If a commutative ring with unity $R$ has an ideal $I$ such that $R/I$ is infinite, then does there exist a maximal such ideal?

Question

Let $R$ be a commutative ring with unity. Consider the collection of ideals $\mathcal F:=\{I \lhd R : R/I $ is infinite $ \}$. If $\mathcal F$ is non-empty , then does $\mathcal F$ necessarily have a maximal element w.r.t. inclusion ?

I can only show that any maximal element of $\mathcal F$ (if exists) is a prime ideal.

Answer 1

Let $k$ be a finite field and let $R=k^I$ for some infinite $I$.

Then ideals of $R$ correspond to filters on $I$. If you are right that such an ideal is prime then this implies that the corresponding filter is prime, so an ultrafilter, hence $R/K$ is actually isomorphic to $k$ so is finite, a contradiction.

Hence if you're right about the primality of such an ideal, then there are examples without a maximal element (and of course examples with a maximal element, such as $\mathbb{Z}$)

Answer 2

Let $k$ be a finite field, let $X=\mathbb{N}\cup\{\infty\}$, and let $R$ be the ring of continuous functions $X\to k$ where $k$ has the discrete topology. (Equivalently, $R$ is the ring of eventually constant sequences of elements of $k$.) Note that ideals in $R$ are in bijection with closed subsets of $X$, by sending a closed subset $A\subseteq X$ to the ideal $I(A)$ of functions vanishing on it. Moreover, $R/I(A)$ can be identified with the ring of continuous functions $A\to k$. In particular, $R/I(A)$ is infinite iff $I(A)$ is infinite.

So, your question for this $R$ is equivalent to asking whether there is a minimal infinite closed subset of $X$. The answer is no, since for any infinite closed subset $A\subseteq X$, you can remove any point of $A$ besides $\infty$ to get a smaller infinite closed subset.

(More generally, you can get similar example with $X$ replaced by any infinite Stone space.)