Let $R$ be a ring (not commutative in general) with identity and let 
$\Psi(x) = x^m-\sum\_{j=0}^{m-1}\psi\_jx^j$ be a monic polynomial over $R$. I want to construct a ring extension $K$ of $R$, which contains a root of $\Psi(x)$.
One construction follows from [non-commutative Hamilton-Caley Theorem](https://mathoverflow.net/questions/237751/polynomial-roots-in-the-ring-extension).


But I also think that there is another construction, which seems to be very interesting.


Let $M = R[x]/R[x]\Psi(x)$ be a left module over $R[x]$ and $A = \mathrm{Ann}\_{R[x]}M\subset R[x]$ be an annihilator of $M$. I want to prove that $\Psi(x)$ has a root in the ring
$$
R[x]/A.
$$


The proof is very simple in the case, where $R$ is a commutative ring. So the main interest is when $R$ is not commutative.



Where can we find polynomial's root?
No, $R[x]/A$ does not always have a root.


Let $R=\mathbb{Z}\langle a,b\rangle$ be the free ring on two noncommuting elements $a$ and $b$. Let $\Psi=x^2+a\in R[x]$, and let $M$ and $A$ be the objects you defined.


In this case, $A=0$. Suppose $f=r\_nx^n+\cdots+r\_0\in A$. In particular, there is some $g=s\_n x^n+\cdots+s\_0\in R[x]$ such that $f\cdot b=g\cdot (x^2+a)$. Matching up coefficients, we find that $r\_0b=s\_0a$, $r\_1b=s\_1a$, $r\_2b=s\_2a+s\_0$,$\dots$, $r\_ib=s\_ia+s\_{i-2}$. In the free ring $R$ the first two equations force $r\_0=s\_0=0$ and $r\_1=s\_1=0$, and by induction the rest of the equations force $f=g=0$. Hence $A=0$.


Finally, the ring $R[x]/A\simeq R[x]$ doesn't have any roots of $\Psi$.


