I Trouble understanding an online solution to an exercise in Dummit & Foote

[elias001]

TL;DR Summary

Below is an online solution to an online solution to an exercise in exercise 10 chapter 9 section 1 of Dummit and Foote

$\textbf{Exercise 10}:$

Prove that the ring $\frac{\mathbb Z[x_{1}, x_{2}, ...]}{(x_{1}x_{2}, x_{3}x_{4},x_{5}x_{6}, ...)}$ contains infinitely many minimal prime ideals.

I came across the following solution online:

Let $R$ be that ring. For example, let $\mathfrak{p} = \langle x_1,x_3,x_5,\dotsc \rangle$. Then $R/\mathfrak{p} = \mathbb{Z}[x_2,x_4,x_6,\dotsc]$ is an integral domain. Hence, $\mathfrak{p}$ is a prime ideal. The prime ideals contained in $\mathfrak{p}$ correspond 1:1 to the prime ideals in the localization $R_{\mathfrak{p}}$. Now in that ring we have $x_n x_{n+1}=0$ for all odd $n$ and $x_{n+1}$ is invertible, so that $x_n=0$. In particular, the image of $\mathfrak{p}$ is just $0$. This means that $R_{\mathfrak{p}} = \mathbb{Z}[x_2,x_4,x_6,\dotsc]_{(0)} = \mathbb{Q}(x_2,x_4,x_6,\dotsc)$ is a field. Fields have exactly one prime ideal.

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 $x_nx_{n+1}=1$ and it should not have anything to do with with whether $x_n x_{n+1}=0$.

3. How does he conclude $R/\mathfrak{p} = \mathbb{Z}[x_2,x_4,x_6,\dotsc]$ given $\mathfrak{p} = \langle x_1,x_3,x_5,\dotsc \rangle$.

I am thinking of using the first, third or fourth isomorphism theorem with the following surjective homomorphic mapping $\varphi:\Bbb{Z}[x_{1}x_{2}, x_{3}x_{4},x_{5}x_{6},\ldots]\to \Bbb{Z}[x_{i_1},x_{i_2},\ldots,x_{i_n},\ldots]$ given by the function $\varphi(f(x_{1}x_{2}, x_{3}x_{4},x_{5}x_{6},\ldots))=\sum_{i=1}^{\infty}x_{i_j}$, where $i_j\in\{j,j+1\}$ with the ideal $(x_1,x_3,x_5,\ldots)$ as kernel. I am not sure if that is correct.

4. Also how does the isomorphism in author's answer about the isomorphism implies that the image of $\mathfrak{p}$ is $0$?

5. Finally can the exercise be solved only using fields of fractions ideas alone instead of bringing in the theory of localization of rings?

Thank you in advance.

 

[martinbn]

Prime ideals of $\mathbb Z[x_1,x_2,\dots]/(x_1x_2,x_3x_4,\dots)$ correspond to prime ideals of $\mathbb Z[x_1,x_2,\dots]$ containing the ideal $(x_1x_2,x_3x_4,\dots)$. Which means that a prime ideal will contain at least one of the $x_n$ and $x_{n+1}$ for all $n$. So the minimal ones are precisely the ones that contain exactly one of each pair.

 

[elias001]

@martinbn why does the author need to bring the idea of localization for solving the problem?

 

[fresh_42]

You have to analyze such proofs line by line. Here we have:

Prove that the ring $\mathbb{Z} [x_{1}, x_{2}, ...]/\bigl\langle x_{1}x_{2}, x_{3}x_{4},x_{5}x_{6},\ldots \bigr\rangle$ contains infinitely many minimal prime ideals.

Let $R$ be that ring.

$\blacktriangleright\;\;$ Means: $R=\mathbb{Z} [x_{1}, x_{2}, ...]/\bigl\langle x_{1}x_{2}, x_{3}x_{4},x_{5}x_{6},\ldots \bigr\rangle.$


For example, let $\mathfrak{p} = \langle x_1,x_3,x_5,\dotsc \rangle$. Then $R/\mathfrak{p} = \mathbb{Z}[x_2,x_4,x_6,\dotsc]$ is an integral domain. Hence, $\mathfrak{p}$ is a prime ideal.

