By Jean Dieudonne

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But lim(. . ← −Z← −Z← − Z) = ← p=2 If we consider mixing these operations in groups, the following remarks are appropriate. i) localizing and then profinitely completing is simple and often gives zero, Gp if ∩ = ∅ (G ) = p∈ ∩ 0 if ∩ = ∅ . g. (G, l, ) = (Z , ∅, p) gives Qp = 0. g. ¯ p = Q ⊗ Zp , the “field of p-adic numbers”, usually a) (Z0 )¯p = Q denoted by Qp . Qp is the field of quotients of Zp (although it 27 Algebraic Constructions is not much larger because only 1/p has to be added to Zp to make it a field).

Is just the localization S1 → S1 studied above. ii) π = Z/pn for π = 0 if p ∈ / π = Z/pn if p ∈ . For general π, take finite direct sums and then direct limits of the first two cases. Step 2. The case (X − →X) = K(π, n) → − K(π , n) . If localizes homology, then it localizes homotopy as in Step 1 because π = Hn X, π = Hn X . If localizes homotopy, then we use induction, Step 1, diagram III in remark b) and remark c) to see that localizes homology. 48 Step 3. The general case X − →X. If localizes homology, apply the Hurewicz theorem for n = 1 to see that localizes π1 .

Proof: We have to check exactness for ( )⊕(0) i−j 0 → Z(p) −−−−→ Zp ⊕ Q −−→ Qp → 0 where i and j are the natural inclusions Zp −−→ Qp , Q −−→ Qp . ⊗Q ⊗Zp Take n ∈ Z and q ∈ Zp then (n/pa ) + q = (n + pa q)/pa can be an arbitrary p-adic number. Thus i − j is onto. It is clear that ( ) ⊕ (0) has zero kernel. To complete the proof only note that a rational number n/m is also a p-adic integer when m is not divisible by p. Thus n/m is in Z localized at p. Corollary The ring of integers localized at p is the fibre product of the rational numbers and the ring of p-adic integers over the p-adic numbers.

### Algebre Lineaire Et Geometric Elementaire by Jean Dieudonne

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