Characterizing the algebras on $\mathbb(Z)/2\mathbb(Z)$

92 Views Asked by At

Is there a characterization or discussion of which algebraic structures may be equipped to the set $\{0,1\}$? For example, $\{0,1\}$ admits a unique group structure, a unique ring structure, $2!=2$ unique orderings (but, of course, only $1$ up to isomorphism).

What about more exotic structures, such as Hopf Algebra, etc?

Perhaps this question is too open-ended for a definite answer (though in theory one ought to exist). In that case, a question which can be directly answered is: are there any published works that investigate general algebras on a finite set?

3

There are 3 best solutions below

0
On BEST ANSWER

Besides the obvious structures, which $\mathbb{Z}/2\mathbb{Z}$ may carry, like being a group or semigroup, a field, a commutative ring, a division ring, or an $R$-module, it is also a topological space, with four possible topologies (the trivial one, the discrete one and two others). But it is also an algebraic group, see here:

Finite groups are algebraic groups

In particular, it is an affine algebraic variety. It is also a projective variety. Equipped with the discrete topology it is a zero-dimensional Lie group. We can continue here.

There are many publications of algebras over $\mathbb{Z}/2\mathbb{Z}$, for example Lie algebras over $\mathbb{F}_2$. Here the classification of simple Lie algebras over $\mathbb{F}_2$ has not been achieved.

0
On

Although this doesn't go exactly in the directions you're asking for, perhaps you can be interested in knowing that there exists $16$ groupoids (aka, magmas) with $\{0,1\}$ as its universe, and $8$ of them are semigroups.

You can also define $2$ quasi-groups on that set.

Less interesting would be the cases of lattices, semi-lattices, Boolean algebras or Heyting algebras because in each of these cases, one would automatically consider $0<1$ and there would be only one of each.

1
On

If your definition of algebra consists of a set together with a collection of operations on that set, then yes there is a complete classification up to a certain equivalence. (Note that this will include groups, rings, modules, and lattices, but not ordered sets, topological spaces, algebraic varieties, or Hopf algebras.)

The classification is due to Post and you can read about it here.

To add some more detail, given an algebra $(A,\{f,g\})$ where $A$ is a nonempty set, $f:A^3\to A$, and $g:A^2\to A$. You can make a new algebra by adding the new operation $h:A^3\to A$ defined by $h(x,y,z)=f(x,g(y,z),y)$. Then the algebra $(A,\{f,g,h\})$ will have the same quotients and substructures as $(A,\{f,g\})$. If you continue adding all the operations you get by composing $f,g$, and projection maps, you get the clone generated by $\{f,g\}$. Two algebras $(A,F)$ and $(A,G)$ are term equivalent if the clone generated by $F$ is equal to the clone generated by $G$. This is similar to the idea behind the phrase "every abelian group is a $\mathbb{Z}$-module."

Post classified all of the clones on a two element set, so, in other words, he classified algebras on a two element up to term equivalence.