# No extension to complex numbers?

Complex numbers are 2D. It is a commonly sited result that there is no 3D or 4D analogue of the complex numbers.

I just want to be clear on exactly what this result says:

• It is impossible to construct a 4D field.

• It is impossible to construct a 4D field which the complex numbers are a subset of.

• It is impossible to construct a 4D field which the real numbers are a subset of.

• Something else?

(I'd also like to know why it's true - but given that I didn't understand the answer the first $2^7$ times, I doubt I'll understand it this time either. :-} Oh well...)

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There are quaternions: en.wikipedia.org/wiki/Quaternion – Kaz May 7 '12 at 19:26
@Kaz Quaternions do not form a field. There are several similar generalisations which are also not fields. – MathematicalOrchid May 7 '12 at 19:29
But note that any set of objects which is a perfect analogue ("does" exactly what complex numbers do, satisfying all their properties) is indistinguishable from complex numbers. – Kaz May 7 '12 at 19:40
Depending on your background. If you adhere to Bourbaki you may speak of fields and commutative fields, I guess. – GEdgar May 9 '12 at 12:57

## 3 Answers

It is impossible to have a field which is $n$ dimensional over $\mathbb{R}$ for any $n\geq 3$. The reason why this is true boils down to the following two statements.

1. Any field $K\supseteq \mathbb{R}$ which is finite dimensional over $\mathbb{R}$ is algebraic over $\mathbb{R}$.
2. The complex numbers are the algebraic closure of $\mathbb{R}$.

Thus is $K\supseteq \mathbb{R}$ is a field which is finite dimensional over $\mathbb{R}$, then it is algebraic over $\mathbb{R}$, and hence is contained in the algebraic closure of $\mathbb{R}$, i.e., $K\subseteq \mathbb{C}$. Since $\mathbb{C}$ has dimension $2$ over $\mathbb{R}$, this implies that $K$ has dimension either $1$ or $2$ over $\mathbb{R}$. In the first case, $K = \mathbb{R}$, and in the second $K = \mathbb{C}$.

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Oh, wow. So it's not merely that $\mathbb{C}$ cannot be a subfield, but that even $\mathbb{R}$ cannot be a subfield. That's stronger than I expected... So is it possible to construct something that might be described as "four dimensional field" in any meaningful sense? (Clearly the reals cannot be a subfield of it.) – MathematicalOrchid May 7 '12 at 20:15
If you work over $\mathbb{Q}$ instead of $\mathbb{R}$ it is possible to have fields $K\supset\mathbb{Q}$ which are $n$-dimensional over $\mathbb{Q}$ for any desired dimension $n$. In the case $n = 4$, you can take $K$ to be the smallest field containing $\mathbb{Q}$ and $2^{1/4}$. This $K$ has as a $\mathbb{Q}$-basis $1, 2^{1/4}, 2^{1/2}, 2^{3/4}$. This sort of thing is part of the subject of field theory. – froggie May 7 '12 at 20:20

Since $\mathbb{C}$ is algebraically closed, every finite extension of $\mathbb{C}$ is $\mathbb{C}$ itself. The quaternions are not a field because they are not commutative; they are what is called a normed division algebra. Hurwitz's theorem gives a complete description of the possible normed division algebras (over $\mathbb{R}$) - the are either the real numbers, the complex numbers, the quaternions or the octonions. Only the first two are fields, and only the first three are associative.

The nonexistence of a three-dimensional real normed algebra can be viewed as a consequence of the hairy ball theorem. If $\mathbb{R}^3$ could be given a normed division algebra structure, the unit sphere $S^2$ could be endowed with a smooth group structure (the elements of norm $1$ form a group). But the hairy ball theorem implies that there is no such thing, because otherwise, by letting an infinitesimal element of the group act on the left on the sphere, we would obtain a nowhere vanishing, continuous vector field on the sphere, which is impossible by the hairy ball theorem.

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This may be worth posting as question on the main site, but I've never understood how one knows "If $\mathbb{R}^3$ could be given a normed division algebra structure, the unit sphere of $S^2$ could be endowned with a smooth group structure.." (emphasis my own). I understand the algebra half - that $S^2$ would be a group under the induced multiplication, but I don't understand why the group multiplication would be smooth. Does it somehow follow from the fact that the addition on $\mathbb{R}^3$ is the standard one? – Jason DeVito May 7 '12 at 20:10
+1 for the hairy ball theorem. Not only is this the most humorous thing I've heard all day, but it is intuitively obvious that you can't comb a hairy ball. From there, I can sort-of see why you can't extend the complex field. – MathematicalOrchid May 7 '12 at 20:11
Dear Jason, that is a very good and legitimate question. I have certainly swept a lot of things under the rug. I will give it some thought and hopefully come back with some sort of answer. @JasonDeVito – Bruno Joyal May 7 '12 at 20:15
Dear @MathematicalOrchid, the hairy ball theorem is definitely the most obscene mathematical theorem out there. :) – Bruno Joyal May 7 '12 at 20:16
Dear @JasonDeVito, I believe it follows more or less immediately from the definition of a normed algebra. The algebra structure is compatible with the norm, and from this it should follow easily that the ring structure is smooth. – Bruno Joyal May 7 '12 at 20:24

The field of rational functions is infinite-dimentional. It includes the complex and real numbers.

The same is correct about the field of analytic(holomorphic) functions.

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