TITLE: $\mathbb R^3$ is not a field
QUESTION [16 upvotes]: I'm trying to prove that $\mathbb R^3$ is not a field with component-wise multiplication and sum defined. I think it's weird, because every properties of a field are inherit from $\mathbb R$.
Anyone can help?
Thanks

REPLY [16 votes]: Suppose that we want to put a field structure on the $\mathbb{R}$-vector space $\mathbb{R}^3$. Suppose $K$ is isomorphic to $\mathbb{R}^3$ as an $\mathbb{R}$-vector space and is a field. Then $K$ contains a copy of $\mathbb{R}$ as a subfield. Thus, $\mathbb{R}^3$ is an algebraic extension of $\mathbb{R}$ of degree $3$. But all algebraic extensions of $\mathbb{R}$ are either or degree $1$ or $2$ because all algebraic field extensions of $\mathbb{R}$ can be embedded into $\mathbb{C}$ and $\mathbb{C}$ has dimension $2$ as an $\mathbb{R}$ vector space. Thus, $\mathbb{R}^3$ can not be equipt with a field structure.
In this answer I have assumed that you are viewing $\mathbb{R}^3$ as an $\mathbb{R}$-vector space. If you only want to view it as an abelian group then this argument won't work (I don't think).