Bourbaki

Exercise 3, § 2

General Topology, VII

Bourbaki marks this one as harder than the rest.

Every pp-adic number xx (Exercise 2) can be written in the form

∑k=−∞+∞αkpk \sum_{k=-\infty}^{+\infty} \alpha_k p^k

where the αk\alpha_k are rational integers, those of index k<0k < 0 being zero for all but a finite number of values of the index kk. Moreover, if

∑k=−∞+∞αkpk=∑k=−∞+∞βkpk, \sum_{k=-\infty}^{+\infty} \alpha_k p^k = \sum_{k=-\infty}^{+\infty} \beta_k p^k,

the rational numbers ∑k=−∞−1αkpk\sum_{k=-\infty}^{-1} \alpha_k p^k and ∑k=−∞−1βkpk\sum_{k=-\infty}^{-1} \beta_k p^k are congruent mod 1. Let φp(x)\varphi_p(x) be the class mod 1 of the rational number ∑k=−∞−1αkpk\sum_{k=-\infty}^{-1} \alpha_k p^k.

a) Show that φp\varphi_p is a continuous homomorphism of the topological group Qp\mathbf{Q}_p into T\mathbf{T}, and that for every continuous homomorphism ff of Qp\mathbf{Q}_p into T\mathbf{T} there exists a unique a∈Qpa \in \mathbf{Q}_p such that f(x)=φp(ax)f(x) = \varphi_p(ax) for all x∈Qpx \in \mathbf{Q}_p. [Note that the knowledge of the elements f(pk)f(p^k) for k≤0k \leq 0 determines ff uniquely; show that each of these elements is the class mod 1 of a fraction whose denominator is a power of pp; hence show that there exists a∈Qpa \in \mathbf{Q}_p such that f(pk)=φp(apk)f(p^k) = \varphi_p(ap^k) for all k≤0k \leq 0.]

b) Show that, if φ:R→T\varphi : \mathbf{R} \to \mathbf{T} is the canonical homomorphism, then φ(x)=∑pφp(x)\varphi(x) = \sum_p \varphi_p(x) for all x∈Qx \in \mathbf{Q}, the sum being over all prime numbers pp (observe that all but a finite number of terms of this sum will be zero).

Solution

No solution has been written yet.

Transcribed from the printed volume. Not translated.