Ore extension

Remarks

For every endomorphism $\sigma$ and $\sigma$-derivation $\delta$, such a ring exists and is unique up to an isomorphism fixing $R$ and $x$. Its elements are written $a_0 + a_1 x + \cdots + a_n x^n$, and rule 3 together with associativity determines all products. Example: let $k$ be a field, $R = k[y]$, $\sigma = \id$ and $\delta = d/dy$. The map $\delta$ satisfies $\delta(ab) = a \delta(b) + \delta(a) b$, the Leibniz rule, so it is an $\id$-derivation, and rule 3 reads $x y = y x + 1$; this ring is the first Weyl algebra. Non-example: take $R = k[y]$, $\sigma = \id$ and $\delta = \id$. Then $\delta(1 \cdot 1) = 1$ but $1 \cdot \delta(1) + \delta(1) \cdot 1 = 2$, and $1 \ne 2$ in every field, so $\delta$ is not an $\id$-derivation. No ring as above exists, since rule 3 would give $x = x \cdot 1 = 1 \cdot x + \delta(1) = x + 1$. When $\delta = 0$ one writes $R[x; \sigma]$, the skew polynomial ring; when $\sigma = \id$ one writes $R[x; \delta]$, a differential polynomial ring. If $R$ is a domain and $\sigma$ is injective, then $R[x; \sigma, \delta]$ is a domain. If $R$ is left Noetherian and $\sigma$ is an automorphism, then $R[x; \sigma, \delta]$ is left Noetherian, a form of the Hilbert basis theorem.