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Given an infinite distributive lattice $L$, does $L$ contain a non-principal prime ideal $I$, or a non-principal prime filter $F$? ($I$ is said to be principal if there is $x\in L$ such that $I=\{y\in L: y\leq x\}$. Dual definition for filters.)

If the answer is "yes" to the question above, what if we replace "prime" by "maximal"?

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  • $\begingroup$ With maximal, not in general: e.g., take any L with a top, and attach to it a new top. $\endgroup$ – Emil Jeřábek Oct 6 '15 at 6:37
  • $\begingroup$ You might be able to prove existence by considering the associated lattices IDL(L) and FIL(L) for a given infinite distributive lattice L. I would not be surprised if this dealt with choice issues, e.g. that in a variant of set theory without choice one finds a model of this theory which believes it has an infinite distributive lattice all of whose prime ideals are principal. Gerhard "Still Capable Of Showing Surprise" Paseman, 2015.10.06 $\endgroup$ – Gerhard Paseman Oct 6 '15 at 18:34
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It’s not clear to me what exactly is the intended definition of prime ideals and filters in distributive lattices. Based on an analogy with other classes of structures, it seems to me that conceptually the best choice should be to make the definition correspond to subdirectly irreducible factors in subdirect products, that is, to surjective homomorphisms to the 2-element lattice. This means that prime ideals and filters are required to be both nonempty and proper.

In that case, $\omega$ with its usual linear order is an infinite lattice in which all prime ideals and prime filters (indeed, all nonempty filters and nonempty proper ideals) are principal.

Different choices in the definition may yield different answers; in particular the empty ideal/filter is always nonprincipal if allowed.

The answer for maximal ideals/filters is unequivocally “no”: for example, one can take any infinite distributive lattice $L$ with a top element $\top$ and a bottom element $\bot$ such that $L\smallsetminus\{\top,\bot\}$ has a top element $t$ and a bottom element $b$. Then the only maximal ideal in $L$ is the one generated by $t$, and dually for filters.

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The bounded distributive lattices which every prime filter is principal can be characterized in terms of Priestley spaces, Priestley duality and certain posets (these are some results that I proved a while ago but I would not be surprised if someone else has already obtained these results already).

Let me review some facts about Priestley duality.

A partially ordered compact space $X$ is said to be a Priestley space if whenever $x\not\leq y$ then there is a clopen upper set $U$ with $x\in U$ and $y\not\in U$. If $X$ is a Priestley space, then let $\mathcal{C}_{d}(X)$ be the collection of all clopen downwards closed subsets of $X$. Then $\mathcal{C}_{d}(X)$.

Suppose that $D$ is a bounded distributive lattice. Let $\mathcal{I}_{p}(D)$ be the set of all prime ideals in $D$ ordered by inclusion. If $a\in D$, then let $C_{a}=\{I\in\mathcal{I}_{p}(D)|a\not\in I\}$. Then $\{C_{a}\setminus C_{b}|a,b\in D\}$ is a basis for a Priestley space topology on $\mathcal{I}_{p}(D)$.

Priestley duality states that if $D$ is a distributive lattice, then $D\simeq \mathcal{C}_{d}(\mathcal{I}_{p}(D))$ and if $X$ is a priestley space, then $X\simeq\mathcal{I}_{p}(\mathcal{C}_{p}(X))$ and that these isomorphisms are natural.

Suppose that $(X,\leq,\mathcal{T})$ be an ordered topological space. Then we shall call $(X,\leq,\mathcal{T})$ an upper limit system if each set of the form $\downarrow x$ is closed and $\{\downarrow x\cap U|x\in X,\text{$U$ is an open upwards closed set}\}$ forms a basis for the topology $\mathcal{T}$.

Take note that if $(X,\leq,\mathcal{T})$ is an upper limit system, then each set $\downarrow x$ is clopen and each lower set is open.

