Suppose that $(\mathbb{R}^2,+,\cdot)$ is a field such that $(\alpha z)\cdot (\beta\omega)=(\alpha\beta)(z\cdot\omega)$, where $\alpha,\beta\in\mathbb{R}$ and $z,\omega\in\mathbb{R}^2$. Prove that there exists $u\in\mathbb{R}^2$ such that $u^2=-e$, where $e$ is the identity for $\cdot$.
My thought: given the identity element $e$, we can find $u\in\mathbb{R}^2$ such that $\{e,u\}$ is an $\mathbb{R}$-basis for $\mathbb{R}^2$. There exist two unique reals $\lambda$ and $\mu$ such that $u^2=\lambda e+\mu u$. If $\lambda=0$, then $u^2=\mu u$, that is, $u=\mu e$, which is not possible, so $\lambda\neq 0$. This shows that $e$ is a linear combination of $u$ and $u^2$, therefore $\{u^2,u\}$ is an $\mathbb{R}$-basis for $\mathbb{R}^2$.
How could I proceed now?
The set $\{e,u,u^2\}$ is linearly dependent, so there exist $a,b\in\mathbb{R}$ such that $u^2+au+be=0$. If the polynomial $x^2+ax+b$ (with real coefficients) has real roots $r$ and $s$, we have $x^2+ax+b=(x-r)(x-s)$, so $$ u^2+au+be=(u-re)(u-se)=0 $$ and so either $u-re=0$ or $u-se=0$, which is impossible. Therefore $a^2-4b<0$. Complete the square from $4u^2+4au+4be=0$: \begin{gather} 4u^2+4au+a^2e-a^2e+4be=0\\[6px] (2u+ae)^2=(a^2-4b)e \end{gather} Can you finish?