Assume that we define the ring of the $p$-adic integers as the projective limit $$\mathbb{Z}_p =\varprojlim \frac{\mathbb{Z}}{p^n\mathbb{Z}}$$
Then $\mathbb{Q}_p$, the field of the $p$-adic numbers is no more than the field of quotients of $\mathbb{Z}_p$.
But, to do so, we must prove that $\mathbb{Z}_p$ is an Integral Domain (with this definition).
In order to do so, I take two sequences $(x_n)_{n\in\mathbb{N}}$ and $(y_n)_{n\in\mathbb{N}}$ in the projective limit which are nonzero (assume that $x_{n_{0}}\not =0 \pmod{p^{n_0}}$ and $y_{m_{0}}\not =0 \pmod{p^{m_0}}$ and I have to prove that the product is nonzero. I am trying to prove that $x_{n_{0}+m_{0}}y_{n_{0}+m_{0}}\not=0 \pmod{p^{n_{0}+m_{0}}}$.
Since both sequences are in the projective limit, I can easily see that $x_{n_{0}+m_{0}}\not=0 \pmod{p^{n_{0}+m_{0}}}$ and $y_{n_{0}+m_{0}}\not=0 \pmod{p^{n_{0}+m_{0}}}$. But how can I be sure that the product will also be nonzero?
Thanks in advance.
Every element $x\in\mathbb{Z}_p\setminus\{0\}$ can be written uniquely in the form $x=p^mu$ with $u$ a p-adic unit and $m\ge 0$. If $\mathbb{Z}_p$ would have zerodivisors, say $xy=0$ with $x,y\in\mathbb{Z}_p$, then we see that up to a unit we have that a power of $p$ is zero in $\mathbb{Z}_p$, which is obviously impossible.
Here's a quick sketch of the proof that every element $x\in\mathbb{Z}_p\setminus\{0\}$ can be written uniquely in the form $x=p^mu$ with $u$ a p-adic unit and $m\ge 0$:
First we show that $x$ is a p-adic unit iff it is not divisible by $p$ (in $\mathbb{Z}_p$).
If $x$ is a p-adic unit, then $x \pmod{p}$ is nonzero, so it is not divisible by $p$. For the converse, if $x=(x_n)_n$ is not divisible $p$, define $y=(x_n^{-1})_n$ where $x_n^{-1}$ denotes the inverse of $x_n\pmod{p^n}$. It is not hard to show $y\in\mathbb{Z}_p$ and $xy=1$, so $x$ is a p-adic unit.
The statement follows from this because if $p^m$ is a divisor of $x$ for all $m\ge 0$ then $x=0$ by the above.