Problem with the ring of p-adic integers

Question

I just tried to figure out the meaning of this equation:

"We can write −1 as a p-adic integer: −1=(p−1)+(p−1)p+(p−1)p^2 +(p−1)p^3 +..."

In which sense is the right side of the equation equal to "-1". For any finite "n" i get the following equation:

(p−1)+(p−1)p+(p−1)p^2 +...+(p−1)p^n = -1+p^n+1 ;

so does the first "equation" mean equal to -1 mod p^n+1 for any finite n?

I just try to find a explicit description for the ring of p-adic integers as set. But i really have trouble with that, because i don't see how we get negative integers with infinite sums of positive integers since the ring of integers seems to be subring.

I apologize for my bad english, also i haven't figured out how this editor works.

Answer 1

By the geometric series we have $$ \frac{1}{1-p}=\sum_{i=0}^{\infty} p^i, $$ so that $$ -1=(p-1)(1+p+p^2+\cdots)=(p-1)+(p-1)p+(p-1)p^2+\cdots $$

Answer 2

The ring $\mathbb{Z}_p$ of $p$-adic integers has a metric $d(x,y) = p^{-\mathrm{ord}_p(x-y)}$ which induces a topology on $\mathbb{Z}_p$ (where $\mathrm{ord}_p(x)$ is the highest power of $p$ which divides $x$, or $\infty$ if $x = 0$). Essentially, this means that a sequence $x_n$ converges to $x$ in $\mathbb{Z}_p$ if and only if $\mathrm{ord}_p(x_n-x) \to \infty$ as $n \to \infty$. Now, the statement says that $$\sum_{n=0}^\infty (p-1) \cdot p^n = -1$$ in this topology, i.e. the sequence of partial sums converges to $-1$.

As for the question about negative sum vs. positive elements: this just means that $\mathbb{Z}_p$ cannot be given a structure of ordered ring, at least not in a way that interacts well with limits. (And in fact, for example for $p \equiv 1 \pmod{4}$, $\mathbb{Z}_p$ turns out to have solutions to $x^2 = -1$ which implies no structure of ordered ring is possible at all in those cases.)