Localization - D&F, Section 15.4, Exercise 12

[Math Amateur]

I am reading Dummit and Foote, Section 15.4 Localization.

Exercise 12 on page 727 reads as follows:

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Let \(\displaystyle R = \mathbb{R}[x,y,z]/(xy - z^2) \), let P be the prime ideal \(\displaystyle P = (\overline{x}, \overline{y})\) generated by the images of x and y in R and let \(\displaystyle R_P \) be the localization of R at P.

Prove that \(\displaystyle P^2 R_P \cap R = (\overline{x}) \) and is strictly larger \(\displaystyle P^2 \).

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I am somewhat overwhelmed by this problem.

I know that we have \(\displaystyle R_P = D^{-1}R \) where D = R - P and that we have a homomorphism \(\displaystyle \pi : \ R \to R_P = D^{-1}R \) where \(\displaystyle \pi (r) = r/1 \).

However I am having trouble getting a clear understanding of the nature of the elements of the problem let alone making a significant start on the problem.

For example, regarding the nature of \(\displaystyle P = (\overline{x}, \overline{y})\) - D&F describe this as the the prime ideal \(\displaystyle P = (\overline{x}, \overline{y})\) generated by the images of x and y in R - but images under what? x and y are both elements of \(\displaystyle \mathbb{R}[x,y,z] \), so again - what mapping and where do the images lie?

I would appreciate some help with the nature of the elements of this exercise - including the elements R, P and \(\displaystyle P^2R_P \cap R \) and some help with making a significant start on the exercise.

Peter

 

[jvicens]

,

First of all, don't be overwhelmed by this problem! It may seem daunting at first, but with some careful thinking and understanding of the definitions and concepts involved, you can definitely make significant progress on it.

Let's start by breaking down the elements involved in this exercise. We have the ring R = \mathbb{R}[x,y,z]/(xy - z^2) , which is the quotient ring obtained by factoring out the ideal (xy - z^2) from the polynomial ring \mathbb{R}[x,y,z] . This means that in R, the elements x and y are identified with each other, as are the elements z and xy. So when we talk about the images of x and y in R, we are really referring to the elements \overline{x} and \overline{y} in R. Similarly, the prime ideal P = (\overline{x}, \overline{y}) is generated by these images, so it consists of all elements in R that can be written as a combination of \overline{x} and \overline{y} with coefficients in R.

Now, let's consider the localization R_P. As you mentioned, this is the ring obtained by inverting all elements in R except those in the prime ideal P. In other words, we are "localizing" the ring R at the prime ideal P, which means that we are only considering elements in R that are "allowed" by the elements in P. In this case, since P consists of \overline{x} and \overline{y}, we are essentially inverting these elements and all their powers. So in R_P, we have elements of the form \frac{r}{s} where r and s are elements in R and s is not in P. Note that this includes elements like \frac{\overline{z}}{\overline{x}} and \frac{\overline{x}^2}{\overline{y}^3}.

Now, let's move on to the main part of the exercise - proving that P^2 R_P \cap R = (\overline{x}) and is strictly larger than P^2. First, let's break down what this notation means. P^2 R_P is the ideal generated by the elements in P^2 (i.e. elements of the form \overline{x}^2, \overline{x