Imaginary Numbers in the Ring of $p$ Adic Integers?

Question

I am currently reading the book "ultrametric calculus: an introduction to $p$ adic analysis" by Schikhof, and I came across this problem:

"Prove that $x^2+1=0$ has no solutions in $\mathbb{Z}_3$ but has two solutions in $\mathbb{Z}_5$."

Here, $\mathbb{Z}_p$ denotes the ring of $p$ adic integers. The issue that I am having is that the book has never given any theorems for irreducibility in $\mathbb{Z}_p$ or any helpful theorems related to finding solutions to polynomials $f(x)\in\mathbb{Z}_p[x]$ up until this point, so I am assuming that there must be some way to do this solely on the construction of $\mathbb{Z}_p$? I am unsure on how I could do this, and could use some help as I have completely run out of ideas. Any help is appreciated.

Answer 1

At the early point where this exercise appears in the text (exercise 3.G, page 9), after defining $p$-adic integers as "infinite expansions to the left", and introducing adding and multiplication on them via "elementary school method (addition with carries, long multiplication) up to infinity", you are probably supposed to do it like this:

a) The $\mathbb Z_3$ case: The number $-1$ written as $3$-adic is

$$\dots 2222$$

So assume there is a $3$-adic number some $x= \dots x_3 x_2 x_1 x_0$ whose square is that, we need for starters $x_0 \cdot x_0 \equiv 2$ modulo $3$ which is already impossible.

b) The $\mathbb Z_5$ case: Now, $-1$ is the $5$-adic number

$$\dots 44444$$

and we need to find two different $ \dots x_3 x_2 x_1 x_0$ which when squared give that. Writing out the long multiplication, we need: $x_0 \cdot x_0 = 4$.

One solution is $x_0=2$. Putting that in and collecting the next column, we need $4x_1 \equiv 4$ modulo $5$, i.e. $x_1=1$. Putting that in and collecting the next column, we need $4x_2+1 \equiv 4$ modulo $5$, i.e. $x_2=2$. Putting that in and collecting the next column, we need $4x_3+4 \equiv 4$ modulo $5$, i.e. $x_3=0$. Coontinuing like this, you'll see you always get an equation of the form $4x_n + b \equiv 4$ modulo $5$ which can be solved uniquely.

Or, we could have started with $x_0=3$. Putting that in and collecting the next column, we need $6x_1+1 \equiv 4$ modulo $5$, i.e. $x_1=3$. Continue from here ...

So there are two solutions,

$$ \dots 0212$$ and $$\dots 4233$$

As a bonus, note they are each other's additive inverse.

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As a final aside, I advise strongly against thinking of these as "imaginary" numbers. Obviously e.g. that $5$-adic number whose expansion is $\dots 0212$ is not more or less imaginary than any other $p$-adic number, or for that matter, the real number whose expansion is $1.4142 \dots$ and which we call $\sqrt{2}$. Thinking of $p$-adic numbers as if they somehow "sit inside" $\mathbb R$ and/or $\mathbb C$ is a painfully common source of confusion. Cf. https://math.stackexchange.com/a/4007515/96384, here especially sections 4 and 5.

Answer 2

There are a profusion of ways of finding an $i$ in $\Bbb Z_5$. For instance, write $-1=4-5$ so that $-\frac14=1-\frac54$. Then find $\sqrt{-\frac14}$ by applying the Binomial Series for $(1+x)^{\frac12}$ to the case $x=-\frac54$ and noticing that the resulting series is $5$-adically convergent, ’cause powers of $5$ go to zero. (There are no fives in the denominators of the expansion.)

Answer 3

One simple way is to use approximation techniques which work for real numbers. In this case, the Wikipedia article Methods of computing square roots refers to the Babylonian method:

Perhaps the first algorithm used for approximating $\sqrt{S}$ is known as the Babylonian method, despite there being no direct evidence beyond informed conjecture that the eponymous Babylonian mathematicians employed this method.$^{[1]}$ The method is also known as Heron's method,

This is a special case of finding square roots (which can also be used for polynomial roots) of Newton's method for $p$-adic numbers.

In p-adic analysis, the standard method to show a polynomial equation in one variable has a $p$-adic root is Hensel's lemma, which uses the recursion from Newton's method on the $p$-adic numbers.

In the specific case of $\sqrt{-1}\,$ in $\mathbb{Z}_5$ the calculations go as follows. Start with an approximation such as $\,x_0 = 2+O(5^5).\,$ Next, $\,x_1 = (x_0 +\frac{-1}{x_0})/2 = (2 + \frac{3124}{2})/2 = 782+O(5^5).\,$ Next, $\,x_2 = (x_1 + \frac{-1}{x_1})/2 = 1432+O(5^5).\,$ Finally, $\,x_3 = (x_2 + \frac{-1}{x_2})/2 = 2057 +O(5^5).\,$ The calculations are all done modulo $5^5$.

Another method is to iterate the Frobenius endomorphism. Again, start with $\,y_0 = 2+O(5^1).\,$ Next, $\,y_1 = y_0^5 = 32+O(5^2) = 7+O(5^2).$ Next, $\,y_2 = y_1^5 = 7^5 + O(5^3) = 57 + O(5^3).$ Next, $\,y_3 = y_2^5 = 182 + O(5^4).\,$ Finally, $\,y_4 = y_3^5 = 2057 + O(5^5).\,$ One advantage of this method is that each iteration yields one more digit and the final precision is not fixed in advance. Also, only multiplications are used. The computations can be easily done with PARI/GP as follows

gp> x = 2+O(5^5); y = 2 + O(5^1);
gp> for(n=1,5, print([x,lift(x)],"\t",[y,lift(y)]); x=(x+(-1)/x)/2; y=y^5)
[2+O(5^5),2]    [2+O(5),2]
[2+5+5^2+5^3+5^4+O(5^5),782]    [2+5+O(5^2),7]
[2+5+2*5^2+5^3+2*5^4+O(5^5),1432]       [2+5+2*5^2+O(5^3),57]
[2+5+2*5^2+5^3+3*5^4+O(5^5),2057]       [2+5+2*5^2+5^3+O(5^4),182]
[2+5+2*5^2+5^3+3*5^4+O(5^5),2057]       [2+5+2*5^2+5^3+3*5^4+O(5^5),2057]
gp>

Of course, PARI/GP can compute the square root directly using sqrt(-1+O(5^5)).

The specific case of $\sqrt{-1}\,$ in $\mathbb{Z}_3$ is easier because there is no quadratic residue congruent to $-1$ modulo $3$ and this implies that there is no solution to $\,x^2+1\equiv 0 \pmod 3$ and hence no solution to $\,x^2+1=0\,$ in $\mathbb{Z}_3$. It is instructive to find out why the two methods used to find solutions $\mathbb{Z}_5$ fail in $\mathbb{Z}_3$.