Equational identities of the $(+,*,0,1)$ reducts of commutative rings

Question

Consider the class of algebras of type $<2,2,0,0>$ which are the $(+,*,0,1)$ reducts of commutative rings. What is a finite basis for the identities of that class of algebras? I conjecture that

1. $x+0=x$
2. $x+y=y+x$
3. $(x+y)+z=x+(y+z)$
4. $x*1=x$
5. $x*y=y*x$
6. $(x*y)*z=x*(y*z)$
7. $x*(y+z)=(x*y)+(x*z)$
8. $x*0=0$

is sufficient. Also, if this list is indeed sufficient, is the last axiom $x*0=0$ redundant? I know most ring theory books prove it from the remaining axioms, but that is with the help of the subtraction operator. I think, in this case, it is not redundant.

Answer 1

Axioms 1-8 are a basis for this variety. This fact follows from 2 observations:

(1) Axioms 1-8 are sufficient to put any polynomial into normal form.
(2) If two polynomials $p, q$ written in normal form agree on $\mathbb Q$, then they are identical.

Let me include a few more details. Let $\mathcal A$ be the variety axiomatized by Axioms 1-8. Let $\mathcal C$ be the variety generated by the $(+,\cdot, 0, 1)$-reducts of commutative rings. Since commutative rings satisfy Axioms 1-8, we have ${\mathcal C}\subseteq {\mathcal A}$. The goal is to show equality. For this, it suffices to show that the $\omega$-generated free algebra $\mathbf F_{\mathcal A}(x_0,x_1,\ldots)$ lies in the subvariety $\mathcal C$. There is a natural surjective homomorphism $h\colon {\mathbf F}_{\mathcal A}(x_0,x_1,\ldots)\to {\mathbf F}_{\mathcal C}(x_0,x_1,\ldots)\colon x_i\mapsto x_i$, and I want to argue that $h$ is injective, and this will complete the argument.

Stage 1. Axioms 1-8 suffice to write each element $p\in {\mathbf F}_{\mathcal A}(x_0,x_1,\ldots)$ in normal form. Here, by the normal form of $p$, I mean ``$p = 0$'' or

p is a left-associated sum of monomials.

each monomial is either 1 or is a left-associated product of $x_i$'s.

within a given monomial that is a product of variables, the variables are ordered by increasing subscript.

within the sum of monomials, the monomials are ordered by increasing degree, and monomials of the same degree are ordered lexicographically by subscript.

Stage 2. To obtain a contradiction, assume that $h(p)=h(q)$ where the normal forms of $p$ and $q$ are different. Choose such a pair $(p,q)$ of this type such that the sum of the lengths of the normal forms, written $|p|+|q|$, is least. Argue that neither $p$ nor $q$ can be $0$, nor can they have a constant monomial $1$, nor can they share a monomial. (If they shared any monomial, then cancelling it from both yields a pair $(p',q')$ with the same properties and shorter total length.)

Stage 3. Now $p-q$ may be considered to be a polynomial with integer coefficients. It cannot be the zero polynomial, since $p$ and $q$ share no monomial, hence there is no cancellation when we form $p-q$. Thus, $p-q$ is a nonzero integer polynomial which vanishes on every commutative ring. To obtain a contradiction, argue by induction on the number of variables appearing that any polynomial that vanishes on $\mathbb Q$ is the zero polynomial.