When is $R\cong \Pi_{Q\in \max(R)}R_Q$?

Question

Problem: Let $A$ be a semilocal Noetherian ring and let $\text{max}(A)$ be the finite set of maximal ideals of $A$. I am trying to prove the following statement (which I hope to be true but am not certain about)

$A\cong \Pi_{Q\in \text{max}(A)}A_Q$. Edit: if it's not true then I am trying to figure out what circumstances this would hold under.

Ideas I have so far: Let $Q_1,...,Q_n$ be the maximal ideals of $A$. We have canonical maps $A\to A_Q$ (for $Q\in \text{max}(A)$) (note that these might not be injective because $A$ might have zero divisors in some of the $R\setminus Q$). Together these give a map $f:A\to \Pi_{i=1}^nA_{Q_i}$ (which is injective if at least one of the maps $A\to A_Q$ is injective). According to a book I am reading (Swanson Huneke integral closure of ideals, rings, and modules) this (or a version of this at least) has something to do with Chinese remainder theorem (which I think would be related to the surjectivity of a map with this domain and codomain if anything).

Answer 1

Let $A$ be a semi-local ring (not necessarily Noetherian) with finitely many maximal ideals $\mathfrak{n}_i$. You are asking about when the morphism $A \rightarrow \prod A_{\mathfrak{n}_i}$ is an isomorphism. The (very broad) class of semi-local rings for which this holds are known as semi-perfect, and there is no particularly interesting classification of them. A ring is semi-perfect if and only if it is isomorphic to a product of local rings.

If the map in question is an isomorphism then clearly $A$ is isomorphic to a finite product of local rings. Conversely, let $A_1, \ldots, A_n$ be a finite number of local rings having maximal ideals $\mathfrak{m}_i$.

Then $A = \prod A_i$ is a semi-local ring with maximal ideals $$\mathfrak{n}_1 = (\mathfrak{m}_1, A_2, \ldots, A_n)$$

$$\mathfrak{n}_2 = (A_1, \mathfrak{m}_2, \ldots, A_n)$$ $$\vdots$$ $$\mathfrak{n}_n = (A_1, A_2, \ldots, \mathfrak{m}_n)$$

and $A_{\mathfrak{n}_i}$ is simply projection onto the $i$th component of $A$, i.e. $A_{\mathfrak{n}_i} \cong A_i$. Thus the map $A \rightarrow \prod A_{\mathfrak{n}_i}$ is readily seen to be an isomorphism.

The other answer to the question derails when it concludes that $(A_P)_Q = 0$ for any pair of maximal prime ideals $P \not= Q$ implies that all maximal ideals are minimal. Indeed, if $A$ is a product of two local rings $A_1 \times A_2$, and $\mathfrak{m}_1, \mathfrak{m}_2$ are its maximal ideals, then ${A_{\mathfrak{m}_1}}_{\mathfrak{m}_2} = 0$ but this tells us nothing about the structure of the $A_i$.