$\blacktriangleright\;\;$Which lemma is used here?

The prime ideals contained in $\mathfrak{p}$ correspond 1:1 to the prime ideals in the localization $R_{\mathfrak{p}}$.

$\blacktriangleright\;\;$Do we have a proof for this statement? Why is this the case?

Now in that ring ...

$\blacktriangleright\;\;$Which ring?

... we have $x_n x_{n+1}=0$ for all odd $n$ ...

$\blacktriangleright\;\;$Why?

... and $x_{n+1}$ is invertible, ...

$\blacktriangleright\;\;$Why?

... so that $x_n=0$.

$\blacktriangleright\;\;$What did we use here?

In particular, the image of $\mathfrak{p}$...

$\blacktriangleright\;\;$Under which mapping?

... is just $0$. This means that $R_{\mathfrak{p}} = \mathbb{Z}[x_2,x_4,x_6,\dotsc]_{(0)} = \mathbb{Q}(x_2,x_4,x_6,\dotsc)$ is a field.

$\blacktriangleright\;\;$ What does the notation with the subscript $(0)$ mean? And why is the second equation correct?

Fields have exactly one prime ideal.

$\blacktriangleright\;\;$ A field $\mathbb{F}$ has two ideals, $\{0\}$ and $\mathbb{F}.$ Which one is prime?

$\blacktriangleright\;\;$ Why does that prove the statement? We wanted to find infinitely many prime ideals, but ended up with only one? How does that match?

_________________

This is how you should approach such a proof. Now you can tell us which steps you have difficulties with, and don't say "all of them".

 

[fresh_42]

I suggest forgetting about the localization and working out @martinbn 's answer.

 

[elias001]

@fresh_42 actually in my original post, where in the quoted solution, the author stated:

'The prime ideals contained in $\mathfrak{p}$ correspond 1:1 to the prime ideals in the localization $R_{\mathfrak{p}}$'

Does he mean in the sense of the 4th isomorphism theorem or does he mean a bijective maps? In either case, how can I translate it into math notation?

Also the last question in my post, (5), can the problem be solved purely using quotient fields/fields of fractions?

 

[martinbn]

.... the author stated:

'The prime ideals contained in $\mathfrak{p}$ correspond 1:1 to the prime ideals in the localization $R_{\mathfrak{p}}$'

... how can I translate it into math notation?

It is written in math notations!

 

[fresh_42]

$$
\{\mathfrak{q}\subseteq R\text{ prime }\,|\,\mathfrak{q}\subseteq \mathfrak{p}\}\stackrel{1:1}{\longleftrightarrow }\{\mathfrak{q}'=(R\setminus\mathfrak{p})^{-1}\mathfrak{q}\subseteq R_\mathfrak{p}\text{ prime }\}
$$
as you already have seen in a different thread.

But again, try to work out post #2. There are still questions to be answered.

 

[elias001]

@martinbn if there is a one-to-one correspondence, it implies there should be bijective maps $f$ and it's inverse $f^{-1}$. I am asking what would the two maps be?

 

[elias001]

@fresh_42 for my last point, seriously, can the question be solved rigoursly, spelled out in symbolic math notations with a minimal amount of math english, without referencing to localization? I am asking because the topic of localization doesn't make its ugly appearance after six more chapters, which is beyond Galois theory.

 

[fresh_42]

You have asked

@martinbn why does the author need to bring the idea of localization for solving the problem?

but this should be addressed to the one who wrote that. I don't see why localization helps here.

Try to understand what @martinbn has written in post #2. This makes more sense.

 

[elias001]

@fresh_42 I want to ask if the solutions that i posted from MSE which involved concepts from localisation, can the question be solved using fields of fractions. The folks on MSE, well you know their power trip ego are only matched by their gate keeping attitude. I am hoping a solution can be foubd using only what is covered in Hubgerford's undergraduate abstract algebra text in the field of fractions section in chapter 10.

