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!
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.