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:


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

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.
