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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

Community
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user42912
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3 Answers3

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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 $F$.

Gerry Myerson
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  • This bijection must be a homomorphism? – user42912 Nov 05 '12 at 23:18
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    You define the operations so the bijection is an isomorphism. – Neal Nov 05 '12 at 23:23
  • @Neal then $\mathbb R^3$ will be isomorphic to $\mathbb R$? – user42912 Nov 05 '12 at 23:31
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    Yes, or to whichever field $F$ you chose. – Gerry Myerson Nov 05 '12 at 23:37
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    The nice trick, as I mentioned in my answer, is that you can keep the addition as it was before. – Asaf Karagila Nov 05 '12 at 23:44
  • So any set can be a field with these operations? – user42912 Nov 06 '12 at 00:30
  • Not sure I understand the question. No set of 6 elements can be a field. Any set $S$ that can be put into one-one correspondence with a field $F$ can be made into a field, by defining the operations on $S$ in such a way as to make the correspondence an isomorphism. – Gerry Myerson Nov 06 '12 at 00:39
  • 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 07 '12 at 22:07
  • $f(a(b+c))=f(a)f(b+c)=f(a)(f(b)+f(c))=f(a)f(b)+f(a)f(c)=f(ab)+f(ac)=f(ab+ac)$ --- doesn't that give distributivity? – Gerry Myerson Nov 08 '12 at 00:05
  • but this map is not only a bijection, it's a homomorphism also, and how can you guarantee that such map exists? – user42912 Nov 08 '12 at 00:36
  • We start with the bijection $f$. We use it (as you did, three comments up) to define operations on ${\bf R}^3$. With those operations, ${\bf R}^3$ is a field, isomorphic to $\bf R$. All we have really done is we have noticed that as sets the two are in one-one correspondence, and we've made the one into a field by grafting onto it the field structure of the other. I'm sorry, I don't know how to say it any better, and I don't see where it is that you are getting stuck. – Gerry Myerson Nov 08 '12 at 03:04
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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.

Asaf Karagila
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  • But how they can isomorphic? they have different dimensions. – user42912 Nov 07 '12 at 13:35
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    @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 07 '12 at 13:39
  • But they will be isomorphic also as multiplicative groups? – user42912 Nov 07 '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 07 '12 at 14:31
  • I'm defining sum and multiplication in $\mathbb R^3$ in the following way: let $f$ be the bijection from $\mathbb R^3$ to $\mathbb R$, so my operations are: $x+y=f^{-1}(f(x)+f(y))$ and multiplication $x\cdot y=f^{-1}(f(x)\cdot f(y))$, where $x,y \in R^3$. Did I defined the operations correctly? The problem with these operations is the $f$ would be linear, then R would be isomorphic to $\mathbb R^3$ as vectorial spaces. – user42912 Nov 07 '12 at 16:34
  • @user42912: Yes. That would work. My answer adds the point that you can require that $f(x+y)=f(x)+f(y)$. This is not the same as requiring that $f$ is linear. Linearity also requires to respect scalar multiplication. You can show that indeed for rational scalars this is true, $f$ is $\mathbb Q$-linear, but it is not $\mathbb R$-linear. – Asaf Karagila Nov 07 '12 at 16:38
  • But how did you define the multiplications in $\mathbb R^3$? – user42912 Nov 07 '12 at 16:39
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    @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 07 '12 at 16:41
  • let us continue this discussion in chat – user42912 Nov 07 '12 at 17:47
  • Let R^3 and R R-vector spaces with the multiplication of vectors by scalars. Then f will be linear, for example we can easily see that f(2x)=2f(x) – user42912 Nov 07 '12 at 17:56
  • @user42912: What are you trying to say?? ALL I said in my answer is that you require the bijection $f$ to keep the same addition operation. Why does that imply that $f$ is linear?? Yes, it cannot respect multiplication and it cannot be linear (at least not for real scalars). I really have no idea what point you are trying to make in the past three comments. – Asaf Karagila Nov 07 '12 at 18:09
  • sorry, I think I don't make myself clear, what I'm trying to say that these operations doesn't make $\mathbb R^3$ a field. For example the distributivity law doesn't work. – user42912 Nov 07 '12 at 22:04
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    Yes. Multiplication is not defined on $\Bbb R^3$ in a way that as a vector space this is a continuous operation. However it is possible to define multiplication in a different way, which does not change addition and makes a filed with the common addition. – Asaf Karagila Nov 07 '12 at 22:45
  • Do you know how we can define multiplication to give a field structure in $R^3$? Because as I said in my question, $R^3$ can NOT be a field. – user42912 Nov 07 '12 at 22:58
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    @user42912: No, there is no explicit way to define the multiplication. This requires the axiom of choice (at least if we want to preserve the addition). – Asaf Karagila Nov 07 '12 at 23:06
  • Do you know how can I do this? – user42912 Nov 07 '12 at 23:08
  • or even any paper online, I would be really grateful – user42912 Nov 07 '12 at 23:09
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    How can you do what? Define multiplication? Not in an explicit way. It is a trivial consequence that $\mathbb R^3$ and $\mathbb R$ are isomorphic as abelian groups, since they are isomorphic as $\mathbb Q$ vector spaces. Fix an isomorphism and use that bijection to define the multiplication. There is no explicit way to do that. – Asaf Karagila Nov 07 '12 at 23:10
  • ha yes, I think now I understood your point. Thank you for the patience. – user42912 Nov 08 '12 at 00:38
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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 $\mathbb{R}$.

Eric M. Schmidt
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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 Knaf Nov 06 '12 at 08:53
  • @Hagen what's the name of this result, do you have a link? – user42912 Nov 06 '12 at 14:12
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    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 Knaf Nov 06 '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 07 '12 at 22:10