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.

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.

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.

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.