Polynomial Rings "Most Efficient" Ring? I'm currently reading though Aluffi's Algebra 0, and within III.2.2 Aluffi shows how  $\mathbb{Z}[x_1,\dots,x_n]$ satisfies a universal property like which free groups satisfy for groups. Specifically, given the category $\mathscr{R}_A$, where $A$ is a fixed set. The objects are pairs $(j,R)$ with $j:A \rightarrow R$ being set-functions from $A$ to some commutative ring $R$. The morphisms 
\begin{equation*}
(j_1,R) \rightarrow (j_2,R)
\end{equation*}
are commutative diagrams such that $\varphi: R_1 \rightarrow R_2$ is a ring homomorphism.
Forgive me if this is obvious, but Proposition 2.1 states that

Proposition III.2.1: $(i, \mathbb{Z}[x_1,\dots,x_n])$ is initial in $\mathscr{R}_A$.

This is compared to free groups which are initial within a similar category for groups. Aluffi also states that free groups are the "most efficient" construction of a group using a fixed set $A$. This brings me to my question (essentially just verification of my intuition):
Are these Polynomial Rings the "most efficient" construction of a commutative ring given some fixed set $A$?
More generally, should I expect universal properties that are analogous to the universal property of free groups to denote the "most efficient" construction of other objects in different categories given some fixed set (or other thing)?
 A: When someone speaks of a free object $c$ in a category $\textbf C$ generated by a set $S$ they mean that there exists an obvious functor $F:\textbf C \rightarrow \textbf{Set}$ which "forgets" the structure on objects of $\textbf C$ and gives you the underlying set. In most cases such a functor has a left adjoint $L: \textbf{Set} \rightarrow \textbf C$. Then $c$ being freely generated by $S$ is taken to mean that $LS =c$. 
$L$ is a left adjoint so preserves colimits, then it must be the case that since $S = \sqcup_{s \in S} \{s\}$, $LS = \sqcup_{s \in S} L\{s\}$. So if you figure out what $LX$ is where X is a one point set is you have computed $L$ for all sets. We can do this pretty easily in most cases, for example in the category of groups when 
$$F :Grp \rightarrow \textbf{Set}$$
is the functor which takes a group $G$ to the set underlying $G$.
We can see that $LX = \mathbb Z$ since group homomorphisms $\mathbb Z \rightarrow G$ are in bijection with elements of $G$. I.e $$Hom(LX,G) = FG = Hom(X,FG)$$ 
since $X$ is a one element set. So the free group generated by a one element set is $\mathbb Z$. To do this in the category of rings we note that homomorphisms $\mathbb Z[x] \rightarrow R$ are in bijection with elements of $R$ so the free ring generated by a one element set is the polynomial ring over $\mathbb Z$ in one variable by the same argument as the one for groups.
