P adic numbers number theory

Question

Why does it make sense for prime adic numbers. I mean why won't be 4 adic numbers possible? ( I do not want any rigorous proof. A simple basic reason in detail will be appreciated) Thank you

Answer 1

The problem is in defining the $p$-adic norm for $p$ composite.

If $p$ is prime, we can write any nonzero rational number $x$ as $x=\frac{p^a r}{s}$ where $r$ and $s$ are not divisible by $p$. The fact that $p$ is prime guarantees that the exponent $a$ is unique. Then we define the $p$-adic norm of $x$ to be $|x|_p=p^{-a}$.

If $p$ is composite, say $p=4$, then the exponent $a$ is not unique. For instance, take $x=8$. We can write $x=\frac{4^1\cdot 2}{1}= \frac{4^2 \cdot 1}{2}$, and in both instances $r$ and $s$ are not divisible by $4$. But this gives us two possible values of $a$ - either $1$ or $2$. There's just no way to fix this to get a $4$-adic norm.

Answer 2

Others are commenting about the inability to define a $n$-adic norm when $n$ is not prime, but it should be noted that you can define $\mathbf{Z}_n$ purely algebraically in the same manner as usual: $$ \mathbf{Z}_n := \varprojlim_k \mathbf{Z}/n^k\mathbf{Z}. $$

One problem with this is that the resulting ring only picks up the distinct prime factors of $n$, and decomposes as a product by the Chinese Remainder Theorem. In particular, it is a domain if and only if $n$ is divisible by at most one prime.

To use your example, $$ \mathbf{Z}_4 = \varprojlim_{k} \mathbf{Z}/4^k \mathbf{Z} = \varprojlim_k \mathbf{Z}/2^k \mathbf{Z} = \mathbf{Z}_2 $$ since the set of subgroups $\{4^k \mathbf{Z}\}$ is cofinal for $\{2^k \mathbf{Z}\}$.