Let's consider the group $(\mathbb{R}^2,+)$ with the usual addition and multiplication by real scalars. Then, $\{1:=(1,0),i:=(0,1)\}$ is a basis, and any $z \in \mathbb{R}^2$ can be uniquely written as $z=x1+yi$.
The most general multiplication $\times$, such that:
- $\times$ is distributive with $+$;
- $1:=(1,0)$ is the neutral element for $\times$;
- $i \times i=\alpha1+\beta i$
is given by:
$$(x_1,y_1 )×(x_2,y_2 ):=(x_1x_2+αy_1y_2,x_1 y_2+y_1x_2+βy_1y_2) \tag 1$$
Now, I note that the condition which ensures:
- the zero-product property to hold, and
- the equation $z^2+w=0$ to have solutions in $(\mathbb{R}^2,+,\times)$ whatever the constant term $w$
is exactly the same, namely:
$$\beta^2+4\alpha<0 \tag 2$$
[The choice $(\alpha,\beta)=(-1,0)$, compatible with $(2)$, gives rise to $\mathbb{C}$, the field of complex numbers; but I don't think this is much relevant for my question hereafter.]
As I believe this is not a mere coincidence, my question is:
What is the seemingly tight relationship between zero-product property and roots of polynomials?