$\mathbf{Lemma}$ Suppose that $X$ is a Priestley space. Let $x\in X$. Then

  1. The ideal $\{C\in\mathcal{C}_{d}(X)|x\not\in C\}$ is principal in $\mathcal{C}_{d}(X)$ if and only if $\uparrow x$ is clopen.

  2. The filter $\{C\in\mathcal{C}_{d}(X)|x\in C\}$ is principal in $\mathcal{C}_{d}(X)$ if and only if the set $\downarrow x$ is clopen.

$\mathbf{Proof}$

We shall only prove part 2 since part 1 is identical.

$\leftarrow$ If $\downarrow x$ is clopen, then clearly $\downarrow x$ is the least element in $\{C\in\mathcal{C}_{d}(X)|x\in C\}$, so $\{C\in\mathcal{C}_{d}(X)|x\in C\}$ is principal in $\mathcal{C}_{d}(X)$.

$\rightarrow$ Suppose now that $\{C\in\mathcal{C}_{d}(X)|x\in C\}$ is principal. Then let $R$ be the least element in $\{C\in\mathcal{C}_{d}(X)|x\in C\}$. Then clearly $\downarrow x\subseteq R$. Suppose now that $\downarrow x$ is a proper subset of $R$. Then let $y\in R\setminus\downarrow x$. Then since $y\not\leq x$, there is a clopen downwards closed subset $S$ with $x\in S$ but $y\not\in S$. Therefore $S\in\{C\in\mathcal{C}_{d}(X)|x\in C\}$ but $R\not\subseteq S$ contradicting the minimality of $R$. We therefore conclude that $\downarrow x=R$ after all. Therefore $\downarrow x$ is clopen.

$\mathbf{QED}$

The following proposition is not too hard to prove.

$\mathbf{Proposition}$ Let $X$ be a Priestley space. Then the following are equivalent.

  1. $X$ is an upper limit space.

  2. Every set of the form $\downarrow x$ is clopen.

  3. Every filter in $\mathcal{C}_{d}(X)$ is principal.

Suppose that $X$ is a poset. Then the minimal upper limit topology on $X$ is the topology generated by the subbasis $\{\downarrow x|x\in X\}\cup\{(\downarrow x)^{c}|x\in X\}$. The minimal upper limit topology is clearly generated by the basis consisting of sets of the form $\downarrow x\cap(\downarrow y_{1})^{c}\cap...\cap(\downarrow y_{n})^{c}$.

Clearly every poset whose minimal upper limit topology is compact is a Priestley space in the minimal upper limit topology and every Priestley space which is an upper limit topology is the minimal upper limit topology. We conclude that by Priestley duality the bounded distributive lattices where every filter is principal are in a one-to-one correspondence with the posets which are compact in the minimal upper limit topology. The following theorem characterizes the posets which are compact in the minimal upper limit topology.

$\mathbf{Theorem}$ Suppose that $X$ is a poset. Then $X$ is compact in the minimal upper limit topology if and only if

  1. for each $B\subseteq X$ there are lower bounds $a_{1},...,a_{n}$ of $B$ such that if $a$ is a lower bound of $B$, then $a\leq a_{i}$ for some $i\in\{1,...,n\}$.

  2. for each $B\subseteq X$ there are $b_{1},...,b_{n}\in B$ where if $a$ is a lower bound of $\{b_{1},...,b_{n}\}$ then $a$ is a lower bound of $B$.

$\mathbf{Proof}$

$\rightarrow$ Suppose that $X$ is compact in the minimal upper limit topology. Suppose that $B\subseteq X$ is a subset.

Then $\downarrow B=\bigcap_{b\in B}\downarrow b$ is closed being the intersection of closed sets. However, $\downarrow B$ is a downwards closed set, so $\downarrow B$ is also an open set. Thus, $\downarrow B$ is clopen.

