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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?

MaiaVictor
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  • You can describe a $3-$adic number with one value? How? Anyway, there isn't even a square root of $-1\pmod 3$. Maybe you meant $5-$adic? – lulu Dec 10 '22 at 21:11
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    See this question for some relevant details. – lulu Dec 10 '22 at 21:13
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    And here is a discussion of the $5-$adic square root of $-1$. – lulu Dec 10 '22 at 21:14
  • @lulu yes, 5-adic, sorry for the confusion. About your question, can't you represent a 5-adic as an infinite string containing its digit expansion? That's a single number, even though it is infinite. That would be a sensible representation in a lazy programming language, like Haskell. How are p-adics represented usually, in programming? I'm curious. – MaiaVictor Dec 10 '22 at 21:16
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    You can easily represent a complex number as a single (usually infinite) string of integers, if you really want to. – lulu Dec 10 '22 at 21:20
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    you can "approximate" $p$-adics by working in $\Bbb Z/p^n\Bbb Z$ for large $n$ – Lukas Heger Dec 10 '22 at 21:20
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    here is an article on implementations of $p-$adics. – lulu Dec 10 '22 at 21:21
  • @lulu great article, thanks – MaiaVictor Dec 10 '22 at 21:26
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    No problem, glad to have helped. – lulu Dec 10 '22 at 21:26
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    To answer the question in the title, it's because the exponent $2$ in $x^2=-1$ is relatively prime to the prime $5$, but not relatively prime to the prime $2$. – Gerry Myerson Dec 10 '22 at 23:00
  • @lulu if I may ask, is there any reason for not representing p-adics as just lazy infinite list of digits? Seems like the idea proposed in the paper is an over-complication. Not sure if I'm missing something. – MaiaVictor Dec 11 '22 at 01:21
  • It all depends what you want to do. Taking an element of $\mathbb Z/p^n\mathbb Z$ for large $n$ as an approximation is fine for most applications. – lulu Dec 11 '22 at 01:23
  • "The square root of -1" is not a thing. Not a thing that exists somewhere as such, a thing in itself or so. Rather: Some fields contain square roots of $-1$ (if they do, they usually contain two), and others don't. The field $\mathbb C$ does, the field $\mathbb R$ does not. The field $\mathbb Q_5$ does (and so do infinitely many other $\mathbb Q_p$), the field $\mathbb Q_2$ does not (and neither do infinitely many other $\mathbb Q_p$). – Torsten Schoeneberg Dec 11 '22 at 01:37

2 Answers2

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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 $a, b \in F$.

KCd
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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.

Oscar Lanzi
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