Suppose one has a filter (a collection of subsets closed under increasing the size of the set and under finite intersection) $F$ on a ring $R$. Say that $F$ is (ring) ideal-like if for every set $U \subset R$ in $F$, there exists a $U'\in F$ such that $U' + U' \subset U$, and there exists a $U'' \in F$ such that $U'' R \subset U$. (This corresponds to the filtered intersection of elements of $F$ giving rise to a pro-set which satisfies the ideal conditions.) Does this imply that $F$ is generated as a filter by actual ring ideals of $R$? In other words, for any $U \in F$, does there exist an ideal $I$ such that $I \in F$ and $I \subset U$?


If such a situation is possible, is it still possible if $R$ is Noetherian? If $R = \mathbb{Z}$?



Ideal-like filter on a ring not generated by ring ideals
This does not hold for all commutative rings, but it does hold for Noetherian rings and for valuation rings (assuming the convention that filters don't contain $\emptyset,$ or else $\mathcal{P}(R)$ is a trivial counterexample for any ring).


Suppose $R$ is a commutative ring, $F$ is an ideal-like filter on $R,$ and $U\_0 \in F$ is such that there is no ideal $I \in F \cap \mathcal{P}(U\_0).$ Recursively choose $U\_n \in F$ such that $U\_{n+1}+U\_{n+1} \subset U\_n$ and $U\_{n+1}=U\_{n+1}R.$ Since no $U\_n$ is empty or an ideal, we immediately have $U\_{n+1} \subsetneq U\_n.$


For $n \ge 1,$ we recursively choose $u\_n \in U\_n$ for $n \ge 1$ as follows: first, let $u\_1$ be any element of $U\_1 \setminus U\_2.$ Suppose $u\_1, \ldots, u\_n$ have been chosen. An inductive argument shows that $I\_n:=\langle u\_1, \ldots, u\_n \rangle \subset U\_0,$ so $I\_n \not \in F.$ Let $u\_{n+1}$ be any element of $U\_{n+1} \setminus I\_n.$ Then $\{I\_n\}$ is a strictly increasing sequence of ideals, so $R$ is non-Noetherian. Furthermore, $\langle u\_1 \rangle$ and $\langle u\_2 \rangle$ are incomparable, so $R$ is not a valuation ring.


Now we'll construct a commutative ring $R$ with an ideal-like filter $F$ not generated by ring ideals. Let $R = C^{\infty}(S^1)$ be the ring of smooth real-valued functions on the circle. Every maximal ideal in $R$ is of the form $\mathfrak{m}\_x=\{f \in R: f(x)=0\}$ for some $x \in S^1$ (see <https://math.stackexchange.com/a/183213/210610>). Let
$A\_n=\{f \in R: \lambda(f^{-1}(0))> 1- \frac{1}{n}\},$ and let $F$ be the filter on $R$ generated by the $A\_n$'s, i.e.
$$F=\bigcup\_{n=1}^{\infty} \{X \subset R: A\_n \subset X\}.$$


It's clear that every $X \in F$ is infinite (in fact, $|X|=|\mathbb{R}|$) and that for every finite $S \subset R \setminus \{0\},$ $R \setminus S \in F.$ We will show that there is no proper ideal in $F.$ Suppose $I \in F$ is a proper ideal on $R.$ There is $x$ such that $I \subset \mathfrak{m}\_x \in F.$ But for every $n,$ there is a bump function in $A\_n \setminus \mathfrak{m}\_x,$ contradiction. Thus, $U=R \setminus \{1\} \in F$ does not contain any ideal in $F.$


Finally, we verify that $F$ is ideal-like. Fix $U \in F.$ Let $n$ be such that $A\_n \subset U.$ Then $V=A\_{2n}$ satisfies $V=VR$ and
$V+V \subset A\_n \subset U.$




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Original answer:


This does not hold for all commutative rings, but if $R$ is a Dedekind domain, and we include the filter convention that $\emptyset \not \in F,$ then an ideal-like filter $F$ is generated by ring ideals.


Suppose $F$ is a counterexample. Let $U \in F$ be such that there is no ideal $I \in F \cap \mathcal{P}(U).$ Let $V, W \in F$ be such that $V+V+V \subset U$ and $WR \subset V.$ We may assume $W=WR.$ Since $W$ is nonempty and not an ideal, we can fix a nonzero non-unit $w \in W,$ and prime ideals $P\_i$ such that $\langle w \rangle = \prod\_{i=1}^n P\_i.$


Let $A \subset \{1,\ldots, n\}$ be maximal such that $\prod\_{i \in A} P\_i \in F.$ Let $X \in F$ be such that $X=XR$ and $\sum\_{i=1}^n X \subset W.$ For $j \in A,$ set $x\_j=0,$ and otherwise let $x\_j$ be an arbitrary element of $X \cap \prod\_{i \in A} P\_i \setminus \prod\_{i \in A \cup \{j\}} P\_i,$ which is nonempty since $X \cap \prod\_{i \in A} P\_i \in F.$ Let $x, y$ be such that $\langle x, y \rangle = \langle x\_i: i \le n \rangle \subset W.$ Consider
$$I:=\langle w, x, y \rangle \subset (W + W + W) \cap \prod\_{i \in A} P\_i \subset U \cap \prod\_{i \in A} P\_i.$$ Since $\prod\_{i=1}^n P\_i \subset I
\subsetneq \prod\_{i \in A} P\_i \in F,$ there is $B \supsetneq A$ such that $I = \prod\_{i \in B} P\_i.$


Fix $j \in B \setminus A.$ Then $A \cup \{j\} \subset B$ and $x\_j \in \langle x, y \rangle,$ so $x\_j \in \prod\_{i \in B} P\_i \setminus \prod\_{i \in A \cup \{j\}} P\_i = \emptyset,$ contradiction.


