If $\mathbb{Z}$ satisfies an identity $\eta$, then every **commutative** ring satisfies $\eta$? And related questions. Assume all rings have unity and that ring homomorphisms preserve unity.
Now by general principles, if every free object in the category of rings satisfies an identity $\eta$, then every object in the category of rings satisfies $\eta.$ However, its not the case that if the free ring on the empty set of generators (namely $\mathbb{Z}$) satisfies an identity $\eta$, then every ring satisfies $\eta$. In particular, take $\eta$ to denote commutativity.
Question 0. Is it true that if $\mathbb{Z}$ satisfies an identity $\eta$, then every commutative ring satisfies $\eta$?
Question 1. Is there an accepted terminology for algebraic theories $T$ such that letting $Z$ denote the initial $T$-algebra, the following holds? "For all identities $\eta$ in the language of $T$, it holds that if $Z$ satisfies $\eta$, then every model of $T$ satisfies $\eta$."
Question 2. Is there a good characterization of such theories $T$?
 A: Question 0: Every identity in the language of rings is equivalent to one of the form $p(\overline{x}) = 0$, where $p(\overline{x})\in \mathbb{Z}[x_1,\dots,x_n]$ for some $n$. So the question amounts to: if a polynomial in $\mathbb{Z}[x_1,\dots,x_n]$ is $0$ at all points of $\mathbb{Z}^n$, is the zero polynomial? 
The answer is yes. There are fancy ways to see this, but one can prove it by a simple induction. It's clearly true for polynomials in $0$ variables (which are just integers). Now suppose $p(\overline{x})\in \mathbb{Z}[x_1,\dots,x_n]$, $n\geq 1$, and $p\neq 0$. Rewrite $p$ as a polynomial in $x_n$ with coefficients in $\mathbb{Z}[x_1,\dots,x_{n-1}]$:  $$p(x_1,\dots,x_n) = \sum_{i = 0}^d q_i(x_1,\dots,x_{n-1})x_n^i.$$
Since $p\neq 0$, $q_i\neq 0$ for some $i$. By induction, there are some $a_1,\dots,a_{n-1}$ in $\mathbb{Z}^{n-1}$ such that $q_i(a_1,\dots,a_{n-1}) \neq 0$. But then $p(a_1,\dots,a_{n-1},x_n)$ is a nonzero polynomial in one variable, $x_n$. A nonzero polynomial can have at most finitely many zeros in an integral domain, so there is some $a_n\in \mathbb{Z}$ such that $p(a_1,\dots,a_{n-1},a_n)\neq 0$, and we're done.
Questions 1 and 2: If we have an algebraic theory $T$ and a collection of models $\{A_i\}_{i\in I}$ such that whenever an identity holds in all $A_i$, it holds in every model of $T$, we say that $\{A_i\}_{i\in I}$ generate the variety of models of $T$. This is equivalent to saying that every model of $T$ is a quotient of a subalgebra of some product of $(A_i)$s, by Birkhoff's theorem. So you're asking about varieties of algebras which are generated by the initial algebra, which can be characterized as those for which every algebra is a quotient of a subalgebra of a power of the intial object. I'm not sure how satisfying this characterization is, though...
By the way, it's nice to observe what this representation amounts to in the case of commutative rings: Given an arbitrary ring $R$, we can choose a set of generators in order to write it as a quotient of some free commutative ring $F$. Now we want to embed $F$ in a power of $\mathbb{Z}$. We can do this as long as for each nonzero element $f\in F$, there is a map $\phi_f:F\rightarrow \mathbb{Z}$ with $\phi(f)\neq 0$. The product of all these maps $\phi_f$ will be an embedding $\phi:F\rightarrow \prod_{f\in F\setminus\{0\}}\mathbb{Z}$. So what's $\phi_f$? $f$ is a polynomial in finitely many variables, so as we saw above there's some tuple of elements from $\mathbb{Z}$ on which it's not $0$. Map $F$ to $\mathbb{Z}$ by sending the relevant variables to this tuple and all the others to $0$.
