Is every "filter" of rings principal?

Question

Let $\mathcal{F}$ be a class of rings* with the following property:

If there is a homomorphism of rings $A \to B$ with $A \in \mathcal{F}$, then $B \in \mathcal{F}$. Moreover, $\mathcal{F}$ is closed under all small products.

Let us call $\mathcal{F}$ a filter of rings in this case. Let me know if this property already has a different name in the literature. It is the same as a filter on the large preorder of isomorphism classes of rings with $A \leq B$ when there is a homomorphism $A \to B$, with the additional assumption that any small set of elements in the filter has a lower bound in the filter. The definition works for every category with products. Since $\mathcal{F}$ is supposed to be closed under the empty product, we necessarily have $0 \in \mathcal{F}$. The trivial filter is $\{\text{zero rings}\}$.

The context for this notion is SE/4803981 where I found out about the (remarkable!) result that for every $n \geq 0$ the class of $n \times n$-matrix rings $\{A : A \cong M_n(R) \text{ for some ring } R\}$ is a filter of rings. One proof uses that the class is equal to $\{A : \exists f,a,b \in A (f,a,b : f^n = 0, \, a f^{n-1} + f b = 1)\}$ by a Theorem of Agnarsson, S. A. Amitsur and J. C. Robson, and this class is clearly a filter. In fact, it can be written as the class of rings $A$ that admit a homomorphism $\mathbb{Z} \langle f,a,b \rangle / \langle f^n,a f^{n-1}+fb-1 \rangle \longrightarrow A$. But is this just a coincidence?

More generally, if $U$ is any ring, then the "principal filter" $$\{A : U \leq A \} = \{A : \exists \text{ hom. } U \longrightarrow A\}$$ is a filter of rings. If we find a presentation $U \cong \mathbb{Z}\langle (X_i)_{i \in I} \rangle / P$, then this filter is equal to $\{A : \exists a \in A^I \forall p \in P ~ (p(a)=0 )\}$. For a basic example, $\{A : \exists a \in A ~ (a^2=2)\}$ is a filter, here we take $U := \mathbb{Z}[X]/\langle X^2-2 \rangle$.

Question. Does every filter of rings have this form? If not, can we "build" all filters from these? And if this is also not the case: how can we then classify all filters of rings?

I am also interested in variations of this theme: what happens in the category of commutative rings? Of course, only categories without zero morphisms are interesting.

It seems that we might run into set-theoretical difficulties here. Namely, filters of rings are clearly closed under large intersections, but the formula $$\bigcap_{i \in I} \{A : \exists \text{ hom. } U_i \longrightarrow A\} = \{A : \exists \text{ hom. } \coprod_{i \in I} U_i \longrightarrow A\}$$ only makes sense when $I$ is a small set (where $\coprod$ signifies the coproduct of rings). But, for example, both $\bigcap_{U \text{ ring}} \{A : U \leq A\}$ and $\bigcap_{0 \neq U \text{ ring}} \{A : U \leq A\}$ are trivial.

When $\mathcal{F}$ is small, then it is easy to check that $U := \prod_{A \in \mathcal{F}} A$ satisfies $\mathcal{F} = \{A : U \leq A\}$; but then we can actually prove $U=0$ and $\mathcal{F} = \{\text{zero rings}\}$, so this case is not interesting.

*All rings and homomorphisms are unital.

Answer 1

To avoid technicalities about what exactly you mean by "class", let me assume you are working in a Grothendieck universe $V_\kappa$ and "class" means any subset of $V_\kappa$. Then Vopěnka's principle (a large cardinal axiom) implies that a very general version of the statement you ask for is true. Here is one possible formulation of this. Let $C$ be any full subcategory of an accessible category. Say a class of objects $F$ of $C$ is a (complete) "filter" if whenever $A\in F$ and there exists a morphism $f:A\to B$ then $B\in F$ and if $S\subseteq F$ is any small set then there exists an element $P\in F$ such that there is a morphism $P\to A$ for each $A\in S$. Then Vopěnka's principle implies that any such filter $F$ is principal (i.e. there exists $A\in F$ such that every $B\in F$ has a morphism from $A$).

To prove this, note that it suffices to show there is a small set $S\subseteq F$ such that every element of $F$ has a morphism from some element of $S$. So, assume no such $S$ exists. We can then recursively choose a sequence $(A_\alpha)_{\alpha<\kappa}$ of elements of $F$ such that $A_\alpha$ has no morphism from $A_\beta$ for each $\beta<\alpha$. But this violates Vopěnka's principle (see for instance Lemma 6.3 in Adamek and Rosicky's Locally Presentable and Accessible Categories).

(I have no idea whether in the special case of rings, this can be proved without large cardinal assumptions.)