Existence of $\sqrt{-1}$ in $5$-adics, show resulting sum is convergent.

Question

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.

Answer 1

The convergence should be in $\mathbb{Z}_5$, that works differently than in $\mathbb{R}$.

I try first an approximation $$2^2 = 4 = -1 + 5$$

Say I have $a$ with $a^2 = -1$. Then

$$\left(\frac{a}{2}\right)^2 = \frac{-1}{-1+5}= \frac{1}{1-5}$$

So take $$\frac{a}{2} = (1-5)^{-1/2}$$

We have the binomial formula $$(1-t)^{-1/2} = 1+\frac{t}{2}+\frac{3 t^2}{8}+\frac{5 t^3}{16}+\frac{35 t^4}{128}+\frac{63 t^5}{256}+\frac{231 t^6}{1024}+\frac{429 t^7}{2048}+\\ +\frac{6435 t^8}{32768}+\frac{12155 t^9}{65536}+\frac{46189 t^{10}}{262144}+ \cdots$$

convergent in $\mathbb{Z}_5$ if $|t|_5<1$. Now, $t=5$ and $|5|_5=1/5 <1$. Therefore: $$a = 2\cdot ( 1+\frac{5}{2}+\frac{3\cdot 5^2}{8}+\frac{5 \cdot5^3}{16}+\frac{35\cdot 5^4}{128}+\frac{63\cdot 5^5}{256}+\frac{231\cdot 5^6}{1024}+\frac{429\cdot 5^7}{2048}+\\ +\frac{6435\cdot 5^8}{32768}+\frac{12155\cdot 5^9}{65536}+\frac{46189 \cdot5^{10}}{262144} + \cdots)$$

with the other root $-a$.

Answer 2

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.