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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.

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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 $\mathbb{Z}$)

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    $\begingroup$ That's a very nice example. $\endgroup$ – Asaf Karagila Jun 4 '18 at 11:46
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    $\begingroup$ @AsafKaragila Thank you (though I still have to see why it would be prime) $\endgroup$ – Maxime Ramzi Jun 4 '18 at 12:26
  • $\begingroup$ These things like ideals of $k^I$ corresponds to filters on $I$ and prime ideals corresponds to ultrafilters and $R/K \cong k$ ... where can I read about these things ? $\endgroup$ – user Jun 4 '18 at 15:24
  • $\begingroup$ Uhm... that $R$ mod an ultrafilter is isomorphic to $k$ just comes from Los's theorem (proofwiki.org/wiki/%C5%81o%C5%9B%27s_Theorem) that you can reprove quite easily in this specific case. That ideals correspond to filters on $I$ is "folklore", I don't have a reference. If you want you can post a question and I would gladly answer. $\endgroup$ – Maxime Ramzi Jun 4 '18 at 15:48
  • $\begingroup$ Note that this a special case of the more general examples coming from Stone spaces I mentioned at the end of my answer, with $X=\beta I$. $\endgroup$ – Eric Wofsey Jun 5 '18 at 1:14
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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.)

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    $\begingroup$ What is the topology on $X$? Discrete? If not, then what? $\endgroup$ – Robert Lewis Jun 4 '18 at 6:35
  • $\begingroup$ Same question as @Robert, .... what is the topology on $X$ ? $\endgroup$ – user Jun 4 '18 at 15:24
  • $\begingroup$ The usual topology, the one-point compactification of $\mathbb{N}$ with the discrete topology. $\endgroup$ – Eric Wofsey Jun 4 '18 at 16:06
  • $\begingroup$ Thanks a lot for your answer ... do you know of any condition on $R$ (apart from the obvious, $R$ being Noetherian) so that the family will have maximal element ? $\endgroup$ – user Jun 6 '18 at 17:29

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