24

Following François's suggestion, I ran alg to find a unital commutative semiring which fails to satisfy
$$
\forall x\, y\, z,\; x + z = y + z \land x \times z = y \times z \Rightarrow x = y.
\tag{1}
$$
The smallest one has size 3. Here is the output of the program, cut off after the first example.
Theory unital_commutative_semiring.
Constant 0 ...

answered Sep 10 '20 at 11:34

Andrej Bauer

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22

All free Jónsson-Tarski algebras on a finite nonempty set of generators are isomorphic. Thus free objects may not know their rank.
Curiously the automorphism group of this free algebra is the famous Thompson simple group $V$.

22

This follows from two Facts:
1) A category monadic over Set/S is always an exact category. That is it has quotient by equivalence relation that are effective and universal. It is in particular a regular category. This is showed for example here.
2) The category of categories is not a regular category. An explicit example of regularity (and hence exactness) ...

21

Sela does not have an objective view of Kharlampovich, Myasnikov's work. We will post a paper
dismissing his statements about "fatal mistakes". It just takes time. There are some typos and inessential errors that were fixed in later works. Sela himself has many such mistakes.
The objects in the two works (Sela's and our work) are similar but not identical. ...

21

Let $\Sigma_0$ be the $\sigma$-algebra of Lebesgue measurable sets on the real interval $[0,1]$. Define $\Sigma$ to be $\Sigma_0$ quotiented by the relation $U \simeq V$ if $U$ and $V$ differ by a Lebesgue negligible set.
$\Sigma$ is a complete boolean algebra, but $\Sigma$ is not a $\sigma$-algebra.
Indeed, assume $\Sigma$ is identified with a $\sigma$-...

20

The answer is No. For $n=2$, there is the Hall identity $[[x,y]^2,z]=0$. Drensky proved in 1981 that these two identities (Hall and standard (=Amitsur-Levitzki)) form a basis of identities of $Mat_2(\Bbb C)$ (i.e. all identities are consequences of these two). For higher $n$, bases of identities are (probably) unknown.
Note that, according to the Kemer ...

20

For what it's worth, the reason I made that comment was because when I gave talks expressing skepticism about the philosophical basis of ZFC, I would often get the reaction "but as long as it's consistent, what's the problem?" Some version of this attitude is also quite prevalent in the philosophical literature on the subject. I wanted to make the point ...

19

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 congruence relations on an algebra of a Mal'cev theory, such as the theory of modules over a ring).
So, a projective plane which is not Desarguesian gives an ...

answered Feb 8 '16 at 13:13

19

I was friends with Frank Adams and Saunders Mac Lane, who invented PROPs in one of the world's most extensive unpublished collaborations. Saunders once showed me a box full of their correspondence. One reason they never published is that they lacked a way of showing the PROPs they were interested in acted on the things they wanted to have actions. Operads ...

19

A door space $X$ is $T_0$, for if $x,y\in X$ are not separated by the
$T_0$ axiom then the set $\{x\}$ is neither open nor closed. A finite $T_0$
space is equivalent to a finite poset $P$ (Enumerative
Combinatorics, vol. 1, second ed., Exercise 3.3). An open set
corresponds to an order ideal of $P$. Thus we want to count posets on
an $n$-element set such ...

answered Oct 8 at 22:10

Richard Stanley

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18

If there are no constant symbols in the language, $\mathcal L$, then the nontrivial varieties where every algebra is free are exactly the varieties term equivalent to sets or to affine spaces over a division ring. Here is why.
If there are no constants in the language, $F_{\mathcal V}(\emptyset)$ is empty, so $F_{\mathcal V}(1)$ must be the 1-element ...

17

The answer to your question is "no", as explained by Anton Klyachko in his answer. Let me refer you to a remarkable statement of Razmyslov and Procesi that describes all identities. They proved (independently) that in fact all identities of $Mat_n(\mathbb{Q})$ follow, in a sense, from the Cayley--Hamilton theorem. To be more precise:
Let us first define the ...

answered Nov 22 '14 at 13:27

Vladimir Dotsenko

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16

Every $\sigma$-complete Boolean algebra is isomorphic to a quotient $\mathcal{M}/I$ where $(X,\mathcal{M})$ is a $\sigma$-algebra and $I$ is a $\sigma$-complete ideal (See the Handbook of Boolean Algebras for a proof or look here). Here by a $\sigma$-complete Boolean algebra, we mean a Boolean algebra where every countable subset has a least upper bound. A $\...

16

One thing you can do with an operad that you cannot do with a prop is write down a monad such that algebras over the monad correspond to algebras over the operad. For example, Hopf algebras have a prop but not an operad, and don't even have a monad: I believe the forgetful functor from Hopf algebras to vector spaces doesn't have a left adjoint, so cannot be ...

answered May 12 '16 at 20:28

Qiaochu Yuan

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16

Theorem: Given $A$ a boolean ring/boolean algebra then there is an equivalence of categories between the category of $A$-modules and the category of sheaves of $\mathbb{F}_2$-vector spaces on Spec $A$. The equivalence sends every sheaf $\mathcal{M}$ of $\mathbb{F}_2$-vector space to its space of section, $\Gamma(\mathcal{M})$ which is a module over $\Gamma(\...

15

McKenzie proved that it is undecidable whether a finite universal algebra has a finite basis of identities.

