Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Based in this question:

$\mathbb R^3$ 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?

Thanks

share|improve this question

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.

share|improve this answer
    
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
1  
@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 $F$.

share|improve this answer
    
This bijection must be a homomorphism? –  user42912 Nov 5 '12 at 23:18
1  
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
1  
Yes, or to whichever field $F$ you chose. –  Gerry Myerson Nov 5 '12 at 23:37
1  
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 $\mathbb{R}$.

share|improve this answer
1  
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
1  
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

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.