How to define a branch cut for the square root function over prime characteristic? I was wondering about the following. Let $k$ denote an algebraically closed field. If it helps, I will take $k = \overline{F}_p$, meaning the algebraic closure of the finite field $F_p$ with $p$ elements, where $p$ is prime. One can also assume that $p$ is odd, for the sake of this question.
Over $\mathbb{C}$, one can define a "branch cut" for the square root function by removing $0$ and the negative real axis. This has the advantage that one can then pick a square root of $z$ in a consistent way.
Working now over $k$ as above, is there a way to define a "branch cut" for the (multivalued) square root function $z \mapsto \sqrt{z}$, for $z \in k$?
I am wondering whether this could be done in a sensible way, so that after removing the cut, the square root function can be defined as a single-valued function, and should be "continuous" in some sense.
Edit 1: in the case where $p \equiv 3 \,(\operatorname{mod} 4)$, in the finite field $\mathbb{F}_q$, where $q = p^r$, and $r$ is odd, so that $-1$ is not a square in $\mathbb{F}_q$ (see What are the finite fields for which -1 is not a square?), let $S_q$ denote all non-zero elements of $\mathbb{F}_q$ which are squares in $\mathbb{F}_q$. Then given any $y \in S_q$, there are two solutions $\pm x$ of $x^2 = y$, exactly one of which lies in $S_q$.
We attempt to generalize this approach to any odd $p$, and any $\mathbb{F}_q$, where $q=p^r$, where $r$ is any positive integer. We also denote by $S_q$ the set of non-zero elements which are squares. Let $\Delta_q$ be a set containing $(q-1)/2$ elements, such that given any non-zero element $z \in \mathbb{F}_q$, precisely one of $z$ and $-z$ lies in $\Delta_q$. Given $y \in S_q$, there are two solutions of $x^2 = y$, precisely one of which lies in $\Delta_q$.
If we can define the $\Delta_q$'s in such a way that they are compatible with the embeddings $\mathbb{F}_q \to \mathbb{F}_{q'}$ when $q'$ is a power of $q$, then I think we would be done, because things would carry over to the direct limit $\overline{\mathbb{F}}_p$ (unless I am mistaken somewhere).
Does there exist such $\Delta_q$'s? Can one construct explicitly such a sequence of $\Delta_q$'s?
Edit 2: I could not find a "good" way to define a branch cut for the square root function over prime characteristic. The discreteness of such fields is the main culprit.
 A: I think there is a much better setting for asking your question. The unfortunate fact is that $\overline{\Bbb F_p}$ is irremediably discrete, and there’s virtually no way of saying that one element is close or closer to any other.
On the other hand, there is a good parallel to the complex numbers $\Bbb C$, namely the various $p$-adic fields and their complete extensions.
The field $\Bbb Q_p$ is the completion of the rationals with respect to a distance function in which two rationals are close if their difference is divisible by a high power of the prime number $p$. Specifically, if $a,b$ are integers both indivisible by $p$, you put
$$
\bigl\vert \frac abp^n\bigr\vert_p=c^{-n}\,,
$$
where $c$ is a fixed real number bigger than $1$. Conventionally, one takes $c=p$. Then, you define, for rational numbers $\lambda$ and $\mu$,
$$
d_p(\lambda,\mu)=\vert\lambda-\mu\vert_p\,.
$$
You can easily show that $d_p$ satisfies the triangle inequality and the other requirements for a metric. It happens that $\Bbb Q$ is not complete, but you can complete it to get an uncountable field $\Bbb Q_p$ that is nowhere near being algebraically closed. Many of the rules you learned in high school apply, like the convergence of a geometric series depending on the size of the common ratio: smaller than $1$ and you get a convergent series whose limit is given by the same formula you learned in high school.
Why have I gone into all this? Because in $\Bbb Q_p$ you have an unambiguous way of defining $\sqrt{z\,}$, as long as the numbers $z$ you’re rooting are close enough to $1$. Think of complex numbers near $1$ for a moment: they have square roots, consistently in the sense of the map $\sqrt{1+z}$ being continuous, and well defined for $z$ now being small enough. In particular, you have trouble if you try to define a complex square root function $\sqrt{1+z}$ when $z$ gets too near to $-1$, ’cause then you get to the weird point of the square-root function.
What’s a basic way of seeing that you have a good, consistent, continuous definition of $\sqrt{1+z}\,$? I offer the Binomial expansion
$$
(1+z)^{1/2}=1+\frac{\frac12}{1!}z+\frac{\frac12(\frac12-1)}{2!}z^2
+\frac{\frac12(\frac12-1)(\frac12-2)}{3!}z^3+\cdots\,,
$$
which is convergent for complex numbers $z$ with $\vert z\vert<1$. And the wonderful fact is that the very same series works in $\Bbb Q_p$, with the same condition on convergence when $p\ne2$, namely $\vert z\vert_p<1$ (for $p=2$, the condition is $\vert z\vert_2<\frac14$).
The unfortunate fact is that when you try to go outside this domain of convergence of the series, you’re essentially in the characteristic-$p$ situation of irremediable discreteness.
So I’d say that in the $p$-adic situation, there is something akin to the branch cut of the complex situation, but that you have to cut out not just something that’s of measure zero (like the negative real line), but something much bigger.
