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Results tagged with universal-algebra
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user 4600
The study of algebraic structures and properties applying to large classes of such structures. For example, ideas from group theory and ring theory are extended and considered for structures with other signatures (systems of basic or fundamental operations).
9
votes
What kind of algebra is the class of ordered pairs equipped with the binary operation which ...
The property (*) is actually equivalent to a set of quasi-identities:
$$(x,y)=(x',y')\rightarrow x=x'$$
$$(x,y)=(x',y')\rightarrow y=y'$$
The converse implication you had ($\leftarrow$) is logically v …
- 24.8k
7
votes
A decision problem for clones
Every finitely generated clone on a finite set is computable.
Indeed, fix $k$. If we want to determine which $k$-ary functions belong to the clone $\mathcal C$, we can start generating functions by c …
- 24.8k
2
votes
Axiomatizing orientation in the complex plane
The ternary relation can be expressed in terms of the triple scalar product:
$$
\mathbf x\cdot (\mathbf y\times \mathbf 1)>0
$$
where $\mathbf x=(x_1,x_2,x_3)$, $\mathbf y=(y_1,y_2,y_3)$, and $\mathbf …
- 24.8k
2
votes
What do you call a lattice whose meet operation preserves disjointness of subsets?
Many lattices do have this property (for example completely distributive, or 5-element nondistributive). Here is a small counterexample.
Let $L$ be the 9-element lattice
$$
L=\{s_1,t_1,s^*,t^*,s_1\we …
- 24.8k
18
votes
4
answers
2k
views
Complete Boolean algebra not isomorphic to a $\sigma$-algebra
Does there exist a complete Boolean algebra that is not isomorphic to any $\sigma$-algebra? If so, what is an easy or canonical example or construction?
- 24.8k
3
votes
What is the name for Boolean algebra's version of $\models$ between sets of identities and i...
It seems the name for this idea is equationally complete theory, see page 30 of Walter Taylor's Equational Logic survey.
Not every theory is like that:
for example in the theory of lattices, which is …
- 24.8k
9
votes
Accepted
Is the equational theory of groups axiomatized by the associative law?
Yes. It suffices to show that any free semigroup embeds in a group.
For this I refer you to MO question 3235:
Let $F$ be a free semigroup (say, $2$-generated) which is embedded in a group $G$, an …
- 24.8k