Functions without complex roots, but with quaternion roots

Question

Many introductions to complex numbers begin with the question "What are the roots of $x^2 + 1 = 0$?"

This function does not have real roots, but does have complex roots.

Are there functions which, in a similar vein, do not have complex roots but do have roots in the quaternions?

Answer 1

This is a good question with an important answer.

The answer is no, because $\mathbb{C}$ has a property called algebraic closure. This means that any degree $n$ polynomial in $\mathbb{C}$ has $n$ factors (though some may be repeated) so the polynomial can always be fully factorised into linear terms. In particular, it means every polynomial has a root, and intuitively, there is nothing 'missing' from $\mathbb{C}$. This is a very important property about $\mathbb{C}$, which is what makes it so useful.

The quaternions don't really have the same relationship to $\mathbb{C}$ as $\mathbb{C}$ has to $\mathbb{R}$, as $\mathbb{C}$ is adding to $\mathbb{R}$ things that are 'missing' in a sense, whereas $\mathbb{C}$ doesn't actually need anything added to it, and the quaternions, $\mathbb{H}$, just add extra roots to polynomials which already have roots.

Answer 2

perhaps surprisingly, the answer here mostly depends on what you consider to be a function (and there are several choices that are more or less valid).

1. a mapping of input points to output points. in this case the question isn't especially interesting because you can construct a function that has or doesn't have roots in any region.

2. a continuous function defined using only complex coefficients. here there still are examples e.g. xy-yx=1

3. analytic functions. this is a broad class of functions which includes all polynomials, logs, and a bunch of other stuff. I'm not sure the answer in this case.

Answer 3

It may be illuminating to point out that a polynomial is not a function. A polynomial is an expression that is built up from a set of fixed variables and constants, addition, subtraction and multiplication. You can then evaluate such a polynomial at some given number. This evaluation process defines a function.

What is the purpose of such a distinction, and why is it related to this question, which doesn't refer to polynomials at all? Well, here it is. Since a function is simply a relation from the input to the output, there are obvious functions that have roots in the quaternions but don't in the complex numbers. For example: $$f(x) = \begin{cases} 0 & (x = 2j + 3k)\\ 1 & (\text{otherwise}) \end{cases}$$ Clearly $f(x) = 0$ has no solutions in $\mathbb C$, but has a unique solution $2j+3k \in \mathbb H$. Without doubt, this is not the intention of the original question because anyone can see this. Then what is the "correct" interpretation of this question, i.e. one that actually answers the intuitive question behind this?

The answer lies exactly in the distinction between "polynomials" and "polynomial functions". The first one is defined by an expression, so even if two expressions give the same function, they are considered different polynomials. The classic example is $x^{p} = x$ which is true in $\mathbb Z/p\mathbb Z$. They are different polynomials but the same polynomial functions. Indeed, in an infinite field, different polynomials give different polynomial functions. (Exercise: Prove it!) But there are other things out there, there are finite fields, there are non-commutative rings, in which case the distinction becomes important.

The same phenomenon appears for other types of so-called "functions". They are actually various classes of expressions that happens to have a natural evaluation operation which turns them into functions. For example, the expression $\exp(x)$ can be evaluated over the complex number using Taylor series. What the OP is really asking, is to give examples of classes of expressions, such that these expressions make sense both over complex numbers and over quaternions, and that they only have roots over quaternions. Functions play very little role in here.

With this in mind, we can proceed to answering. Indeed, the expression $(xy - yx)^2 + 4$, as pointed out in the comments, have no roots over the complex number, but has a root $(j, k)$. Since we often only consider polynomials over commutative rings, the definition of polynomials usually "builds-in" the commutativity, and dictates that $(xy - yx)$ is the same polynomial as $0$. This defines the commutative ring of polynomials. But since quaternions are not commutative, such a definition of polynomials are of very little use. We are better off using the free algebra such as $\mathbb H\langle X,Y \rangle$, in which $XY \ne YX$. The concept of algebraic-closedness must be altered for the non-commutative case. So I object to @ZoeAllen, for the complex numbers do not form algebraically closed skew fields. With that said, there are actually different definitions of algebraically closed skew fields, and my previous sentence may be true or false depending on the definition. This is a very interesting subject, and you can take a read of this.