26 votes
Accepted

Are there axioms satisfied in commutative rings and distributive lattices but not satisfied in commutative semirings?

Following François's suggestion, I ran alg to find a unital commutative semiring which fails to satisfy $$ \forall x\, y\, z,\; x + z = y + z \land x \times z = y \times z \Rightarrow x = y. \tag{1} ...
Andrej Bauer's user avatar
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20 votes
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Representation theorem for modular lattices?

There are lots of relations satisfied in lattices of submodules besides the ones implied by modularity. For example, there is the Desarguesian identity mentioned here (which holds in any lattice of ...
Todd Trimble's user avatar
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18 votes
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Are these two quotients of $\omega^\omega$ isomorphic?

Very nice question! They are not isomorphic. What I claim is that when we take the quotient with respect to density, there is a countably infinite antichain above $0$ having a minimal upper bound, ...
Joel David Hamkins's user avatar
17 votes

Are arbitrary nonempty intersections of principal filters principal?

Let $L$ be the lattice of finite and cofinite subsets of $\mathbb N$. Let $\mathcal F_n$ be the principal filter of elements of $L$ that contain the element $n$ (i.e., $\mathcal F_n$ contains the ...
Keith Kearnes's user avatar
17 votes
Accepted

Are the Boolean algebras ${\cal P}(\omega)/(\text{fin})$ and ${\cal P}(\omega)/(\text{thin})$ isomorphic?

The answer is no. In the Boolean algebra $P(\omega)/\text{Fin}$ every strictly descending sequence $A_0>A_1>A_2>\cdots$ has a nonzero lower bound, by the famous construction of Hausdorff. ...
Joel David Hamkins's user avatar
15 votes

Class of lattices that excludes $M_3$?

This is a (slightly edited) copy of the answer I posted on math.SE to the question What do we call a lattice that does not have a sublattice the shape of the diamond $M_3$?: Let $\mathbf K$ be the ...
bof's user avatar
  • 11.1k
14 votes

Representation theorem for modular lattices?

The lattice of submodules of a module satisfies a stronger identity, namely the Arguesian law: $$ (x_0\vee y_0)\wedge (x_1\vee y_1)\wedge (x_2\vee y_2)\leq ((z\vee x_1)\wedge x_0) \vee ((z\vee y_1)\...
Pedro Sánchez Terraf's user avatar
14 votes
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Group structure for distributive lattices

No: If it's natural, it should be invariant under the automorphism group of the original lattice. let $X$ be the free distributive lattice on 2 generators $x,y$: it has 6 elements, $$0\quad<\quad x\...
YCor's user avatar
  • 60k
14 votes
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Determining whether a lattice is the face lattice of a polytope - NP hard or undecidable?

There's no contradiction: I don't know the correct complexity, but I recall hearing several times that it is at least as hard as NP. It is not difficult to show (using Tarski's algorithm, as ...
Geva Yashfe's user avatar
  • 1,356
14 votes
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Are modular lattices shallow?

Terminology. If $f$ is a fundamental operation of an algebra $A$, then a polynomial of the form $f(c_1,\ldots,c_k,x,c_{k+2},\ldots,c_n)$ is often called a 'basic translation' or a '$1$-translation' ...
Keith Kearnes's user avatar
13 votes
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Is this Wikipedia article linking to the wrong notion of coherent space

Yes, the notion of "coherent space" at the page you linked is completely unrelated to spectral spaces. The coherent spaces of that article were introduced by Jean-Yves Girard as a denotational ...
Damiano Mazza's user avatar
13 votes
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What is known about ideal and divisibility lattices of GCD domains and their generalizations?

Given a ring $R$, let us denote by $L(R)$ the lattice of two-sided ideals of $R$ for which the infimum and supremum are given by $\inf(I, J) = I \cap J$ and $\sup(I, J) = I + J$. Such lattices are ...
Luc Guyot's user avatar
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13 votes

Are arbitrary nonempty intersections of principal filters principal?

There are two good answers already; but I’ll add a little motivation for how one might find the way to them. If $L$ is complete, then as you say, it’s easy to see that any intersection of principal ...
Peter LeFanu Lumsdaine's user avatar
13 votes

A note on orders in quaternion algebras

Two orders need not be isomorphic. First of all, in number fields $K$ other than $\mathbf Q$ not all orders are isomorphic rings (even if they are isomorphic abelian groups): the full ring of integers ...
KConrad's user avatar
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12 votes
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Automorphisms of power set lattice mod finite

This is known as Rudin-Shelah problem. Note that, by Stone duality, this is equivalent to determine the self-homeomorphism group of the Stone-Cech boundary of $N$. Notably, consider the group induced ...
YCor's user avatar
  • 60k
12 votes
Accepted

Fibers of the morphism from the free Heyting algebra to the free Boolean algebra

$\let\eq\leftrightarrow$Notice that $\psi(A)=u$ iff $\vdash_\mathrm{CPC}A\eq u$ iff $\vdash_\mathrm{IPC}\neg\neg(A\eq u)$. (I will write just $\vdash$ for $\vdash_\mathrm{IPC}$.) Thus: $\bot$ has a ...
Emil Jeřábek's user avatar
12 votes
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A meet-semilattice with top element that is not a lattice?

