) > in particular, its dimension would be independent of p . ) ... ) ) Why? a basis of A over Z maps (bijectively) to a basis of A/pA over Z/pZ . thus the dimension of A/pA over Z/pZ equals the rank of A , which is independent of p .

...

2A = { (a_n) in P | v(a_n) --> infinity and each a_n is even }

so there is an obvious countable basis for A/2A over Z/2Z , namely { e_i } , where e_i is 0 in all coordinates, except for a single 1 in the i-th coordinate. thus the dimension of A/2A over Z/2Z is countably infinite.

The e_i, regarded either as group generators or as a vector space basis, generate only finite sums, i.e. sequences that are 0 for all but finitely many coordinates. On the other hand, A/2A consists of arbitrary 0,1 sequences. Thus A/2A is uncountably infinite, and so its dimension over Z/2Z is also uncountably infinite.

no, the condition on A (namely, v(a_n) --> infinity as n --> infinity ) shows that A/2A consists only of even-odd sequences that have only finitely many odd terms. mike