Who first realized that the field of complex numbers is isomorphic to the set of real matrices following the form $\left(\begin{array}{cc} a & -b \\ b & a\end{array}\right)$? Hamilton? What is the first written register of such isomorphism?

Also, who first realized that the field of complex numbers is isomorphic to the quotient field $\textbf{R}[X]/(X^{2} + 1)$, where $\textbf{R}[X]$ is the ring of polynomials with real coefficients?

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    $\begingroup$ The quotient field representation was certainly known to Kronecker, but I don’t know if he’s the first. For the matrix representation, I’d look at the creators of matrix theory, Cayley and Sylvester. But that’s just a guess. $\endgroup$ May 20 '20 at 13:03

I really don't know who discovered this isomorphism, but I think I can feel intuition behind them.

For the first isomorphism, we see that $$\begin{pmatrix} 0 & -1\\ 1& 0\end{pmatrix}\begin{pmatrix} 0 & -1\\ 1& 0\end{pmatrix}=\begin{pmatrix} -1 & 0\\ 0& -1\end{pmatrix}=-I$$ i.e. the equation $x^2+1=0$ has at least one answer in this field, so, why we don't try to find the isomorphism.

For the $$\Bbb{R}[X]/(X^2+1)\cong \Bbb{C},$$ in fact after taking quotient, we will have $$X^2+1=0\quad \text{or}\quad X^2=-1$$ so it makes scene to look for an isomorphism $$\Bbb{R}[X]/(X^2+1)\to \Bbb{C}.$$

We may note that, in solving the polynomial equations, specially the quadratics $$ax^2+bx+c=0 \tag{*}$$ before discovering the complex number, it was said that the equation $(*)$ has no solution if $b^2-4ac<0$.

But then, as we know, the sum and product of roots of $(*)$, calculated respectively by $$-\frac{b}{a}\quad \text{and}\quad \frac{c}{a}$$ Now for example the equation $x^2+x+1=0$ which has no solution, has its sum of roots equal to $-1$ and prosuct of roots as $1$, well how is this possible? It was the reason to investigating on the "roots behind the plane" (as you know, ancient mathematicians solved equations by considering the intersection point(s) of curves in plane) and the simplest equation with no roots was $x^2+1=0$.


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