Field structure on $\mathbb{R}^2$ I have the following question:
Is there a simple way to see that if we put a multiplication $*$ on $\mathbb{R}^2$ (considered as a vector space over $\mathbb{R}$) such that with usual addition and this multiplication $\mathbb{R}^2$ becomes a field, then there exists a nonzero $(x,y)$ such that $(x,y)*(x,y)=-1$?
Remark:


*

*What I mean by "a simple way to see" is that I really don't want to refer to Frobenius's Theorem on real finite dimensional division algebras.

*I haven't said this in the problem but I'm also assuming that with this multiplication $\mathbb{R}^2$ becomes an algebra meaning $x*(\alpha y)=\alpha(x*y).$
 A: Sorry if I make it too elementary: If $1\in\mathbb R^2$ denotes $1$   of your field, and if $x\in\mathbb R^2$ is not its real multiple: $1,x,x^2$ are linearly dependent (over $\mathbb R$), i.e. $ax^2+bx+c=0$ for some $a,b,c$, and $a\neq 0$ (as $x$ is not a multiple of $1$), so we can suppose it's 1. If we complete squares, we get $(x+p)^2+q=0$ for some $p,q\in \mathbb R$. Now $q$ must be positive - otherwise $(x+p+\sqrt{-q})(x+p-\sqrt{-q})=0$, so you don't have a field (we found divisors of $0$). So finally $(x+p)/\sqrt{q}$ is the element you want.
A: The usual addition forces $\Bbb R$ to be a two dimensional subfield of your field $F$ (Consider $\Bbb R1_F)$. If you assume the fundamental theorem of algebra, and have some background in field theory, then it is relatively straightforward. Assume no root of $x^2+1=0$, then $$F(i)=F[x]/(x^2+1)\ \text{is degree $2$ over $F$}$$
so $F(i)$ is a degree four finite (therefore algebraic) extension of $\Bbb R$ as $$[F(i):F][F:\Bbb R]=[F(i):\Bbb R] $$ But this cannot happen as any finite extension of $\Bbb R$ is contained inside a field isomorphic to $\Bbb C$. (since $\Bbb C$ is the algebraic closure of $\Bbb R$ and is of degree two over $\Bbb R$). 
A: You are looking at a field extension of $\Bbb R$ of degree two, but by the fundamental theorem of algebra, that field sitting above $\Bbb R$ is unique up to isomorphism and is isomorphic to $\Bbb C$.
