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 unitary 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 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 unitary polynomial $f(x)$ over $R$ there exists an extension $S$ of $R$ such that $f(x)$ has a root in $S$.

  • $\begingroup$ The issue with the polynomial ring $R[x]$ is that $x$ commutes with elements of $R$, and there is no reason a root of $f$ should commute with $R$. Would it work to start with the free non-commutative algebra in one generator over $R$ (I think we can construct this as the set of finite sums of elements $r_1 x^{n_1}\ldots r_k x^{n_k}$, modulo the necessary relations), and quotient by the two-sided ideal generated by $f(x)$? $\endgroup$ Apr 30 '16 at 22:32
  • $\begingroup$ @JulianRosen, I understand the backbone of your comment. But I don't understand how we can realize your idea to construct the ring extension and the polynomial root. $\endgroup$ May 1 '16 at 6:44

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 $\tilde{R} = \{\mathrm{diag}(r,r,\ldots,r): r\in R\}\in M_m(R)$.

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.


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