Is the square root of a complex number over a field always well-defined?

Question

A complex number over a field $F$ is defined as $a+b\text{i}$ where $a,b\in F$ and $\text{i}$ is the square root of the inverse of the multiplicative identity of $F$, denoted as $\text{i}^2 = -1$. I have several questions about this definition.

1. Is there any restriction on $F$ in addition to that the square roots of $-1$ must exist? Should $F$ be a quadratically closed field, i.e. every number in $F$ has a square root?

2. How square root of a complex number is defined? The square root of a complex number over $\Bbb R$ can be defined by De Moivre's formula, but I don't see how this is extended to an arbitrary field.

For question 2, I can imagine a simple definition. For a complex number $c=\alpha+\beta\text{i}$, we can define the square root of $c$ as a complex number $a+b\text{i}$ st. $a^2-b^2=\alpha,2ab=\beta $. But I don't know if this definition is appropriate.

Any help and reference is appreciated. Thank you!

Answer 1

I believe the definition is best described as a quotient structure. Let $F$ be our field and $F[X]$ the polynomial ring. Then the "complex numbers" over the field is the structure $F[X]/\langle X^2+1\rangle$, at which the idea of squareroot becomes entirely superflous. To check that the quotient is a field we only need to check that $\langle X^2+1\rangle$ is maximal which isn't a too terribly difficult task.

As such we do not need the function of squareroot to be meaningful, we can have $F=\mathbb{Q}$ and as such there is no way to extend it to any field. If you place additional restrictions on the field in question you can start making the definition meaningful.

Answer 2

First, if $F$ already contains something whose square is $-1$ -- for example, if $F=\mathbb C$ or $F=\mathbb F_{37}$. -- then trying to make complex numbers over it goes wrong, and you end up with zero divisors, and not a field.

For example, with complex numbers over $\mathbb F_{37}$ we have $(6+i)(6-i)=0$.

Second, in general there is no canonical way to pick out a single square root of an element of a field. If we have $x$ and some $y$ such that $y^2=x$, then both $y$ and $-y$ will be possible square roots of $x$, and there is no general systematic way to pick exactly one of them to be "the" square root of $x$.

In order to find $y$ given $x$ one can use your computation at the end of the question to find a quadratic equation for $a$ and $b$, and then solve that in $F$ -- which generally requires you to be able to find square roots in $F$ already somehow.