Ore extensions of noetherian $k$-algebras are again noetherian (proof explanation)

Question

Let $k$ be a field. All occuring $k$-algebras are required to be associative and unital. By noetherian I always mean left noetherian.

---

In a lecture I’m currently taking the notion of an Ore extension of a $k$-algebra was introduced (the definition used here can be found at the very end of this post). I’m having trouble understanding a crucial step in the proof of the following theorem:

Theorem: Suppose $R[t,\alpha,\delta]$ is an Ore extension of a $k$-algebra $R$. If $R$ is noetherian, then $R[t,\alpha,\delta]$ is also noetherian.

Proof: Let $I \trianglelefteq R[t,\alpha,\delta]$ be a left ideal. We show that $I$ is finitely generated. For $d \in \mathbb{Z}_{\geq 0}$ let $$ I_d := \left\{a \in R \,\middle|\, \exists P \in I, \deg P = d, P = \sum_{i=0}^d a_i t^i : a_d = a\right\} \cup \{0\}. $$ $I_d$ is a left ideal in $R$. If $a$ is the leading coefficient of $P \in I$ then $\alpha(a)$ is the leading coefficient of $t \cdot P$. Hence $\alpha(I_d) \subseteq I_{d+1}$. It follows that $$ I_0 \subseteq \alpha^{-1}(I_1) \subseteq \alpha^{-2}(I_2) \subseteq \dotsb \subseteq \alpha^{-n}(I_n) \subseteq \dotsb $$ is an increasing chain of left ideals in $R$. Because $R$ is noetherian this chain stabilizes for some $N$. Then $\color{red}{I_{N+i} = \alpha^i(I_N)}$ for all $i \geq 0$.
For $0 \leq d \leq N$ choose generators $a_{d,1}, \dotsc, a_{d,\ell(d)}$ of $I_d$ and polynomials $P_{d,j} \in I$ of degree $d$ such that $a_{d,j}$ is the leading coefficient of $P_{d,j}$. Then $\{P_{d,j} \mid 0 \leq d \leq N, 1 \leq j \leq \ell(d)\}$ generates $I$ (details left as an exercise).

I see that this proof is a generalization of the usual proof of Hilbert’s basis theorem, and by using that $I_{N+i} = \alpha^i(I_N)$ for all $i \geq 0$ I’m able to finish the proof. My problem is that I don’t see why $I_{N+i} = \alpha^i(I_N)$ for all $i \geq 0$. It seems to me that we only get that $$ \alpha^{-N}(I_N) = \alpha^{-(N+i)}(I_{N+i}) \quad \text{for all $i \geq 0$}. $$ As $\alpha$ is not necessarily surjective this is a weaker statement.

I would like to know:

1. Is it even true that $$ \text{there exists $N \in \mathbb{N}$ with $\alpha^i(I_N) = I_{N+i}$ for all $i \geq 0$?} $$
2. If it is not true, then does the above statement $$ \alpha^{-N}(I_N) = \alpha^{-(N+i)}(I_{N+i}) \quad\text{for all $i \geq 0$} $$ suffices to prove the theorem?

---

While it is not the focus of this question I also include a sketch of the rest of the proof, to show how I use that $\alpha^i(I_N) = I_{N+i}$ for all $i \geq 0$.

Rest of the proof (Sketch): Let $J := \langle P_{d,j} \mid 0 \leq d \leq N, 1 \leq j \leq \ell(d) \rangle \subseteq I$. It is easy to see that $P \in J$ for every $P \in I$ with $\deg P \leq N$.
Suppose that $P = \sum_{i=0}^d a_i t^i \in I$ with $P \notin J$, and that $P$ is of smallest degree with this property. Then $d > N$, so $\color{green}{I_d = \alpha^i(I_N)}$ for $i = d-N > 0$.
Because $a_d \in I_d$ there exists some $a \in I_N$ with $a_d = \alpha^i(a)$. Let $r_1, \dotsc, r_n \in R$ with $$ a = \sum_{j=1}^{\ell(N)} r_j a_{N,j} $$ and set $$ Q := \sum_{j=1}^{\ell(N)} r_j P_{N,j} $$ So $Q \in J$ is of degree $N$ and has $a$ as its leading coefficient. By induction it follows that $t^m Q \in J$ is of degree $N+m$ with leading coefficient $\alpha^m(a)$ for all $m \geq 0$.
In particular $t^i Q \in J$ is of degree $d$ and has leading coefficient $\alpha^i(a) = a_d$. As $P \notin J$ it follows that $P - t^i Q \notin J$. Because $P - t^i Q \in I$ with $\deg(P - t^i Q) < d = \deg(P)$ this contradicts the choice of $P$.

---

The used definition of an Ore extension:

Proposition/Definition: Suppose that $R$ is a $k$-algebra without zero divisors, $\alpha \colon R \to R$ is an injective $k$-algebra morphism and $\delta \colon R \to R$ is an $\alpha$-derivation, i.e. $\delta$ is $k$-linear with $$ \delta(ab) = \alpha(a)\delta(b) + \delta(a)b \quad \text{for all $a,b \in R$}. $$ Then there exists a unique $k$-algebra structure on $R[t] = \bigoplus_{i \geq 0} R t^i$ such that

1. the inclusion $R \hookrightarrow R[t]$, $r \mapsto r t^0$ is a morphism of $k$-algebras,
2. $\deg(PQ) = \deg(P) + \deg(Q)$ for all $P,Q \in R[t]$,
3. $t \cdot a = \alpha(a) \cdot t + \delta(a)$ for every $a \in R$.

This $k$-algebra is called an Ore extension of $R$ and denoted by $R[t,\alpha,\delta]$.

(I actually put a few statements and definitions from the lecture together, so in this form there might be some redundancy.)

Answer 1

Indeed, the conclusion $\alpha^{i}(I_N)=I_{N+i}$ does not follow. In fact, the claim of the theorem is incorrect in this generality. The theorem is true if $\alpha$ is an automorphism, and then the questionable implication also becomes true.

To see that it is not necessarily true if $\alpha$ is not surjective, consider the following example. Let $R=k[x]$ be a polynomial ring over a field, and let $\alpha$ be the injective endomorphism defined by $\alpha(x)=x^2$. Thus $\alpha(f(x)) = f(x^2)$ for all $f \in R$.

Claim: $S=R[t;\alpha]$ is not left Noetherian.

Proof: For $n\ge 0$, let $J_n$ be the left ideal of $S$ generated by $x$, $xt$, $\ldots\,$, $xt^n$. Then of course $J_n \subset J_{n+1}$. Moreover, $J_n \subsetneq J_{n+1}$ since $xt^{n+1} \notin J_n$, as $x \notin \alpha(R)$.

To see where exactly the proof in your question goes wrong, consider the left ideal $I= \bigcup_{n \ge 0} J_n$. Denoting, as in your question, the ideal of leading coefficients of $I$ of degree $n$ by $I_n \subset R$, we have simply $I_n=(x)$. However $\alpha^i(I_n) = \{p(x^{2^i}) \mid p \in k[x], p(0)=0 \} \subsetneq (x)$ for $i \ge 1$.

As a final remark, the ring $S$ is in fact also not right Noetherian.