MHB Noetherian Rings - Dummit and Foote - Chapter 15 - exercise 10

  • Thread starter Thread starter Math Amateur
  • Start date Start date
  • Tags Tags
    Exercise Rings
Math Amateur
Gold Member
MHB
Messages
3,920
Reaction score
48
I am reading Dummit and Foote Chapter 15, Section 15.1: Noetherian Rings and Affine Algebraic Sets.

Exercise 10 reads as follows:

--------------------------------------------------------------------------------------------------------------------

Prove that the subring: k[x, x^2y, x^3y^2, ... ... ... \ , x^iy^{i-1} ... ... ] of the polynomial ring k[x,y] is not a Noetherian ring and hence not a finitely generated k-algebra.

-----------------------------------------------------------------------------------------------------------------------

Can someone please help me get a start on this exercise.

Peter[Note: This has also been posted on MHF]
 
Last edited:
Physics news on Phys.org
It's more or less obvious that the chain of ideals should be

$$(x) \subseteq (x, xy) \subseteq (x, xy, xy^2) \subseteq (x, xy, xy^2, xy^3) \subseteq \, \cdots$$

But the strict inclusions need to be settled. Define the ideals $I_n = (x, xy, xy^2, \cdots, xy^n)$ of $k[x, xy, xy^2, \cdots ]$. Clearly, $I_0 \not = I_1$, as $xy \notin I_0$. Furthermore, $I_2 \not = I_1$ as $xy^2$ can't be written as a $k$-linear combination of $x$ and $xy$.

Can you convince yourself in this way that $I_n \not = I_{n-1}$?


Or you can just show (in the above approach) that the ideal $(x, xy, xy^2, \cdots)$ of $k[x, y]$ is not finitely generated, thus showing that $k[x, xy, xy^2, xy^3, \cdots ]$ is not finitely generated, which is equivalent to being non-Noetherian.
 
I asked online questions about Proposition 2.1.1: The answer I got is the following: I have some questions about the answer I got. When the person answering says: ##1.## Is the map ##\mathfrak{q}\mapsto \mathfrak{q} A _\mathfrak{p}## from ##A\setminus \mathfrak{p}\to A_\mathfrak{p}##? But I don't understand what the author meant for the rest of the sentence in mathematical notation: ##2.## In the next statement where the author says: How is ##A\to...
##\textbf{Exercise 10}:## I came across the following solution online: Questions: 1. When the author states in "that ring (not sure if he is referring to ##R## or ##R/\mathfrak{p}##, but I am guessing the later) ##x_n x_{n+1}=0## for all odd $n$ and ##x_{n+1}## is invertible, so that ##x_n=0##" 2. How does ##x_nx_{n+1}=0## implies that ##x_{n+1}## is invertible and ##x_n=0##. I mean if the quotient ring ##R/\mathfrak{p}## is an integral domain, and ##x_{n+1}## is invertible then...
The following are taken from the two sources, 1) from this online page and the book An Introduction to Module Theory by: Ibrahim Assem, Flavio U. Coelho. In the Abelian Categories chapter in the module theory text on page 157, right after presenting IV.2.21 Definition, the authors states "Image and coimage may or may not exist, but if they do, then they are unique up to isomorphism (because so are kernels and cokernels). Also in the reference url page above, the authors present two...
Back
Top