Intuition behind the relationship between defining a homomorphism of commutative rings and finding two elements in the range that commute What is the intuition behind the following statement? 

Given a ring $R$, defining a homomorphism of rings from $\mathbb Z[x,y] \to R$ is equivalent to giving an ordered pair $(r_1,r_2)$ in $R \times R$ such that $r_1$ and $r_2$ commute. 

I know the forward direction is always true but what makes this true in this particular case in the backwards direction?
More specifically, given two commuting elements in RxR, what allows me to define a homomorphism of Rings?
 A: HINT: Suppose $f$ is a homomorphism from $\mathbb{Z}[x, y]$ to $R$. How many specific values of $f$ do I need to know, before I know all of $f$? (For instance, you don't need to tell me "$f(1)=1$," since that's guaranteed by the assumption that $f$ is a homomorphism; similarly, if I know $f(x-2)$, then I know $f(2x+3)$, etc.)
A: The polynomial ring is the free commutative $\mathbb Z$-algebra, generated by $x$ and $y$. Thus the only relation, that $x$ and $y$ satisfy, is commutativity: $xy-yx=0$.
For defining an homomorphism from such an algebra, it is sufficient and neccessary that the images of $x$ and $y$ also satisfy commutativity: $f(x)f(y)-f(y)f(x)=0$.
That is the whole point of a free object: Giving a morphism from an free object is the same as giving an element for each generator. If the generators satisfy some relations, we are little bit more restricted: the images have to satisfy the same relations.
On the other hand, you can just compute it: Give commuting $r_1,r_2 \in R$, we can just define $$f: \mathbb Z[x,y] \to R, f(x,y) \mapsto f(r_1,r_2)$$
Verify that this is an homomorphism and you will quickly realize, that commutativity of $r_1,r_2$ is neccessary and sufficient to do so.
