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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

3 Answers 3

5

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 $\leq 2$.2012-11-06
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    @Hagen what's the name of this result, do you have a link?2012-11-06
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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 $\mathbb{R}$.2012-11-06
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    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.2012-11-07
11

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?2012-11-05
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    You define the operations so the bijection is an isomorphism.2012-11-05
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    @Neal then $\mathbb R^3$ will be isomorphic to $\mathbb R$?2012-11-05
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    Yes, or to whichever field $F$ you chose.2012-11-05
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    The nice trick, as I mentioned in my answer, is that you can keep the addition as it was before.2012-11-05
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    So any set can be a field with these operations?2012-11-06
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    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.2012-11-06
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    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.2012-11-07
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    $f(a(b+c))=f(a)f(b+c)=f(a)(f(b)+f(c))=f(a)f(b)+f(a)f(c)=f(ab)+f(ac)=f(ab+ac)$ --- doesn't that give distributivity?2012-11-08
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    but this map is not only a bijection, it's a homomorphism also, and how can you guarantee that such map exists?2012-11-08
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    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.2012-11-08
6

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.2012-11-07
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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.2012-11-07
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    But they will be isomorphic also as multiplicative groups?2012-11-07
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    @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.2012-11-07
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    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.2012-11-07
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    @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.2012-11-07
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    But how did you define the multiplications in $\mathbb R^3$?2012-11-07
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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 $\mathbb R$.2012-11-07
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    let us [continue this discussion in chat](http://chat.stackexchange.com/rooms/6368/discussion-between-user42912-and-asaf-karagila)2012-11-07
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    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)2012-11-07
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    @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.2012-11-07
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    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.2012-11-07
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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.2012-11-07
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    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.2012-11-07
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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).2012-11-07
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    Do you know how can I do this?2012-11-07
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    or even any paper online, I would be really grateful2012-11-07
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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.2012-11-07
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    ha yes, I think now I understood your point. Thank you for the patience.2012-11-08