I haven't been following too closely, but let me add a few remarks. Suppose N > 1 is an integer. Multiplication on the ring of N-adic integers is continuous (as already discussed below). The purported issue:

If multiplication is continuous, you can then extend multiplication to the n-adics by continuity. But continuity of multiplication depends on the fact that | a | * | b | = | (a * b) | which only holds for n prime.

is not really an issue. Continuity does not require equality there; indeed it follows from the inequality | a | * | b | >= | a * b | , which holds trivially. In fact, continuity would follow from much less than that. ) So as I had read somewhere, there is indeed a natural multiplication on the N-adic integers which is continuous relative to the topology induced by the N-adic metric. ) ) Even though for N composite the N-adic integers do have zero-divisors. Not necessarily. If N is a power of a prime, say p^a (with a > 0 ) then the N-adic integers are the same as the p-adic integers, which have no zero-divisors. In general, if N = p_1^a_1 p_2^a_2 ... p_r^a_r is its prime factorization, then the ring of N-adic integers is isomorphic to the direct product of the q_i-adic integers , for i = 1, 2, ... r , where q_i = p_i^a_i , by the Chinese remainder theorem. So if N has (at least) two different prime factors, then there are zero-divisors. The construction can be done in much greater generality. Atiyah-MacDonald (in "Introduction to Commutative Algebra") define the a-adic completion of a commutative ring A having identity, where a is an ideal of the ring A . It's quite possible that these completions work equally well for non-commutative rings. Michael Reid