I got two questions about $p$-adic numbers:

1. I often read that the field $\mathbb Q_p$ is much different than the field $\mathbb R$.

An element of $\mathbb Q_p$ is of the form $\sum_{i=-k}^{\infty}a_ip^i$ where $a_i\in \{0,...,p-1\}$.

But isn't this just a real number? So at least the elements of $\mathbb Q_p$ are a subset of $\mathbb R$? That would mean that these fields are especially different in terms of their operation?

1. Let $x\in \mathbb Q_p^*$. Why $x$ can be written uniquely like this: $x=p^na$ where $a$ is an element of the $p$-adic integers?

• $\Bbb{Q}_p$ is not isomorphic to a subfield of $\Bbb{R}$ for any $p$. Dietrich's (+1) argument can be extended to all $p$. See here. – Jyrki Lahtonen May 15 '15 at 17:57

As for the first question, the answer is no. For example, what real number should represent this $$\sum_{n=0}^{\infty} 5^{n!}$$ $5$-adic number? Note that in real numbers this series is obviously divergent.

As for the second question: you should know that $\Bbb{Z}_p$ is a local ring whose unique maximal ideal is $p\Bbb{Z}_p$. This means that every $p$-adic integer which is not divisible by $p$ is invertible. Moreover it is a UFD, and every element can be factorized as $up^k$ for some unit $u$, some $k \ge 0$.

So every element of $\Bbb{Q}_p$ has the form $$\frac{up^k}{vp^h} = (uv^{-1})p^{k-h}$$

EDIT: The confusion comes to your mind, since you are thinking these numbers as they were real numbers: but they are not! Let's consider for example the sequence of integers (actual integers in $\Bbb{Z}$) $$1, \ \ 1+5, \ \ 1+5+5^2, \ \ 1+5+5^2+5^3, \dots$$ in $\Bbb{R}$ these sequence diverges. However, if we think it inside $\Bbb{Q}_5$, this sequence converges to the $5$-adic number $$A=\sum_{n=0}^{\infty} 5^n$$ actually, this is the inverse of $-4$ in $\Bbb{Q}_5$ since $$-4A=A-5A = (1+5+5^2+5^3+5^4+ \dots)-(5+5^2+5^3+\dots) = 1$$ (all of this is not true in $\Bbb{Q}_p$ for $p \neq 5$, where the sequence diverges). So you have $$\sum_{n=0}^{\infty} 5^n = -\frac{1}{4} \ \ \ \ \mbox{ in } \Bbb{Q}_5$$ This is possible because of the strange topological structure of $p$-adic integers.

• This "!" in $n!$ is a typo, right? – Duke May 15 '15 at 11:00
• No, it is not a typo. I didn't want to write $\sum_n 5^n$ since this is $1/4$, which could be confused with the real number $1/4$. – Crostul May 15 '15 at 11:18
• Sorry, but I am very confused now. $\sum_{n=0}^{\infty}5^{n!}$ is not of the form which I wrote above in my question, but $\sum_{n=0}^{\infty}5^n$ is in that form, with $p=5,k=0$ and $a_i=1$. And how we have $\sum_{n=0}^{\infty}5^n=\frac{1}{4}$. (This sequence diverges) Sorry for this (probably stupid question) but I am very new to p-adic numbers. – Duke May 15 '15 at 13:40
• @Duke $\sum_{n=0}^\infty 5^{n!}$ is of that same form, with $a_i=1$ if $i=0,1,2,6,24, \ldots$, and $a_i=0$ otherwise. – mathmandan May 15 '15 at 16:11
1. No, for example $x=\sqrt{-1}=i\in \mathbb{Q}_5$, but $x\not\in \mathbb{R}$.

2. See here.