Proof that $\mathbb{C}[x,(x-\lambda)^{-1}]$ is a finitely-generated left $A_1(\mathbb{C})$-module I want to prove  that $\mathbb{C}[x,(x-\lambda)^{-1}]$ is a finitely-generated left $A_1(\mathbb{C})$-module where $\lambda \in \mathbb{C}$ and $\mathbb{C}[x,(x-\lambda)^{-1}]$ is a subring of $\mathbb{C}(x)$. So far I know that for a left $R$-module $M$ over a left noetherian ring; $M$ is noetherian $\iff$ $M$ is a finitely-generated $R$-module, and that $M$ is noetherian $\iff$ $N$ and $M/N$ are noetherian where $N$ is a submodule of $M$.
I was hoping to use this with $M=\mathbb{C}[x,(x-\lambda)^{-1}]$ and $N = \mathbb{C}[x]$ as I know $\mathbb{C}[x]$ is noetherian by HBT and $A_1(\mathbb{C})$ is also noetherian. The question also mentions that I can assume that $\mathbb{C}(x)$ is a left $A_1(\mathbb{C})$-module but I'm not sure how that helps. Any help will be appreciated :)
 A: You are making this way more complicated than necessary.  To prove $\mathbb{C}[x,(x-\lambda)^{-1}]$ is finitely generated, you can just find a finite set of generators.  To do this, you just need to understand the action of $A_1(\mathbb{C})$: $A_1(\mathbb{C})$ is generated as a $\mathbb{C}$-algebra by elements which act on $\mathbb{C}(x)$ by multiplication by $x$ and by differentiation.  So you just have to find finitely many elements of $\mathbb{C}[x,(x-\lambda)^{-1}]$ which generate everything with respect to the operations of multiplication by $x$, differentiation, and taking $\mathbb{C}$-linear combinations.
Since every element of $\mathbb{C}[x,(x-\lambda)^{-1}]$ is a $\mathbb{C}$-linear combination of a power of $x$ times a power of $(x-\lambda)^{-1}$, we just need a finite set that generates all the powers of $(x-\lambda)^{-1}$.  Now we observe that if we differentiate $(x-\lambda)^{-1}$ $n$ times, we get a nonzero scalar multiple of $(x-\lambda)^{-1-n}$.  So, $(x-\lambda)^{-1}$ alone already generates all the positive powers of $(x-\lambda)^{-1}$, and so $\{1,(x-\lambda)^{-1}\}$ generates the entire module. (In fact, it is not necessarily to include $1$, since it is $x\cdot (x-\lambda)^{-1}-\lambda\cdot(x-\lambda)^{-1}$.)
