# Quotient of the polynomial ring over a local noetherian ring is finitely generated?

I am going through these lecture notes, and I am confused about the proof of Corollary 10.10 there.

Corollary 10.10. Let $$A$$ be a local noetherian domain with maximal ideal $$\mathfrak{p}$$, let $$g \in A[x]$$, and let $$B := A[x]/g(x)$$. Every maximal ideal $$\mathfrak{m}$$ of $$B$$ contains the prime $$\mathfrak{p}B$$.

In the proof of this, the author writes

The ring $$B$$ is finitely generated over the noetherian ring $$A$$, hence a noetherian $$A$$-module, ...

I don't understand this part, because I think if $$g$$ is not monic, then $$B$$ will not be finitely generated? For example, take $$A = \mathbb{Z}_{(3)}$$, and $$g(x) = 3 \in A[x]$$. Then $$B := A[x]/g(x) = \mathbb{Z}_{(3)}[x]/3\mathbb{Z}_{(3)}[x] = (\mathbb{Z}/3\mathbb{Z})[x]$$ which is certainly not finitely generated over $$A$$? I'm not sure what I am missing here. If there is another source for the proof of this corollary, please let me know. Thank you so much.

• You cannot use g=3=0 for the quotient. If g is not zero, because you are in a filed, you can assume that g is monic. Jun 10 at 8:39
• @Bongo Local ring Z_(3) does have unique maxinal ideal 3Z_(3), though. (Edit) I also do not get what this material is trying to convey.. Jun 10 at 8:42
• My hunch is that deg g > 0 is implied. Jun 10 at 9:02
• This happens when one learns mathematics from lecture notes instead of books. Jun 10 at 10:10
• @user26857 I tried to learn from Cassels and Frohlich but found it to be too dry for self-studying. Would you recommend me some books that cover similar material as these notes? Jun 10 at 17:08