The reasons, as discussed in [1, Remark 2.2], are as follows. In the final displayed equation in the proof of Theorem 6.9, to conclude that θ j/N k = N… Click to show full abstract
The reasons, as discussed in [1, Remark 2.2], are as follows. In the final displayed equation in the proof of Theorem 6.9, to conclude that θ j/N k = N kθ j+k , we must treat θ j as an element of R, not of S (there are many solutions to N kγ = θ j in S). Then later, throughout §8, the statements include ‘let r j := β/N jθ j ’, which makes sense only if θ j ∈ R; and, in particular, in the displayed calculation below equation (8.3) in the proof of Lemma 8.1, which is a calculation about real numbers, it is crucial that N 2θ j+1 = θ j in R, not just in S. Note that this reduces the generality of the results substantially: for a given θ1 ∈ S there are infinitely many sequences (θn) in S that satisfy N θn+1 = θn for all n; but, given θ1 ∈ R, there is just one sequence (θn) in R that satisfies N θn+1 = θn for all n.
               
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