Proving the p-adic numbers $\mathbb{Q}_p$ form a field

Question

I am trying to prove that $\mathbb{Q}_p$ forms a field. However, I am unsure of the best way to go about proving it. If I work with the power series representation of p-adic numbers I run in to trouble when showing the existence of multiplicative inverses and closure under multiplication. What is the best way to go about proving this?

Answer 1

$\mathbb Q$ is a field, $|\cdot|_p$ is a norm compatible with the field operations, $\mathbb Q_p$ is the completion with respect to this norm. That should suffice to show everything -and in fact show it just as one shows that $\mathbb R$ is a field.

For $a,b\in\mathbb Q_p$ let $a_n\to a$, $b_n\to b$ be converging sequences. Show that $a_nb_n$ converges. We define $a\cdot b=\lim_{n\to\infty} a_nb_n$.

For $a\ne 0$ let $a_n\to a$ be a converging seqeunce. As $|a|_p>0$, for almost all $n$, $|a-a_n|_p<|a|_p$ and hence $a_n\ne0$. Thus we can invert almost all terms, i.e. let $b_n=\frac1{1-n}$ for almost all $n$. Then $a_nb_n=1$ for almost all $n$, hence $a\cdot b=1$, where $b=\lim_{n\to\infty}b_n$ (why does the limit exits?).

Answer 2

For a complete answer, I refer you to that of @HagenvonEitzen. But I personally like to think of things in a maximally computational way. Just for the question of knowing the $\mathbb Q_p$ is closed under the operation of multiplicative inverse, let me offer the following, which uses strongly, however, the fact that $\mathbb Q_p$ is complete in the $|\bullet|_p$-topology.

Any nonzero $z\in \mathbb Q_p$ can be written in form $z=p^nz_0$, where $|z_0|_p=1$, or as we say, $z_0$ is a $p$-adic unit. Clearly it’s enough to show that $z_0$ has a reciprocal in $\mathbb Q_p$. I’ll do this first for the case that $\big|z_0-1\big|_p<1$. But $$ 1/z_0=\frac1{1-(1-z_0)}=\sum_{i-0}^\infty(1-z_0)^i\,, $$ a convergent series. The remaining case is that $z_0\equiv m\pmod p$, for an integer $m$ prime to $p$. Then there’s $n$ with $mn\equiv1\pmod p$, and it’s enough to find the reciprocal of $z_1=nz_0\equiv1\pmod p$, since $z_0^{-1}=nz_1^{-1}$.