 

[martinbn]

@fresh_42 I want to ask if the solutions that i posted from MSE which involved concepts from localisation, can the question be solved using fields of fractions. The folks on MSE, well you know their power trip ego are only matched by their gate keeping attitude. I am hoping a solution can be foubd using only what is covered in Hubgerford's undergraduate abstract algebra text in the field of fractions section in chapter 10.

I don't understand you! You keep complaning about irrelevent things and you keep asking for solutions and explanations in one way or another, but you never seem satisfied.

There is a solution in posit #2, what do you not like abouy it?

 

[elias001]

Screenshot 1

@fresh_42 In the MSE author's solution, where he said

'The prime ideals contained in $\mathfrak{p}$ correspond 1:1 to the prime ideals in the localization $R_{\mathfrak{p}}$'

Does he mean theorem 3.3.15 in the first screenshot?

Also when i said if the problem can solve without mentioning anything from localization. I mean if one sticks to the the content of section 10.4 in the next few screenshots from Hungerford's abstract algebra text, can the problem be solved only using what is in the seven pages below?

 

[martinbn]

You ask a lot of questions, but you don't answer any!

 

[elias001]

@martinbn you said I keep complaining because I never seem satisfied. I got the impression because I was not being precise enough with my questions or I did not provide enough details.

 

[martinbn]

@martinbn you said I keep complaining because I never seem satisfied. I got the impression because I was not being precise enough with my questions or I did not provide enough details.

And why do you ignore post #2?

 

[elias001]

@martinbn I am not ignoring your post. To fill in the details of your post requires me to solve some of the questions of the original MSE author's answers. Your solution allows one to weasel their way out of having to make direct of use of localization,but that is towards the very end. Your suggestion also allow one to bypass the step about the one to one correspondence and whether that meant the correspondence theorem or straight up bijection.

Just like you who is obsessive and loves to gate keep about who or what is consider to be a reliable source of math exposition. I like to understand things which are new to me where I am not clear about.

 

[fresh_42]

@martinbn I am not ignoring your post. To fill in the details of your post requires me to solve some of the questions of the original MSE author's answers. Your solution allows one to weasel their way out of having to make direct of use of localization,but that is towards the very end. Your suggestion also allow one to bypass the step about the one to one correspondence and whether that meant the correspondence theorem or straight up bijection.

Just like you who is obsessive and loves to gate keep about who or what is consider to be a reliable source of math exposition. I like to understand things which are new to me where I am not clear about.

Cross-posting makes little sense. If you don't understand what has been said on MSE, you should ask on MSE. What you have posted doesn't really match the question you have posted. Post #2 did. It still requires some details to be filled in, which is a good exercise to do. To learn, you must do. Or in this case, recognize why a suggested solution isn't one. The MSE post seems to answer a different question.

 

[martinbn]

Cross-posting makes little sense. If you don't understand what has been said on MSE, you should ask on MSE. What you have posted doesn't really match the question you have posted. Post #2 did. It still requires some details to be filled in, which is a good exercise to do. To learn, you must do. Or in this case, recognize why a suggested solution isn't one. The MSE post seems to answer a different question.

I've just realized that MSE stands for mathstackexchange. I found a question there about this problem and it has only one answer, which is the same as post #2 here. So, which question and solution on MSE are we talking about?

 

[martinbn]

Just like you who is obsessive and loves to gate keep about who or what is consider to be a reliable source of math exposition.

Can you please stop saying this about me or anyone else for that matter?

 

[fresh_42]

So, which question and solution on MSE are we talking about?

I have no idea. I ended up asking this while reading the OP ...

Why does that prove the statement? We wanted to find infinitely many prime ideals, but ended up with only one? How does that match?

... in post #4.

@elias001, please forget about what you quoted so far and try to figure out the details of the solution directed to in post #2. Let's settle this first before we get to your other threads.