Based in this question:
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
Based in this question:
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
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 $\mathbb{R}$.
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 $F$.
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.