Abelian Groups - László Fuchs [Springer]

2 Complete Groups 193
A
⏐
μ A A
α
−−−−→ B
⏐
⏐μB
B
commute.
Proof. The proof is given in the comments after Lemma 7.6 in Chapter 2. ut
We hasten to point out that the completion functor: A 7! QA is a pure-exact
functor in the sense that it carries short pure-exact sequences into pure-exact
sequences. Moreover, a stronger result holds:
Theorem 2.11. If 0 ! A ˛!B ˇ!C ! 0 is a pure-exact sequence, then the
induced sequence
0 ! QA Q˛! QB Qˇ! QC ! 0 (6.7)
of completions is splitting exact.
Proof. If the given sequence is pure-exact, then the induced sequence 0 ! A=nA !
B=nB ! C=nC ! 0 is exact for every n; see Theorem 3.1 in Chapter 5. The
completion functor (as inverse limit) is left-exact (Theorem 5.6 in Chapter 2), so for
the exactness of (2) it suffices to show that Qˇ is a surjective map. By Lemma 7.2 in
Chapter 2, Im Qˇ is complete and contains ˇB as a dense subgroup, so necessarily
Im Qˇ D QC. What remains to be proved is only the purity of Im Q˛ in QB. Themap
a C A 1 7! ˛a C B carries A=A 1 onto a pure subgroup of B=B 1 , which along with the
purity of B .B/ in QB shows that B .˛A/ is pure in QB. Inviewof B˛ D Q˛ A and
the divisibility of Q˛ QA= Q˛ A .A/, we infer that Q˛ QA must be pure in QB. The algebraic
compactness of QA implies the splitting.
ut
Corollary 2.12. Under the canonical map A W A ! QA, a p-basic subgroup of A
maps upon a p-basic subgroup of QA.
Proof. Since Ker D A 1 , Sect. 6(F) in Chapter 5 implies that maps p-basic
subgroups of A isomorphically upon p-basic subgroups of A. Owing to the
divisibility of QA=A and the purity of A in QA, the claim follows.
ut
Example 2.13. The p-adic (as well as the Z-adic) completion of Z .p/ is J p ,andtheZ-adic
completion of Z is Q p J p with p running over all primes.
Example 2.14. The Z-adic completion of ˚p Z.p/ is Q p Z.p/.
Example 2.15. Let B D˚n B n where B n Š˚Z.p n / with an arbitrary number of components.
The p-adic (Z-adic) completion of B is C,whereC=B denotes the divisible part of the factor group
. Q n B n /=B.