15

In characteristic $0$, if $Y$ is a Hopf subalgebra of $U(\mathfrak{g})$, then $Y = U(\mathfrak{g}')$ for some Lie subalgebra $\mathfrak{g}'$ of $\mathfrak{g}$ (and conversely, every such subalgebra is a Hopf subalgebra). This is a consequence of Proposition 6.13 and Theorem 5.18 of John W. Milnor and John C. Moore's 1965 paper "On the structure of Hopf ...

answered Dec 13 '16 at 2:36

Christopher Drupieski

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15

It was proved by Andreas Blass in Words, free algebras, and coequalizers that free infinitary algebras are not constructible neither in topoi nor in ZF. It is easy to see that the existence of free algebras for all theories is equivalent to the existence of initial algebras of all theories.
Even though initial algebras do not exist in "the basic ...

15

The answer is no. What you call a generalized cancellation rule is called a quasi-identity in universal algebra. Malcev proved in 1939 that there is no finite basis of quasi-identities defining group embeddable monoids or equivalently defining the quasi-variety of monoids generated by groups.
You can find details in Volume 2 of Clifford and Preston's ...

answered Dec 7 '20 at 15:38

Benjamin Steinberg

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14

The lattice of submodules of a module satisfies a stronger identity, namely the Arguesian law:
$$
(x_0\vee y_0)\wedge (x_1\vee y_1)\wedge (x_2\vee y_2)\leq ((z\vee x_1)\wedge x_0) \vee ((z\vee y_1)\wedge y_0).
$$
You might check the short paper
Day, Alan; Jónsson, Bjarni. The structure of non-Arguesian lattices. Bull. Amer. Math. Soc. (N.S.) 13 (1985), no....

13

A deep theorem of Oates and Powell shows that any finite group has a finite basis for its identities. One might think that the same is true for semigroups. But Perkins showed the 6-element semigroup consisting of the $2\times 2$ identity matrix, the zero matrix and the four matrix units $E_{ij}$ is not finitely based.
Mark Sapir, in a tour-de-force work ...

13

No.
Indeed, let $\mathcal{V}_n$ be the variety of magmas generated by the relating identities with variable $y$ saying that for every $k\le n$, all products of $k$ copies of $y$ are equal. Since the variety of power-associative magmas is $\bigcap_n \mathcal{V}_n$, a negative answer to the question is equivalent to showing that for every $n$ the relatively ...

13

OEIS A306445: 2, 4, 13, 74, 732, 12085, 319988, 13170652, 822378267, 76359798228, 10367879036456, 2029160621690295, 565446501943834078, 221972785233309046708, 121632215040070175606989, 92294021880898055590522262, 96307116899378725213365550192, 137362837456925278519331211455157, 266379254536998812281897840071155592
Number of collections of subsets of $\{1, 2,\...

12

Following George Bergman's recent preprint
http://arxiv.org/abs/1309.0564
(btw, he is very good at finding strange (counter-)examples in universal algebra!)
we recently found out that the universal group of the subsemigroup from the free monoid $\{a,b,c\}^*$ generated by $\{bc,abcabc,bcabca,bcabcabcabc\}$ is isomorphic to $(\mathbb{Z}\times\mathbb{Z})\ast\...

12

The answer is no.
Let $A$ be the algebra with universe $\{0,1\}$ and fundamental operations $f(x,y,z)=x+y+z \pmod{2}$ and $g(x)=x+1\pmod{2}$. Then $f$ and $g$ commute with each other and with themselves, so the variety generated by $A$ is commutative. This variety has a weird property: on every $B\in \mathcal V(A)$ the operation $g$ interprets either as a ...

ct.category-theory ra.rings-and-algebras semigroups-and-monoids monoidal-categories universal-algebra

12

You might be interested in the whole (mostly Russian) literature on "Banach-Kantorovich" or "lattice-normed" spaces, which are:
"a triplet $(\mathcal U,E,\lambda)$ consisting of a vector space $\mathcal U$, a Dedekind complete vector lattice $E$ and a map $\lambda:\mathcal U\to E_+$ satisfying some natural conditions that allow one to consider $\lambda$ ...

answered May 28 '14 at 0:41

Francois Ziegler

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12

What you need is the keyword "polyadic groups". A polyadic group is a non-empty set $G$ equipped with an associative $n$-ary operation $f:G^n\to G$ such that for all $a_1, \ldots, a_{n}$ and $b\in G$, the equations
$$f(a_1, \ldots, a_{i-1}, x, a_{i+1}, \ldots, a_n)=b, (1\leq i\leq n)$$
have (unique) solution for $x$. The simplest examples are polyadic ...

12

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 complete and modular. If $R$ is a principal ideal domain (see original MSE question/answer) or if $R$ is the ring of integers of a number field [3, page 135], ...

12

Lawvere theories can be thought of as "cartesian operads." That is, we have an analogy
$$\text{Lawvere theories} : \text{cartesian monoidal categories} :: \text{operads} : \text{symmetric monoidal categories}.$$
Consider the 2-category of symmetric monoidal cocomplete categories (where the monoidal structure distributes over colimits in both variables), ...

answered Jun 28 '18 at 13:51

Qiaochu Yuan

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12

An infinite Shelah semigroup must be a Jonsson semigroup (meaning that it is an infinite semigroup whose proper subsemigroups have lesser power). Therefore the following paper answers the question asked on this page:
McKenzie, Ralph
On semigroups whose proper subsemigroups have lesser power.
Algebra Universalis 1 (1971), no. 1, 21-25.
McKenzie proves ...

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