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

In summary, the exercise in Dummit and Foote Chapter 15, Section 15.1 asks readers to prove that the subring of the polynomial ring $k[x,y]$ generated by the terms $x^iy^{i-1}$ is not a Noetherian ring and therefore not a finitely generated $k$-algebra. This can be shown by considering the chain of ideals $(x) \subseteq (x, xy) \subseteq (x, xy, xy^2) \subseteq (x, xy, xy^2, xy^3) \subseteq \, \cdots$ and proving that the strict inclusions hold, or by showing that the ideal $(x, xy, xy^2
  • #1
Math Amateur
Gold Member
MHB
3,998
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: [TEX] k[x, x^2y, x^3y^2, ... ... ... \ , x^iy^{i-1} ... ... ] [/TEX] 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
  • #2
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.
 

FAQ: Noetherian Rings - Dummit and Foote - Chapter 15 - exercise 10

What is a Noetherian ring?

A Noetherian ring is a commutative ring with the property that every ideal can be generated by a finite number of elements. In other words, every ideal in a Noetherian ring is finitely generated.

What is the significance of Noetherian rings?

Noetherian rings have important applications in algebraic geometry and commutative algebra. They are also used in the study of number theory and algebraic topology.

How is the concept of Noetherian rings related to the work of mathematician Emmy Noether?

Emmy Noether was a pioneering mathematician who made significant contributions to abstract algebra and theoretical physics. She proved the Noetherian normalization lemma, which states that every Noetherian ring is a finitely generated algebra over a field.

What is the connection between Noetherian rings and Hilbert's basis theorem?

Hilbert's basis theorem states that any ideal in a polynomial ring over a field is finitely generated. This is a special case of the Noetherian property, which applies to any commutative ring.

Are all rings Noetherian rings?

No, not all rings are Noetherian. For example, the ring of polynomials with infinitely many variables is not Noetherian. However, most commonly studied rings, such as polynomial rings over a field and rings of integers, are Noetherian.

Back
Top