$\mathbb R^3$ can not be a field 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
 A: 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.
A: 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 $\mathbb{R}$.
A: 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 $F$. 
