TITLE: Existence of $\sqrt{-1}$ in $5$-adics, show resulting sum is convergent.
QUESTION [6 upvotes]: I know that to prove the existence of a square root of $-1$ in $\mathbb{Z}_5$, I can just plug $x = -5$ and $a = 1/2$ into the Taylor expansion$$(1 + x)^a = \sum_{n=0}^\infty \binom{a}{n} x^n.$$However, I am confused on how I would make sure that the resulting sum is convergent. Could anyone help me with that? Thanks very much in advance.

REPLY [4 votes]: There are many ways of getting convergent processes for finding roots $p$-adically, and in addition, there is a special method for finding a $(p-1)$-th root of unity in $\Bbb Q_p$.
First, for the binomial expansion of $(1+t)^{1/2}$, in which we need the coefficients all to be $p$-adic integers when $p\ne2$. Consider any $p$-adically convergent sequence of positive integers, with limit $1/2$. For $p=5$, for instance, the expansion is $\dots2223;$ That is, $\frac12=3+2\cdot5+2\cdot5^2+2\cdot5^3+\cdots$. This exhibits $1/2$ in just this way. Call the approximating integers $n_j$, and look at the coefficient of $t^m$ in $(1+t)^x$, that is, $\binom xm=x(x-1)(x-2)\cdots(x-m+1)/m!$. Even when the denominator has powers of $5$ in it, it’s still a polynomial, and continuous, so that when you substitute the integers $n_j$ into this polynomial, the limiting value is the coefficient of $t^m$ in $(1+t)^{1/2}$. Thus all coefficients are $5$-adic integers, and you can safely plug in $5$ for $t$ and get a $5$-adically convergent series.
Second, you can always start with an approximate square root of your number, $2$ will do if you want $5$-adic $\sqrt{-1}$, and use Newton-Raphson. Just as in the real situation, you double your accuracy at every repetition of the process. If you have a pocket $p$-adic calculator, this is really fast.
The special method for finding $(p-1)$-th roots of unity is this: if $\zeta$ is such a root in $\Bbb Z_p$ and $z\in\Bbb Z_p$ with $v_p(z-\zeta)=n>0$, that’s the (additive) $p$-adic valuation normalized
so that $v_p(p)=1$, then $v_p(z^p-\zeta)=n+1$ (unless $p=2$, when the claim is good only when $n>1$, but you already know the roots of unity in $\Bbb Q_2$!). So for $i\in\Bbb Q_5$, just start with $2$ and repeatedly raise to the fifth power.