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Let $R$ be a ring. We know the well-known Chinese Reminder Theorem and we also know that it is not true for arbitrary index set.

My question:

If the Krull dimensin of $R$ is $0$ (i.e. every prime ideal is maximal), let $\DeclareMathOperator{\Spec}{Spec} m \in \Spec R$ and $\alpha =\bigcap_{p\in \Spec(R)-m}p$, is $m+\alpha=R$?

In some sense, if the Krull dimension of $R$ is $0$ and the radical of $R$ is $0$, does it follow that the intersection of arbitrary proper subset of $\Spec R$ is $0$?

Thanks!

Viktor Vaughn
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Jian
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1 Answers1

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Towards your first question: This is false in general, but true in the noetherian case.

In the non-noetherian case, we might have $\alpha = 0$ and then it fails obviously. For an example take $R$ to be the product of infinitely many copies of a field, for instance $\mathbb R$. If you intersect all maximal ideals but one, you get the zero ideal.

In the noetherian case, we have that there are only finitely many maximal ideals and then you can use the following:

If $I+I_j = R$ for any $j =1, \dotsc, n$, then also $I + \bigcap_{j=1}^n I_j=R$.

The proof is very easy:

$$R = (I+I_1) \dotsb (I+I_n) \subset I + I_1 \dotsb I_n \subset I + \bigcap_{j=1}^n I_j.$$ Note that the first inclusion holds because if you expand the LHS, then any summand admits a factor $I$, with the exception of the summand $I_1 \dotsb I_n$.

Towards your second question: Again, this is only true in the noetherian case and it immediately follows from the answer of the first question.

MooS
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  • You should read this https://math.stackexchange.com/questions/87981/infinite-product-of-fields to understand the spectrum of my counter example. Of course you cannot take $k=\mathbb F_2$, but you should take the reals for instance. – MooS Sep 01 '17 at 06:22
  • Just take $m$ to be a "non-trivial" maximal ideal and then the intersection of all the others will be automatically zero, because the intersection of all "trivial" maximal ideals is zero. – MooS Sep 01 '17 at 06:24
  • sorry，I still don't understand.when I ask questions next time,I will show my efforts.Thanks for your help,when I understand this ,I will talk with you. – Jian Sep 01 '17 at 06:36
  • I don't realize this biggest question before.when I consider my question,I have considered infinite product of fields,but I never realize the non-trivial maximal ideal it has.after I read the article you advise maybe we consider $\mathbb Q$ is better?since $\mathbb R$ is complete.Thank you very much! I will delete my ashamed comments.Please forgive me. – Jian Sep 01 '17 at 07:14
  • $\mathbb R$ is just as fine: You can also look at this thread (check the answer with one upvote) for a simple argument that there are "non-trivial" maximal ideals. https://math.stackexchange.com/questions/687465/maximal-ideals-of-mathbbr-infty – MooS Sep 01 '17 at 07:28
  • Thanks a lot.It's a great harvest today – Jian Sep 01 '17 at 07:42