For a commutative ring $R$ with identity, it is well known that if a finite intersection of ideals is contained in a prime ideal $\frak{p}$, then one of them is contained in $\frak{p}$. I am looking for an equivalent condition on $R$ under which if an arbitrary intersection of ideal is contained in a minimal prime ideal $\frak{p}$, then one of them is contained in $\frak{p}$. Or is there any research paper related to that?
[Math] When an intersection is contained in a minimal prime ideal
ac.commutative-algebraag.algebraic-geometryprime-idealsra.rings-and-algebras
Related Solutions
Suppose I have a set of disjoint, nonempty sets, and I want to choose one element from each. Consider the free polynomial ring generated by all the elements of all the sets, then take the quotient by the ideal generated by $xy$ for each pair $x$ and $y$ different elements in the same set. Any prime ideal must contain all but one element from each set.
We need to show that a minimal prime ideal does not contain all the elements from any set. Then a minimal prime ideal will give us a choice function. We can take the minimal prime to be generated by the elements, since every prime ideal contains a prime ideal generated by elements. Then remove one element from a set entirely contained in the prime ideal. The ideal will still be prime, and smaller, this is a contradiction.
So minimal primes give a choice function.
In the example in my comment, the ideal $I$ is not finitely generated. Here is an example with $I$ finitely generated (this example also illustrates why the Prime Avoidance Theorem only holds for finite sets of primes). Let $A$ be the commutative ring $k[x,y,z]$ with $k$ a field. Let $M$ be the ideal $\langle x,y,z\rangle$. Let $\mathcal{P}$ be the set of all height two primes $\mathfrak{p}$ of $A$ that are contained in the ideal $M$, i.e., $\mathcal{P}$ indexes the irreducible curves in $\mathbb{A}^3$ that contain the origin. For each $\mathfrak{p}$ in $\mathcal{P}$, denote by $A_{\mathfrak{p}}$ the corresponding localization with maximal ideal $\mathfrak{p}A_{\mathfrak{p}}$. Let $A_{\mathcal{P}}$ denote the infinite product, $$ A_{\mathcal{P}} = \prod_{\mathfrak{p}\in \mathcal{P}} A_{\mathfrak{p}},$$ together with the diagonal ring homomorphism, $$\Delta: A \to \prod_{\mathfrak{p}\in \mathcal{P}}A_{\mathfrak{p}}.$$ Denote by $R$ the subring of $A_{\mathcal{P}}$ generated by $\text{Image}(\Delta)$ and the ideal $F$ of all elements in the product ring that have zero components for all but finitely many $\mathfrak{p}\in \mathcal{P}$. Since every homomorphism $A\to A_{\mathfrak{p}}$ is injective, the induced ring homomorphism, $$\rho: A \to R/F,$$ is an isomorphism.
Denote by $I$ the ideal of $R$ generated by $\Delta(J)$, i.e., the finitely generated ideal $\langle \Delta(x), \Delta(y) \rangle$. For every $\mathfrak{p}$ in $\mathcal{P}$, projection on the corresponding factor ring, $$\pi_{\mathfrak{p}}:R \to A_{\mathfrak{p}},$$ is a surjective ring homomorphism. Thus the inverse image ideal is a maximal ideal, $$ \mathfrak{m}_{\mathfrak{p}} = \pi_{\mathfrak{p}}^{-1}\left( \mathfrak{p}A_{\mathfrak{p}}\right).$$ Via $\rho$, every element of $R$ has a unique decomposition as $\rho(a) + b$ with $a\in A$ and with $b\in F$. In particular, every element of $I$ is of the form $\rho(a)+b$ with $a\in M$. The element $a$ is contained in infinitely many height two primes $\mathfrak{p}$ that are contained in $M$, i.e., every hypersurface in $\mathbb{A}^3$ that contains the origin contains infinitely many irreducible curves that contain the origin. Thus, there exists a height two prime $\mathfrak{p}$ in $\mathcal{P}$ containing $a$ and such that the component of $b$ in the corresponding factor $A_{\mathfrak{p}}$ is zero. Thus, $\rho(a)+b$ is contained in $\mathfrak{m}_{\mathfrak{p}}$. So $I$ is contained in the union of the maximal ideals $\mathfrak{m}_{\mathfrak{p}}$ for all $\mathfrak{p}\in \mathcal{P}$.
On the other hand, for every finite set of height two primes $\mathfrak{p}$ in $\mathcal{P}$, there exists $a\in M$ such that $a$ is contained in none of these finitely many primes, i.e., for every finite set of irreducible curves containing the origin, there exists an irreducible surface containing the origin and containing none of the curves. Thus $\rho(a)$ is not contained in the union of the finitely many corresponding maximal ideals $\mathfrak{m}_{\mathfrak{p}}$.
Best Answer
I would extend Jason's comment to say that this condition trivially holds in any Artinian ring since in those rings every intersection of ideals is a finite intersection. It seems that indeed this is the only case.
Let $R$ be an arbitrary noetherian ring and $\mathfrak m$ an arbitrary maximal ideal in $R$. If $R$ is not an integral domain, let $\mathfrak p\subset R$ be a minimal prime ideal contained in $\mathfrak m$ and let $A:=R/\mathfrak p$. Clearly $A$ is an integral domain and $\overline{\mathfrak m}:=\mathfrak m/\mathfrak p$ is a maximal ideal in $A$. By the Krull intersection theorem $\cap_n \overline{\mathfrak m}^n=0$.
It follows that then $\cap_n\mathfrak m^n\subseteq \mathfrak p$ in $R$. If $R$ satisfied the desired condition, then this would imply that $\mathfrak m^n\subseteq \mathfrak p$ for some fixed $n\in \mathbb N$. But then $\mathfrak m= \mathfrak p$ is both maximal and minimal. This is true for every maximal ideal, so $\dim R=0$ and hence $R$ is Artinian.
I leave it for you to decide about the non-noetherian case.