Bourbaki

Exercise 23, § 6

General Topology, III, 322

Bourbaki marks this one as harder than the rest.

If pp is any prime number and xx is any non-zero rational number, let vp(x)v_p(x) denote the exponent of pp in the decomposition of xx into prime factors. The mapping vpv_p of Q∗\mathbf{Q}^* into Z\mathbf{Z} is called the pp-adic valuation on Q\mathbf{Q}. The set of all x∈Qx \in \mathbf{Q} such that either x=0x = 0 or vp(x)≥mv_p(x) \geq m (for some m∈Zm \in \mathbf{Z}) is the fractional ideal (pm)(p^m) of Q\mathbf{Q}.

a) Show that the ideals (pm) (m∈Z)(p^m)\ (m \in \mathbf{Z}) form a fundamental system of neighbourhoods of o in Q\mathbf{Q} for a locally retrobounded topology (Exercise 22) Tp\mathfrak{T}_p on Q\mathbf{Q}, called the pp-adic topology. The completion Qp\mathbf{Q}_p of Q\mathbf{Q} in this topology is a field called the pp-adic field, whose elements are called pp-adic numbers. The closure Zp\mathbf{Z}_p of Z\mathbf{Z} in Qp\mathbf{Q}_p is a compact open principal ideal domain in Qp\mathbf{Q}_p, in which p=pZp\mathfrak{p} = p\mathbf{Z}_p is the only non-zero prime ideal; Zp/p\mathbf{Z}_p/\mathfrak{p} is isomorphic to the prime field Fp=Z/(p)\mathbf{F}_p = \mathbf{Z}/(p), and more generally the additive quotient group pm/pn\mathfrak{p}^m/\mathfrak{p}^n is isomorphic to

Z/(pn−m) if m<n. \mathbf{Z}/(p^{n-m}) \text{ if } m < n.

The elements of Zp\mathbf{Z}_p are called pp-adic integers.

b) Show that every continuous homomorphism of the additive group Qp\mathbf{Q}_p into itself is of the form x→axx \to ax, where a∈Qpa \in \mathbf{Q}_p [if ff is such a homomorphism, show that f(rx)=rf(x)f(rx) = rf(x) for all r∈Qr \in \mathbf{Q}, and then pass to the limit in Qp\mathbf{Q}_p].

c) Show that every compact subgroup G≠{0}G \neq \{0\} of the additive group Qp\mathbf{Q}_p is identical with one of the pn\mathfrak{p}^n (n∈Zn \in \mathbf{Z}), and that there is no non-compact closed subgroup of the additive group Qp\mathbf{Q}_p, except Qp\mathbf{Q}_p itself. [Let mm be the largest integer such that G⊂pmG \subset \mathfrak{p}^m; by considering the quotient group (G+pn)/pn(G + \mathfrak{p}^n)/\mathfrak{p}^n for n>mn > m, show that G+pn=pmG + \mathfrak{p}^n = \mathfrak{p}^m, and then use formula (1) of § 3, no. 1].

Solution

No solution has been written yet.

Transcribed from the printed volume. Not translated.