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?