Аннотация:
Let $R$ be a ring and $\sigma$ an automorphism of $R$. We prove that if $R$ is a 2-primal Noetherian ring, then the skew polynomial ring $R[x;\sigma]$ is 2-primal Noetherian. Let now $\delta$ be a $\sigma$-derivation of $R$. We say that $R$ is a $\delta$-ring if $a\delta\in P(R)$ implies $a\in P(R)$, where $P(R)$ denotes the prime radical of $R$. We prove that $R[x;\sigma,\delta]$ is a 2-primal Noetherian ring if $R$ is a Noetherian $\mathbb{Q}$-algebra, $\sigma$ and $\delta$ are such that $R$ is a $\delta$-ring, $\sigma(\delta(a))=\delta(\sigma(a))$, for all $a\in R$ and $\sigma(P)=P$, $P$ being any minimal prime ideal of $R$. We use this to prove that if $R$ is a Noetherian $\sigma(*)$-ring (i.e. $a\sigma(a)\in P(R)$ implies $a\in P(R)$, $\delta$ a $\sigma$-derivation of $R$ such that $R$ is a $\delta$-ring and $\sigma(\delta(a))=\delta(\sigma(a))$, for all $a\in R$, then $R[x;\sigma,\delta]$ is a 2-primal Noetherian ring.
Ключевые слова и фразы:2-primal, minimal prime, prime radical, nil radical, automorphism, derivation.