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While studying a calculation in Koblitz's P-adic Numbers, P-adic Analysis, and Zeta-Functions, he remarks the following:

...we were sloppy when we wrote $4/3 = (1 + 7/9)^{1/2}$. In both $\mathbb{R}$ and $\mathbb{Q}_7$ the number $16/9$ has two square roots $\pm 4/3$. In $\mathbb{R}$, the series for $(1 + 7/9)^{1/2}$ converges to $4/3$, i.e., the positive value is favored. But in $\mathbb{Q}_7$ the square root congruent to $1 \text{ mod } 7$, i.e., $-4/3 = 1 - 7/3$, is favored.

So, my question is why is, in $\mathbb{Q}_7$, the root $-4/3$ "favoured"?

Naweed G. Seldon
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    It's nothing more than noting that $-4/3\equiv 1\pmod{7}$, while $4/3 \equiv -1\pmod{7}$. Unlike $\mathbb{R}$, the space $\mathbb{Z}_7$ is the limit of a series of finite rings, and there's no canonical notion of positivity in them that carries over. – anomaly Nov 12 '18 at 00:33
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    "favoured" means finding a function $f$ continuous (and analytic) for $|x-1| < 1$ such that $f(x)^2 = x$ and $f(1) = 1$. In $\mathbb{Q}_7$ the square root of $1+7/9$ that is in $|x-1| < 1$ is $-4/3$. – reuns Nov 12 '18 at 00:51
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    Let’s face it, it’s just a matter of taste. Don’t forget that there is no positivity in the $p$-adic world. In this case, I would go with the square root congruent to $1$ rather than that congruent to $-1$, too. – Lubin Nov 24 '18 at 21:39

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Consider the polynomial $1+X$ in the ring $A=\mathbb Q[[X]]$ of power series over $\mathbb Q$. The degree 2 equation $T^2 = 1+X$ has no solution over $\mathbb Q[X]$ but it has two solutions over the domain $A$. Mainly, $T = \pm\left( 1+\frac{X}{2}-\frac{X^2}{8}+\frac{X^3}{16}-\frac{5 X^4}{128}+\frac{7 X^5}{256}-\frac{21 X^6}{1024}+\frac{33 X^7}{2048}-\frac{429 X^8}{32768}+\dots\right)$

The "positive" or "favoured" one is $$ B_{1/2}(X)= 1+\frac{X}{2}-\frac{X^2}{8}+\frac{X^3}{16}-\frac{5 X^4}{128}+\frac{7 X^5}{256}-\frac{21 X^6}{1024}+\frac{33 X^7}{2048}-\frac{429 X^8}{32768}+\dots $$ This is the Taylor expansion of $\sqrt{x+1}$ around $0$ in the classical-euclidean sense, seen purely as a formal power series. It is easy to check that $B_{1/2}\in \mathbb Z[1/2][[X]] \subset \mathbb Z_p[[X]$ for every odd prime $p$. In particular, for every odd $p$ one can consider $B_{1/2}$ as a continuous function $$ B_{1/2} : p\mathbb Z_p \rightarrow 1+ p\mathbb Z_p $$ and also $$ B_{1/2} : (-1,1)\subset \mathbb R\rightarrow [0,\infty) \subset\mathbb R. $$

Koblitz points out that $7/9$ is in both $(-1,1)$ and $7\mathbb Z_7$. Hence $B_{1/2}(7/9) = 4/3$ as function of $(-1,1)\subset \mathbb R$ while $B_{1/2}(7/9)=-4/3$ as function of $7\mathbb Z_7$.

eduard
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