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$\mathbb R^3$ is not a field

I'm wondering how to prove that $\mathbb R^3$ is not a field no matter which operations you choose. I'm trying to prove using Field theory, anyone knows how to prove it?


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3 Answers 3

up vote 4 down vote accepted

The cardinality of $\mathbb R^3$ is the same as the cardinality of $\mathbb R$ or $\mathbb C$.

In fact as additive groups they are the same as well. This means that one can define multiplication on $\mathbb R^n$ which makes it isomorphic to $\mathbb R$ or even $\mathbb C$.

But you shouldn't stop there. You could find a bijection of $\mathbb R$ with $\mathbb Q_p$, the $p$-adic field; or with fields of positive characteristics, then you can use this bijection to define a new structure on the set $\mathbb R^3$ which will be isomorphic to the selected field.

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But how they can isomorphic? they have different dimensions. –  user42912 Nov 7 '12 at 13:35
@user42912: Not isomorphic as vector spaces, just as additive groups. Forget about scalar multiplication; forget about topological properties; forget about metrics; forget about anything except the addition. Then $\mathbb R$ and $\mathbb R^3$ are isomorphic. –  Asaf Karagila Nov 7 '12 at 13:39
But they will be isomorphic also as multiplicative groups? –  user42912 Nov 7 '12 at 14:19
@No, because $\mathbb R^3$ is not a multiplicative group, neither is $\mathbb R$ if we don't remove $0$; but even if we remove $(0,0,0)$, $(0,0,1)\cdot(0,1,0)=(0,0,0)$ so it's not a group. –  Asaf Karagila Nov 7 '12 at 14:31
@user42912: The multiplication is defined as you suggested, $x\odot y=f^{-1}(f(x)\cdot f(y))$, where $\cdot$ is the usual multiplication in $\mathbb R$. –  Asaf Karagila Nov 7 '12 at 16:41

What you are trying to prove is impossible. Take any bijection from ${\bf R}^3$ to a field $F$ (say, $F={\bf R}$), and define operations on ${\bf R}^3$ by pulling them back from $F$.

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This bijection must be a homomorphism? –  user42912 Nov 5 '12 at 23:18
You define the operations so the bijection is an isomorphism. –  Neal Nov 5 '12 at 23:23
@Neal then $\mathbb R^3$ will be isomorphic to $\mathbb R$? –  user42912 Nov 5 '12 at 23:31
Yes, or to whichever field $F$ you chose. –  Gerry Myerson Nov 5 '12 at 23:37
The nice trick, as I mentioned in my answer, is that you can keep the addition as it was before. –  Asaf Karagila Nov 5 '12 at 23:44

As you have written it, the statement is false, as you can transfer the field structure of, say, $\mathbb{R}$ to $\mathbb{R}^3$ via a bijection. However, a famous result says that the only (finite-dimensional, associative) division algebras over $\mathbb{R}$ are the real numbers, the complex numbers, and the quaternions. (You may wish to look at http://mathworld.wolfram.com/DivisionAlgebra.html for references.) In particular, it is not possible to make $\mathbb{R}^3$ a field in such a way that $\mathbb{R}^3$ is of degree 3 over $\mathbb{R}$.

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To get your last statement you don't need to refer to the deep results you mention before: the algebraic closure of $\mathbb{R}$ has degree $2$ over $\mathbb{R}$, hence all algebraic extensions of $\mathbb{R}$ have degree $\leq 2$. –  Hagen Nov 6 '12 at 8:53
@Hagen what's the name of this result, do you have a link? –  user42912 Nov 6 '12 at 14:12
In the result about division algebras, that Eric mentions, it is assumed that the reals are a subring of the division algebra. But if you assume that $\mathbb{R}^3$ carries a field structure, such that $\mathbb{R}$ is a subring of it, then $\mathbb{R}^3$ is an algebraic extension of $\mathbb{R}$. We know that the complex numbers $\mathbb{C}$ are algebraically closed by the "fundamental theorem of algebra", hence every algebraic extension field of $\mathbb{R}$ is contained in $\mathbb{C}$. And $\mathbb{C}$ has dimension $2$ over $\mathbb{R}$. –  Hagen Nov 6 '12 at 15:16
If we define the operations in $\mathbb R^3$: $x+y=f^{-1}(f(x)+f(y))$ and $x\odot y=f^{-1}(f(x)\cdot f(y))$ I think is the operations you mentioned. The distributivity law doesn't work. –  user42912 Nov 7 '12 at 22:10

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