It is well-known that a finite meet-semi-lattice with a maximum element is a lattice. The reason is that we can define $a \vee b := \wedge \{c\colon \textrm{$c$ is an upper bound for $a,b$}\}$, where ...
Sam Hopkins's user avatar
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12 votes
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Ultrafilter lemma for arbitrary lattice

It is equivalent to AC. Consider any collection $A$ of nonempty sets, and let $\newcommand\P{\mathbb{P}}\P$ be the set of partial choice functions, so that $p\in\P$ if and only if $p$ is a partial ...
Joel David Hamkins's user avatar
11 votes

Duality between topology and bornology

It was not clear to me at first what your question has to do with bornologies, but now (EDITED) I see it. Any collection $\nu$ of subsets of $X$ (I assume that $X$ is nonvoid through the remainder of ...
Pedro Lauridsen Ribeiro's user avatar
11 votes
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Homotopy type of some lattices with top and bottom removed

Let $C(P)$ be the poset obtained by removing top and bottom from $P$ (where $P$ is a poset having a top and a bottom, not equal). Then $C(P\times Q)$ is homotopy equivalent to $\Sigma(C(P)\ast C(Q))$, ...
Tom Goodwillie's user avatar
11 votes
Accepted

Which lattices are quotients of finite powerset lattices?

The class of finite powerset lattices is closed under quotients, up to isomorphism. That is, the quotients are exactly the lattice reducts of finite Boolean algebras. In particular, any quotient $L$ ...
Emil Jeřábek's user avatar
11 votes
Accepted

Is every homogeneous poset a lattice?

Counterexample. Let $$P=\{(x,i)\in\mathbb Q\times\{0,1\}:0\le x\le1,\ x\ne i\}$$ be ordered so that $$(x,i)\lt(x',i')\iff x\lt x'.$$
bof's user avatar
  • 11.1k
10 votes

What is the meaning of this analogy between lattices and topological spaces?

EDIT: I mentioned in the comments that Nik's list of definitions and theorems looks to me strikingly similar to Wallman's generalization of Stone duality. After getting some feedback from Nik and ...
Will Brian's user avatar
  • 17.3k
10 votes
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Order-preserving surjection ${\mathbb N}^{\mathbb N}\to [0,\infty)$

Yes. Let us isomorphically identify the poset of functions $\omega \to \omega$ (under the pointwise order) with the set of functions $\omega \to \mathbb{N}_2 = \{n \in \mathbb{N}: n \geq 2\}$, again ...
Todd Trimble's user avatar
  • 52.2k
10 votes
Accepted

Does the lattice of all topologies embed into the lattice of $T_1$-topologies?

Is there an injective lattice homomorphism $\varphi: \text{Top}(\kappa)\to \text{Top}^{T_1}(\kappa)$? The answer is Yes, there is such an embedding. I will argue that if $\kappa$ is an infinite ...
Keith Kearnes's user avatar
10 votes

Class of lattices that excludes $M_3$?

As bof says, the class $\mathcal K$ of lattices omitting $\mathbf M_3$ as a sublattice is a quasivariety that is not a variety. As bof also says, this implies that $\mathcal K$ is closed under the ...
Keith Kearnes's user avatar
10 votes
Accepted

A definition in poset theory

I recall seeing in various sources the terminology "cover preserving embedding" and "cover preserving subposet". Googling it now (https://www.google.com/search?q=poset+%22cover+...
Vladimir Dotsenko's user avatar
9 votes
Accepted

Pseudocomplements in the lattice of topologies

Yes, and in fact, most familiar topologies do not have a pseudo-complement. To see this, notice that that it often happens with a topology $\tau$ on a set $X$ that there are non-open sets $A$ and $B$ ...
Joel David Hamkins's user avatar
9 votes
Accepted

Uniquely complemented but not Boolean

As mentioned in the comments, there are and this follows from Dilworth's construction of free uniquely complemented lattices. A simpler construction was later given by Chen and Grätzer [On the ...
François G. Dorais's user avatar
9 votes
Accepted

$2^\omega$ vs $({\omega+1})^\omega$

No. The lattice $2^\omega$ is complemented, but the lattice $\omega+1$ is not complemented. Since a homomorphic image of a complemented lattice is complemented, there is no surjective lattice ...
bof's user avatar
  • 11.1k

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