TITLE: Polynomial roots in the ring extension
QUESTION [7 upvotes]: Let $R$ be a ring with identity (not necessarily commutative) and $R[x]$ be a ring of polynomials over $R$. We say that a ring $S$ is an extension of $R$ if there is a subring $\tilde{R}$ in $S$ isomorphic to $R$.
Let $S$ be an extension of $R$, and $$\phi: R\to \tilde{R}\subset S$$
be a ring isomorphism.
We say that a polynomial $f(x) = \sum\limits_{j\geq 0}f_jx^j\in R[x]$ has a root $\alpha\in S$ if
$$
\sum\limits_{j\geq 0}\phi(f_j)\alpha^j = 0.
$$
In the case, where $R$ is a commutative ring every monic polynomaial $f(x)\in R[x]$ has a root $[x]_f$ in the extension $S = R[x]/R[x]f(x)$ of $R$. In the case, where $R$ is not commutative the set $R[x]/R[x]f(x)$ is a left $R[x]$-module but not a ring, because an ideal $R[x]f(x)$ is not two-sided ideal, but only one-sided. Also in non-commutative case there are examples such that two-sided ideal, containing $f(x)$ that is an ideal $R[x]f(x)R[x]$ is equal to $R[x]$ and in this case $R[x]/R[x]f(x)R[x]$ isomorphic to zero ring.
I want to prove that for every ring with identity $R$ and every monic polynomaial $f(x)$ over $R$ there exists an extension $S$ of $R$ such that $f(x)$ has a root in $S$.

REPLY [4 votes]: I have found a very simple and demonstrative construction of the ring extension, which came from non-commutative generalization of Hamilton-Caley's Theorem.
Let $R$ be a ring and $f(x) = x^m-\sum\limits_{j=0}^{m-1}f_jx^j\in R[x]$ be a monic polynomial. 
We identify ring $R$ with a subring $\tilde{R} = \{\mathrm{diag}(r,r,\ldots,r): r\in R\}\subset M_m(R)$. 
Then $f(x)$ has a root of the form
$$
\alpha=\left(\begin{array}{llllll}
0& e& 0&\ldots& 0& 0 \\
0& 0& e&\ldots& 0& 0 \\
. & . &  . & . & .& . \\
0& 0&0&\ldots& 0& e \\
f_0& f_1& f_2&\ldots& f_{m-2}& f_{m-1} \\
\end{array}
\right)
$$
That is $\alpha^m-\sum\limits_{j=0}^{m-1}f_j\alpha^j = 0\in M_m(R)$.
But we note that in general:
$f(\alpha^T)\neq 0$.
See article for proof  of non-commutative generalization of Hamilton-Caley's Theorem.