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 $x*(\alpha y)=\alpha(x*y).$

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 $\Bbb R$). 
