Theorem 2

Let A be a left Noetherian ring, and let $\sigma$ be an automorphism of A and $d$ an endomorphism of the additive group of A satisfying relation (1). The ring $A[X]_{\sigma ,d}$ is left Noetherian.

Set $B = A[X]_{\sigma ,d}$. For any integer $n\geqslant 0$, we denote by $B_n$ the set of elements of B of the form $a_0+a_1X +\cdots +a_nX^n$. It is a left A-submodule of B. The mapping $\varphi_n: B_n\rightarrow A_s$ defined by $\varphi_n(a_0+a_1X +\cdots +a_nX^n) =a_n$ is A-linear.

Let $\mathfrak{b}$ be a left ideal of B. For every integer $n\geqslant 0$, the set $\mathfrak{a}_n=\varphi_n(\mathfrak{b}\cap B_n)$ is a left ideal of A. Since we have $Xa=\sigma (a)X +d(a)$ for every $a\in A$, we have

(6) $\varphi_{n+1}$(XQ) $=\sigma (\varphi_n(Q))$

for every $Q\in B_n$ and therefore $\sigma (\mathfrak{a}_n)\subset \mathfrak{a}_{n+1}$. Consequently, the sequence $\mathfrak{a}'_n=\sigma^{-n}(\mathfrak{a}_n)$ of ideals of A is increasing. Since the ring A is left Noetherian, there exists an integer $m\geqslant 0$ such that we have $\mathfrak{a}'_n=\mathfrak{a}'_{n+1}$ for $n\geqslant m$. Since $\sigma$ is surjective, we have the relation

for every integer $n\geqslant m$.

Let $\mathfrak{c}$ be the left ideal of B generated by $\mathfrak{b}\cap B_m$. Since the left A-module $B_m$ is finitely generated and the ring A is left Noetherian, the left A-module $\mathfrak{b}\cap B_m$ is finitely generated (VIII, p. 7, Proposition 4 a)). The left ideal $\mathfrak{c}$ is therefore generated by a finite subset of B. It is clear that it is contained in $\mathfrak{b}$. Let us prove that it is equal to $\mathfrak{b}$ by proving by induction that for every integer $n\geqslant 0$, we have

Relation (8) is true by construction for $n\leqslant m$. From now on, we assume that $n$ is an integer $\geqslant m$ such that $\mathfrak{b}\cap B_n\subset \mathfrak{c}$. Let P be an element of $\mathfrak{b}\cap B_{n+1}$. Then $\varphi_{n+1}(P)$ belongs to $\mathfrak{a}_{n+1}=\sigma (\mathfrak{a}_n)$, and there consequently exists an element Q of $\mathfrak{b}\cap B_n$ such that $\varphi_{n+1}(P) =\sigma (\varphi_n(Q))$. Set $R = P-$ XQ. Because of relation (6), we have $\varphi_{n+1}(R) = 0$, that is, $R\in B_n$. Since P and Q belong to the left ideal $\mathfrak{b}$ of B, the same is true for R; hence, R and Q belong to $\mathfrak{b}\cap B_n$, which is contained in the ideal $\mathfrak{c}$ by the induction hypothesis. Consequently, P belongs to $\mathfrak{c}$. This proves that we have $\mathfrak{b}\cap B_{n+1}\subset \mathfrak{c}$.

It follows that $\mathfrak{b}$ is equal to $\mathfrak{c}$; it is therefore a finitely generated ideal of B. This proves that the ring B is left Noetherian.

If the endomorphism $\sigma$ of the ring A is not an automorphism, then the ring $A[X]_{\sigma ,d}$ is not necessarily left Noetherian, even when A is a Noetherian commutative ring (VIII, p. 22, Exercise 26).