2

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

user185887
  • 425
  • 1
    One way to think of $p$-adic numbers is by analogy with the real numbers, only you are completing a metric topology using a different notion of distance. Values are "close" if their difference is divisible by a high power of $p$. In that respect a $4$-adic number system might be exactly the same as a $2$-adic number system. You should let Readers know how you think of the $p$-adic numbers if you want a more cogent response. – hardmath Mar 04 '18 at 17:25

2 Answers2

7

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 $\{2^k \mathbf{Z}\}$.

bzc
  • 8,622
  • This does also define it topologically, since the "limitands" are finite, thus having unique Hausdorff topologies, and thus giving the projective limit a canonical topology. Since it is a countable limit of complete metrizable spaces, it is complete metrizable. – paul garrett Mar 04 '18 at 18:15
0

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.

kccu
  • 20,808
  • 1
  • 22
  • 41
  • But wait isn't the condition for r and s to be co prime with p rather than being not divisible by p?? – user185887 Mar 04 '18 at 17:31
  • Those are equivalent if $p$ is prime (the only factor $r$ and $s$ could possibly share with $p$ is $p$ itself since $p$ is prime). If we phrase it that way and demand that $r$ and $s$ are coprime to $4$, then you can see for $x=4$ there is no way to choose $r$ and $s$ - one of them will have to be divisible by $2$ and thus will share a factor with $4$. – kccu Mar 04 '18 at 17:33
  • 1
    $2$-adic norms are commonly defined on number fields where $2$ isn't a prime number. Even on number fields where $2$ isn't even prime in the ring of integers of the completion! $2$ has the same problem in these cases that $4$ does in this case. This answer asserts too much; the only problem defining the $p$-adic norm for $p$ composite is that you can't do it with this specific method that only works for $p$ prime. –  Mar 04 '18 at 18:08
  • @Hurkyl You have a good point, I am surely not an expert on this. The other answer may be more accurate, but I am doubtful that OP can understand it. – kccu Mar 05 '18 at 02:08