Let's now show that statement 1 holds. Since $\downarrow B$ is open and $\{\downarrow x|x\in\downarrow B\}$ is an open cover of $X$, there are $a_{1},...,a_{n}\in B$ so that $\downarrow B=\downarrow a_{1}\cup...\cup\downarrow a_{n}$. Therefore, $a_{1},...,a_{n}$ are lower bounds of $B$ so that if $a$ is a lower bound of $B$ then $a\leq a_{i}$ for some $i\in\{1,...,n\}$.

We shall now show statement 2 holds. We have $(\downarrow B)^{c}=\bigcup_{b\in B}(\downarrow b)^{c}$. Therefore, since $(\downarrow B)^{c}$ is compact, there are $b_{1},...,b_{n}$ with $(\downarrow B)^{c}=(\downarrow b_{1})^{c}\cup...\cup(\downarrow b_{n})^{c}$, so we have $\downarrow B=\downarrow b_{1}\cap...\cap \downarrow b_{n}$. Therefore, $b_{1},...,b_{n}$ are elements in $B$ such that if $a$ is a lower bound of $\{b_{1},...,b_{n}\}$ then $a$ is also a lower bound of $B$.

$\leftarrow$ To prove this direction, we shall use Alexander's subbase theorem. In particular, we shall show that $X$ is compact by showing that every cover from the basis $\{\downarrow x|x\in X\}\cup\{(\downarrow x)^{c}|x\in X\}$ has a finite subcover.

Suppose that $A,B\subseteq X$ are subsets such that $\mathcal{U}=\{\downarrow a|a\in A\}\cup\{(\downarrow b)^{c}|b\in B\}$ covers $X$.

Let $\mathcal{V}=\{(\downarrow b)^{c}|b\in B\}\cup\{\downarrow x|x\in\downarrow B\}$. Then $\mathcal{V}$ covers $X$. I claim that $\mathcal{V}$ refines $\mathcal{U}$. Suppose that $V\in\mathcal{V}$. If $V=(\downarrow b)^{c}$, then $V\subseteq V\in\mathcal{U}$. If $V=\downarrow x$ for some $x\in\downarrow B$, then $x\in U$ for some $U\in\mathcal{V}$. However, since $x\in\downarrow B$, if $b\in B$, then $x\in\downarrow b$, so $x\not\in(\downarrow b)^{c}$. Therefore $U\neq(\downarrow b)^{c}$ for $b\in B$. Therefore, $U=\downarrow a$ for some $a\in A$. We conclude that $V=\downarrow x\subseteq\downarrow a=U$. We conclude that $\mathcal{V}$ refines $\mathcal{U}$.

We shall now show that $\mathcal{V}$ has a finite subcover.

By 2, there are $b_{1},...,b_{n}\in B$ where $a$ is a lower bound of $b_{1},...,b_{n}$ if and only if $a$ is a lower bound of $B$. In other words, we have $\downarrow b_{1}\cap...\cap\downarrow b_{n}=\downarrow B$. Therefore, we have $(\downarrow b_{1})^{c}\cup...\cup(\downarrow b_{n})^{c}=\bigcup_{b\in B}(\downarrow b)^{c}=(\downarrow B)^{c}$.

Similarly, there are lower bounds $a_{1},...,a_{m}$ of $B$ so that if $a$ is a lower bound of $B$, then $a\leq a_{i}$ for some $i\in\{1,...,m\}$. Therefore, we have $\downarrow a_{1}\cup...\cup\downarrow a_{m}=\downarrow B$. Therefore, we have $X=\downarrow a_{1}\cup...\cup\downarrow a_{m}\cup(\downarrow b_{1})^{c}\cup...\cup(\downarrow b_{n})^{c}$. Thus $\{\downarrow a_{1},...,\downarrow a_{m},(\downarrow b_{1})^{c},...,(\downarrow b_{n})^{c}\}$ is a finite subcover of $\mathcal{V}$. Since $\mathcal{V}$ refines $\mathcal{U}$, the cover $\mathcal{U}$ has a finite subcover as well. Therefore, by Alexander's subbase theorem, the space $X$ is compact. $\mathbf{QED}$

Therefore we conclude that the bounded distributive lattices where every prime filter is principal are in a one-to-one correspondence with the posets satisfying the conditions of the above theorem. We conclude with a proof of a proposition that Dominic van der Zypen has claimed in his answer.

