How to recognise an algebra form a ring?

Question

Suppose $\mathcal{V}$ be a variety of algebras of some signature $\Sigma =\{f_1 ,f_2\dots f_n\}$. Let $\mathfrak{A}=\{A, f_1, f_2,\dots ,f_n\}$ be an algebra in $\mathcal{V}$.

Sometimes we are able to define operations $t_+, t_*$ by means of $f_1, f_2,\dots,f_n$ such that $\mathfrak{A} $ form a ring.

Sometimes it is possible to define operations $t_+, t_*$ by means of $f_1, f_2,\dots,f_n$ in uniform way such that every algebra of $\mathcal{V}$ form a ring.

Well known example of such is the variety of Boolean algebras. An arbitrary Boolean algebra form Boolean ring with respect to meet $t_*(a,b)=a\wedge b$ and simmetric differene $t_+(a,b)=(a\wedge \neg b)\vee(\neg a \wedge b)$ operation terms.

Is it possible to recognise somehow that either the algebra from a variety or every algebra from variety from a ring (possibly non-commutative) with respect to some terms $o,i\in \mathfrak{A}$ (either for zero-element and unity, or without unity) and $t_{+},t_{\times}:\mathfrak{A}^2 \to \mathfrak{A} $ in signature $\Sigma$?

Answer 1

I think this question is hopelessly broad, but I will give some partial answers that may be helpful.

Sufficient conditions for an algebra to be term-equivalent to a ring

The only way I can see to do this would be algorithmically. To expand upon my comment above, if $\mathbf{A}=\langle A;f_1,\dots, f_n\rangle$ is a finite algebra with finite signature, you can list out all the binary term operations of the algebra, and then check to see if any two of them correspond to addition and multiplication. The Universal Algebra Calculator can list out the binary term operations, but you would have to figure out whether any two of them could be addition/multiplication yourself. Finally, if you did find addition and multiplication, you'd then have to check if $f_1,\dots,f_n$ could be obtained from addition, multiplication, (unary) negation, and (constants) 0 and 1.

Necessary conditions

If you have a good understanding of the congruences of the algebra, then here are some things you could check:

1. Congruence-permutability: if $\mathbf{A}$ is term-equivalent to a ring and $\theta,\psi\in\mathrm{Con}(\mathbf{A})$, then $\theta\circ\psi:=\{(x,z)\in A^2:\exists y\in A, (x,y)\in\theta,(y,z)\in\psi\}$ is equal to $\psi\circ\theta$. This also implies $\theta\vee\psi=\theta\circ\psi$.

2. Congruence modularity: if $\mathbf{A}$ is term-equivalent to a ring, then $\mathrm{Con}(\mathbf{A})$ is a modular lattice.

3. Congruence-uniformity: if $\mathbf{A}$ is term-equivalent to a ring, $a,b\in A$, and $\theta\in\mathrm{Con}(\mathbf{A})$, then there is a bijection between $a/\theta$ and $b/\theta$. In other words, each of the congruence classes of a congruence are the same size.

4. Congruence-regularity: if $\mathbf{A}$ is term-equivalent to a ring, $a\in A$, and $\theta,\psi\in\mathrm{Con}(\mathbf{A})$, then $a/\theta=a/\psi$ implies $\theta=\psi$. In other words, if two congruences have a congruence class in common, then they are the same congruence.

This list may seem helpful, but these four conditions can't even differentiate between groups and rings. If you are further willing to restrict to rings with identity, then I can add this to the list:

5. If $\mathbf{A}$ and $\mathbf{B}$ are both term-equivalent to rings with identity, then $\mathrm{Con}(\mathbf{A}\times\mathbf{B})$ is isomorphic to $\mathrm{Con}(\mathbf{A})\times\mathrm{Con}(\mathbf{B})$. This is definitely not true for groups.