Why is it actually important for $p$ to be prime in the ring of $p$-adic numbers?

Question

In the proof that the set $Z_{[p]}$ of all $p$-adic numbers is a ring, it is not necessary for $p$ to be prime. Is it actually important for $p$ to be prime for the ring of $p$-adic numbers?

Answer 1

Let $R$ be any ring and $\mathfrak a\subset R$ be any ideal. Then for all $n\in \Bbb N$, $\mathfrak a^n$ is also an ideal, hence we have quotient rings $R/\mathfrak a^n$, and of course we have natural homomorphisms $R/\mathfrak a^n\to\ R/\mathfrak a^m$ whenever $m>n$. From this, we arrive at the projective limit $$ \lim_\leftarrow R/\mathfrak a^n.$$ This does not require $\mathfrak a$ to be prime.

Answer 2

For $a=\prod_{p|a} p^{e(p)}$ non-prime we have $$\varprojlim \Bbb{Z}/(a^n)=\varprojlim \Bbb{Z}/(\prod_{p|a}p^{e(p)n})\cong \varprojlim \prod_{p|a}\Bbb{Z}/(p^{e(p)n})$$ $$\cong \prod_{p|a}\varprojlim \Bbb{Z}/(p^{e(p)n})=\prod_{p| a} \Bbb{Z}_p$$ Which is the completion of $\Bbb{Z}$ for the norm $$\| \frac{c}{d} \prod_{p| a} p^{r(p)}\|_a = 2^{-\min_p r(p)}, \qquad \gcd(a,cd)=1$$ This norm is not an absolute value when $a$ has more than one prime divisor.

The completion of $\Bbb{Q},\|.\|_a$ is isomorphic to $\prod_{p|a} \Bbb{Q}_p$.