∞ or |α| < |σn α|c∞ , the similar argument would give the same contradiction.