Why, intuitively, `i` exists on 5-adics, but not on 2-adics?

Question

The square root of -1 exists in 5-adics, but not on 2-adics. I understand this since there is clearly a number that, multiplied by itself, results in -1, on p-adics. I also understand why, algebraically, we can't construct such number on 2-adics. What I do not understand is the meaning and intuition behind that. Does that mean that 3-adics work like complex numbers? If we used 5-adics for programming, would we be able to represent complex numbers over the entire complex plane as a single value, rather than two? Why would that be the case?

Answer 1

Nothing "works like the complex numbers" in the $p$-adics when you are looking at the fields $\mathbf Q_p$: the complex numbers are algebraically closed, but $\mathbf Q_p$ is very far from being algebraically closed: its algebraic closure is an infinite-degree extension. Do not try to think of $\mathbf C$ as being like some finite extension of $\mathbf Q_p$.

On the other hand, you could say "every quadratic extension in characteristic $0$ looks like the complexes over the reals" if you are just thinking about the field automorphisms: if $E/F$ is a field extension of degree $2$ then $E = F(\sqrt{d})$ where $d \in F^\times$ is not a square in $F$ and the unique field automorphism of $E$ fixing $F$ is $a + b\sqrt{d} \mapsto a-b\sqrt{d}$ for $a, b \in F$.

Answer 2

In the equation $x^2=-1$ the exponent is a unit in $5$-adics but not $2$-adics. You can lift the additive inverse ordered pair $(2,3\bmod 5)$ by the Hensel process; but if you try it with $(1,1\bmod 2)$ the non-unit exponent, becoming a coefficient upon differentiation, gives a wrong form for the derivative killing the Hensel lift.

The derivative in the $2$-adics case can be fixed if you can ground the solution $\bmod 4$, but this works only when the radicand has residue $1\bmod 8$. Thus you can get $\sqrt{-7}$ in $2$-adics.