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Let $m$ be an integer such that $m \ge 2$. We define $R$ as the countable direct product of the ring $\mathbb{Z}/m\mathbb{Z}$ $$R:=\prod_{n=1}^\infty \mathbb{Z}/m\mathbb{Z}$$ I am trying to prove that the dimension of $R$ is $0$. Which means I have to prove that all the prime ideal of $R$ is also a maximum ideal of $R$ ($*$). It is obviously that if $m$ is prime then it only have $1$ element in $\operatorname{Spec} R$, then it's maximal. But I don't know how to show ($*$) when m is not a prime number.

The ideals of $\mathbb{Z}/m\mathbb{Z}$ have the form $(d)$, where $d$ divides $m$. And the maxium ideal of $\mathbb{Z}/m\mathbb{Z}$ is look like $(p)$ where $p$ is a prime dividing $m$.

Let$$R_i=\prod_{n=1}^i \mathbb{Z}/m\mathbb{Z}$$

The ideal of $R_2$, looks like $(d_1)\times (d_2)$, $(d_1)\times R_1$ or $R_1\times (d_2)$ , but the ideal of $R_2$ is a prime ideal iff $(p)\times R_1$ or $R_1 \times (p)$ and $p$ is prime (1) and is also a maximal ideal(2) (if (1)(2) are true I think I can prove it). But I am not sure if (1) and (2) are true. And what I could do when $i$ is $\infty$?

KReiser
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yukimaze
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  • I think you make a mistake on (1) the pirme ideal shoud be$

    \times \mathbb{z}/m\mathbb{z}$ or $\mathbb{z}/m\mathbb{z}\times

    $

     – kingzone Jul 29 '20 at 06:13
  • @kingzone yes I have make this mistake, I have edited it thank you. – yukimaze Jul 29 '20 at 06:18
  • @KentaS thank you for your comment, I was wondinring if this work when n is $\infty$? – yukimaze Jul 29 '20 at 06:57
  • It doesnt. The disjoint union is not quasi compact while it should be if it would be the spectrum of a ring. –  Jul 29 '20 at 07:18
  • https://math.stackexchange.com/questions/1529988/spectrum-of-infinite-product-of-rings

    This might be of interest?

     –  Jul 29 '20 at 07:19
  • @PaulK Thank you for your comment, I will go and reed this questions. – yukimaze Jul 29 '20 at 07:42

1 Answers1

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This was posted as an answer to essentially the same question about by Angina Seng a few weeks ago, but the question was deleted for other reasons. I thought the answer was instructive, so I'm reproducing it here, as a community-wiki post:

You need to show that all prime ideals are maximal. Let $R$ denote the ring in question, then the prime ideals of $R$ are the kernels of ring homomorphisms $\phi:R\to K$ where $K$ is a field. As $R$ has characteristic $n$, the characteristic of $K$ must be a prime $p$ dividing $n$. This means that the kernel of $\phi$ contains $\prod(p\Bbb Z/\Bbb Z)$ and so $R$ factors through $R'= \prod(\Bbb Z/p\Bbb Z)$. In effect we can suppose that $n=p$ with $p$ prime.

Each element of $R'$ satisfies $x^p=x$. The same is true for the image of $\phi$. So the image of $\phi$ must be contained in the prime subfield $K_0$ of $K$ with $p$ elements. Therefore $\phi(R)=K_0$ which is a field. Therefore $\ker \phi$ is a maximal ideal of $R$.

KReiser
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  • Thank your for your reproducing. May I ask why $\phi(R)$ must be contained in the prime subfield $K_0$? – yukimaze Jul 29 '20 at 07:39
  • The prime subfield of a field of characteristic $p$ is the unique subfield where all elements satisfy $x^p=x$. All elements in $\phi(R)$ satisfy this equation, so they must belong to that subfield. – KReiser Jul 29 '20 at 07:42