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

*

*$x+0=x$

*$x+y=y+x$

*$(x+y)+z=x+(y+z)$

*$x*1=x$

*$x*y=y*x$

*$(x*y)*z=x*(y*z)$

*$x*(y+z)=(x*y)+(x*z)$

*$x*0=0$
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.
 A: 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.