$\mathbf{Proposition}$ Every infinite bounded distributive lattice has a non-principal prime filter or a non-principal prime ideal.

$\mathbf{Proof}$ Suppose that $X$ is the Priestley space for a bounded distributive lattice where every ideal and filter is principal. Then each set of the form $\downarrow x$ and $\uparrow x$ is clopen, so $\{x\}=\downarrow x\cap\uparrow x$ is clopen. Therefore, since every point in $X$ is isolated, we conclude that $X$ is finite, so its corresponding bounded distributive lattice is also finite.

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A counterexample is $\omega+1$, turned upside down. However, it is true that every infinite distributive lattice contains either a non-principal prime ideal or a non-principal prime filter. This is due to the following argument:

Any infinite distributive lattice contains at least a non-principal ideal or a non-principal filter (not necessarily prime). We may assume that $J$ is a non-principal ideal, so that $j^* = \bigvee J \notin J$. Let $G = \{y\in L: y \geq j^*\}$ be the principal filter generated by $j^*$. As $J\cap G = \emptyset$ we can use the Prime Ideal Theorem and get a prime ideal $P$ such that $J\subseteq P$ and $P\cap G = \emptyset$, which implies $j^* \notin P$.

Next we show that $P$ is not principal: if we had $p^*:=\bigvee P \in P$ then $J\subseteq P$ would imply $p^* \geq j^* = \bigvee J$ and $j^* \in P$ because $P$ is a down-set. But by construction $j^* \notin P$. Therefore $P$ is a non-principal prime ideal.

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    $\begingroup$ Does not your argument depend on completeness? What if $\bigvee J$ does not exist at all? $\endgroup$ – მამუკა ჯიბლაძე Oct 7 '15 at 14:18
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    $\begingroup$ მამუკა ჯიბლაძე. The proof still works if $\bigvee J$ does not necessarily exist. If $\bigvee J$ does not exist, then one simply defines $G$ to be the set of all upper bounds of $J$ and the proof goes through with no problem. $\endgroup$ – Joseph Van Name Oct 9 '15 at 3:24
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    $\begingroup$ Dominic van der Zypen. I must ask what argument do you use to conclude that every infinite distributive lattice has an non-principal ideal or a non-principal filter? $\endgroup$ – Joseph Van Name Oct 9 '15 at 3:26
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    $\begingroup$ Clearly not every infinite lattice has a non-principal ideal or a non-principal filter (for example, take the Dedekind-Macneille completion of any infinite antichain). Let me prove that every infinite distributive lattice has a non-principal ideal or a non-principal filter. Suppose that $X$ is an infinite distributive lattice with no non-principal ideals and no non-principal filters. Let $C$ be the set of all join-prime elements in $X$. Then take note that since $X$ has no infinite descending chains, every element in $X$ is the least upper bound of finitely any join-irreducible elements. $\endgroup$ – Joseph Van Name Oct 9 '15 at 18:05
  • $\begingroup$ However, since $X$ is inifnite, the set $C$ must also be infinite. Recall that as a consequence of Ramsey's theorem, every poset must have an infinite ascending chain, an infinite descending chain, or an infinite antichain. Therefore, the set $C$ must have an infinite antichain $(c_{n})_{n\in\mathbb{N}}$. Take note that since each $c_{n}$ is join-prime, we have $c_{n}\not\leq c_{1}\vee...\vee c_{n-1}$. Therefore, we have $c_{1}\vee...\vee c_{n-1}\leq c_{1}\vee...\vee c_{n}$. We therefore conclude that $(c_{1}\vee...\vee c_{n})_{n\in\omega}$ is an infinite ascending chain after all. $\endgroup$ – Joseph Van Name Oct 9 '15 at 18:05

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