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** polynomial $f(x)\in R[x]$ 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 the 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, the ideal $R[x]f(x)R[x]$, is equal to $R[x]$, and in this case $R[x]/R[x]f(x)R[x]$ is isomorphic to zero ring.

**I want to prove that for every ring with identity $R$ and every monic polynomial $f(x)$ over $R$ there exists an extension $S$ of $R$ such that $f(x)$ has a root in $S$.**