**Note:** Please let me know if this question is too basic for MathOverflow. It is about a subject commonly taught in graduate school (commutative algebra), and is based in large part on a (very elementary) research paper from 2018, and seems to be about a concept ("super-arithmetic" rings) which may not have been studied in the literature, so it seemed like it might fit, but I wasn't sure. If the question is too basic, please let me know so I can close and delete it and re-ask it on Math.SE.

Let $\mathfrak{a, b, a_i, b_i}$ denote ideals of an arithmetic ring $R$.

Question:Is italwaystrue that $\sum_i\mathfrak{a}\cap \mathfrak{b_i=a\cap(\sum_ib_i)}$ and $\mathfrak{a(\bigcap_i b_i)=\bigcap_iab_i}$ in arithmetic rings, even for infinite sums or infinite intersections? If not, what are some counterexamples?

Since one has that $\mathfrak{a \cap b_1 + a\cap b_2 = a \cap (b_1 + b_2)}$ and $\mathfrak{a(b_1 \cap b_2) = ab_1 \cap a b_2}$ for arithmetic rings, it follows by associativity of addition of ideals and of intersection, plus induction, that the above equalities *are* always true for *finite* intersections and *finite* sums. But of course such a proof technique doesn't extend to the infinite case. I suspect the result is false in the infinite case.

Bonus Question 1:If the above isnottrue in arithmetic rings forinfinitesums and intersections, then what is the term for rings for which it does hold for index sets of arbitrary cardinality? "Super-arithmetic" rings? Can anyone point to a reference?

Bonus Question 2:If "super-arithmetic" rings are a strict subset of arithmetic rings, is it true that for "super-arithmetic" rings $\mathfrak{\sum_i (a_i : b) = (\sum_i a_i : b)}$ and $\mathfrak{\sum_i (a : b_i) = (a: \bigcap_i b_i)}$ for ideals $\mathfrak{b, b_i}$ which are infinitely generated and/or for infinite sums?

How do the answers to these questions relate to localization and/or Noetherian conditions? For example, is it the case that any Noetherian arithmetic ring is also "super-arithmetic", since for any localization $R_{\mathfrak{m}}$ of $R$ by a maximal ideal $\mathfrak{m}$, all ideals are contained in an ascending chain ($A_{\mathfrak{m}}$ being uniserial for any maximal $\mathfrak{m}$ is another equivalent characterization of arithmetic rings, see again here), and because $R$ is Noetherian that chain is finite, so all infinite sums and intersections are actually equal to finite sums and intersections (I think? I'm being sloppy obviously).

Bonus Question 3:Is there any example of a "super-arithmetic" ring which isnotNoetherian?

Also I guess the ideal quotient properties aren't likely to apply even in "super-arithmetic" rings when the $\mathfrak{b, b_i}$ aren't finitely generated since their being finitely generated is the most common sufficient condition for their ideal quotients localizing of which I'm aware.

**Definitions:** An arithmetic ring is a "Prüfer domain" that isn't necessarily an integral domain. A Prüfer domain is an arithmetic ring which is also an integral domain.

Arithmetic rings are characterized by the property that:

For all ideals $\mathfrak{a, b_1, b_2}$, $$\mathfrak{a \cap (b_1 + b_2) = a \cap b_1 + a \cap b_2} \,.$$ (I.e. the modular law holds with equality, i.e. $\mathfrak{a \cap b_1 + a \cap b_2 \supseteq a \cap (b_1 + b_2)}$ since the other direction is always true for arbitrary rings.)

In Prüfer domains (i.e. when the ring can be assumed to be an integral domain) the above characterizing property is equivalent to the following:

For all ideals $\mathfrak{a, b_1, b_2}$, $$ \mathfrak{a (b_1 \cap b_2) = ab_1 \cap ab_2} \,.$$ (I.e. $\mathfrak{ab_1 \cap ab_2 \subseteq a(b_1 \cap b_2)}$, since the other direction is always true for arbitrary rings.) However in the full generality of arithmetic rings (i.e. not necessarily integral domains) this property is strictly weaker/more general than the above property (see this paper for details).

multiplication ringsmay bring further ideas. $\endgroup$ – Luc Guyot Jul 7 '19 at 21:49invertible, I do meaninvertible ideal, i.e., $\mathfrak{a} \mathfrak{a}^{-1} = R$. $\endgroup$ – Luc Guyot Jul 8 '19 at 8:33Bonus Question 2. $\endgroup$ – Luc Guyot Jul 8 '19 at 8:37Bonus Question 2needs to be fixed. $\endgroup$ – hasManyStupidQuestions Jul 8 '19 at 15:471more comment