How to "Visualize" Ring Homomorphisms/Isomorphisms? I understand that the formal definition of a ring homomorphism is some function $f$ that maps $R$ to $S$  ($f: R \rightarrow S$) s.t. $f(ab) = f(a)f(b)$ and $f(a + b) = f(a) + f(b)$. But how do I "visualize" this? I guess I'm asking for a "dumbed down" explanation.
For instance, I came across this problem:
Describe all ring homomorphisms of $\mathbb{Z} \times \mathbb{Z}$ into $\mathbb{Z}$. 
How do I go about "visualizing" problems like this? Really, any help would be great. Thank you!
 A: This is a difficult question to satisfactorily answer.
I don't often try visualizing it, but one can imagine the multiplication and addition tables being "transported" from one ring to another.  How "perfectly" this transport occurs depends on the size of the kernel (or, put in an equivalent way, how "injective" the map is).  If the map is perfectly injective, then the kernel is trivial, and there will exist a literal copy of the multiplication table of the first ring inside the multiplication table of the second ring (and likewise for addition).  In such an event, the homomorphism is an isomorphism onto its image.  If the map isn't injective (with the size of the kernel revealing how badly there is "overlap" onto target elements), then the operation tables of the image of the map will be the identical to the original operation tables, except that some columns and rows will suddenly be considered "the same" (modulo the elements in the kernel)**.
For example, there is the inclusion homomorphism $\phi: \mathbb{Z} \hookrightarrow \mathbb{R}$ that is injective (with trivial kernel), and indeed you can see that there is a perfect copy of the operation tables for $\mathbb{Z}$ inside of the operation tables for $\mathbb{R}$.  On the other hand, there is a homomorphism $\psi: \mathbb{Z} \rightarrow \mathbb{Z}_n$ defined by $x \mapsto x \pmod{n}$ that is not injective (with $\ker(\psi)$ consisting of the multiples of $n$).  If we wanted to visualize the operation tables in $\mathbb{Z}_n$, this is telling us to start with the operation tables in $\mathbb{Z}$ and suddenly regard the $k^{\text{th}}$ column / row as the same as the $(k+mn)^{\text{th}}$ column / row for every multiple $m$ of $n$ (and every $k$).
Finally, a "worst case scenario" for injectivity is given by the trivial homomorphism $\phi:R \rightarrow S$ defined such that $x \mapsto 0_S$ for all $x \in R$.  For this map, the kernel is the entire ring, and every column / row in the original operation table has been made "equivalent".
An idea we're touching on here is expressed in the first isomorphism theorem: the image of any homomorphism can be thought of as a quotient of the original ring in which we mod out by the kernel.  Visually, one can imagine using $|\ker(\phi)|$ different colors to highlight equivalent elements (and their corresponding rows/columns) wherein every element/row/column colored $\color{red}{ \textbf{red}}$ (or what have you) is regarded as the same.  It's worth mentioning that the quotients appear throughout mathematics, as in group theory, set theory, topology, etc.
Moral of the story:  If you know how to add and multiply in a ring $R$, then you will automatically know how to add and multiply in $\phi(R)$ for any explicitly-defined homomorphism $\phi$.

Finding homomorphisms $\phi: \mathbb{Z} \times \mathbb{Z} \rightarrow \mathbb{Z}$.
Notice that we can write every element in $\mathbb{Z} \times \mathbb{Z}$ as a sum of some number of copies of $(0, 1)$ together with some number of copies of $(1, 0)$.  Therefore, $\phi((1, 0))$ and $\phi((0, 1))$ determine the entire binary operation table for the target ring.
What are the possibilities for the images of these?  Well, notice that $(1,0) \cdot (1,0) = (1, 0)$, so we must have $\phi \Big( (1, 0) \cdot (1, 0) \Big) = \phi \Big( (1, 0) \Big)$, and now we can take advantage of the multiplicativity of $\phi$: the left hand side of this becomes $\phi \Big( (1, 0) \Big) \cdot \phi \Big( (1, 0) \Big)$.  Thus, $\phi \Big( (1, 0) \Big)$ must satisfy $x^2 = x$ in $\mathbb{Z}$.  Really narrows down the possibilities, huh?  The same argument holds for $\phi \Big( (0, 1) \Big)$.
Since the multiplicative identity is preserved under homomorphisms, $\phi \Big( (1, 1) \Big) = 1$, and you can use this together with the above to prove that exactly one of $(0, 1)$ or $(1, 0)$ maps to $0$.
Since each homomorphism is determined by the image of these two elements, you now have everything to explicitly describe every possible homomorphism.



**To be perfectly clear, when the rows/columns are regarded as equivalent, this is true also of the elements that label them.  As an example, with the homomorphism $\mathbb{Z} \rightarrow \mathbb{Z}_5$ where $x \mapsto x \pmod{5}$, we have $2, 7, 12$, and so forth suddenly being regarded as "the same" in $\mathbb{Z}_5$, and by extension, their rows/columns in the operation table for $\mathbb{Z}$.
