TITLE: Where can we find polynomial's root?
QUESTION [5 upvotes]: 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.
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.

REPLY [3 votes]: 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 $\Psi$.