The first Noether isomorphism theorem I'm studying for my Intro to Algebra class. I've reached the point where I want to understand the first Noether isomorphism theorem.
Here is the definition I was given in class:
Let $f :R \rightarrow S$ be a surjective ring homomorphism. Then we have a "commutative diagram" of ring homomorphisms. 
Here is the scanned diagram I received: Could someone please explain this diagram to me in basic terms? In addition, I'd deeply appreciate examples or proof of examples.

Ultimately, can someone answer what it means to say something is equivalent to another thing "up to isomorphism?"
 A: The first thing in the diagram are the solid arrows: those are maps that you "know."
You "know" the map $f\colon R\to S$, because you are assuming that you are given this map.
You also "know" the map $\pi\colon R\to R/\mathrm{ker}(f)$. This is simply the canonical projection onto the quotient. Remember that whenever you have a ring $R$ and an ideal $\mathcal{I}$ of $R$, you can form the quotient ring $R/\mathcal{I}$; and there is a canonical projection map $\pi\colon R\to R/\mathcal{I}$ given by $\pi(r) = r+\mathcal{I}$. In this case, since $f$ is a homomorphism, you know that $\mathrm{ker}(f)$ is an ideal, so you automatically get the quotient ring $R/\mathrm{ker}(f)$ and the quotient map $\pi\colon R\to R/\mathrm{ker}(f)$.
The second thing is that the "dashed arrow" represents a homomorphism which the theorem asserts exists; so one thing the theorem asserts is that there is a ring homomorphism $R/\mathrm{ker}(f)\to S$, and which is an isomorphism (assuming $f$ is onto).
Finally, saying the diagram "is commutative" means the following: whenever you have two paths from one ring in the diagram to another (in this case, there are two ways to get from $R$ to $S$: either through the arrow $f$, or by first going to $R/\mathrm{ker}(f)$ via the map I called $\pi$, and the map $R/\mathrm{ker}(f)\to S$ that we are asserting exists), then the results you get are the same no matter which path you take (in this case: if $r\in R$, then $f(r)$ will be the same as first taking $\pi(r)$, and then applying the map $R/\mathrm{ker}(f)\to S$ to $\pi(r)$.
For an example, take $R=\mathbb{Z}\times\mathbb{Z}$, $S=(\mathbb{Z}/10\mathbb{Z})\times(\mathbb{Z}/3\mathbb{Z})$, and $f(a,b) = (a\bmod 10, b\bmod 3)$.
This is a ring homomorphism:
\begin{align*}
f\bigl((a,b) + (c,d)\bigr) &= f(a+c,b+d) = \bigl( a+c\bmod 10, b+d\bmod 3\bigr)\\
&= (a\bmod 10,b\bmod 3) + (c\bmod 10, d\bmod 3) = f(a,b) + f(c,d),\\
f\bigl((a,b)(c,d)\bigr) &= f(ac,bd) = \bigl(ac\bmod 10, bd\bmod 3\bigr)\\
&= (a\bmod 10, b\bmod 3)(c\bmod 10,d\bmod 3) = f(a,b)f(c,d).
\end{align*}
It is also onto: given $(x\bmod 10,y\bmod 3)\in S$, we can always find an element $(a,b)\in R$ such that $f(a,b)=(x\bmod 10, y\bmod 3)$. For instance, we can take $a=x$, $b=y$. 
what is the kernel of this map? $f(a,b) = (0\bmod 10,0\bmod 3)$ if and only if $a\equiv 0 \pmod{10}$ and $b\equiv 0 \pmod{3}$. So the kernel is $10\mathbb{Z}\times 3\mathbb{Z}$. 
Now, we have a map from $R$ to $R/\mathrm{ker}(f)$; namely,
\begin{align*}
\pi\colon \mathbb{Z}\times\mathbb{Z} &\longmapsto \frac{\mathbb{Z}\times\mathbb{Z}}{10\mathbb{Z}\times 3\mathbb{Z}}\\
(a,b) &\longmapsto (a,b) + 10\mathbb{Z}\times 3\mathbb{Z}
\end{align*}
What the theorem states is that there is an isomorphism
$$\varphi\colon \frac{\mathbb{Z}\times\mathbb{Z}}{10\mathbb{Z}\times3\mathbb{Z}} \to \left(\frac{\mathbb{Z}}{10\mathbb{Z}}\right)\times\left(\frac{\mathbb{Z}}{3\mathbb{Z}}\right)$$
which makes "the diagram commute". That is, for every $(a,b)\in R$, we will have
$$f(a,b) = \phi(\pi(a,b)).$$
What is $\phi$? 
$$\phi\bigl((a,b) + 10\mathbb{Z}\times 3\mathbb{Z}\bigr) = (a\bmod 10,b\bmod 3).$$
Does this work? First, $\phi$ is well defined: if $(a,b)+10\mathbb{Z}\times 3\mathbb{Z} = (c,d)+10\mathbb{Z}\times3\mathbb{Z}$, then $(a,b)-(c,d)\in 10\mathbb{Z}\times 3\mathbb{Z}$; this means that $a-c\in 10\mathbb{Z}$ and $b-d\in 3\mathbb{Z}$, so $a\bmod 10 = c\bmod 10$ and $b\bmod 3$ = $d\bmod 3$. Thererfore, $(a\bmod 10,b\bmod 3) = (c\bmod 10, d\bmod 3)$. Thus,
$$\phi\bigl( (a,b)+10\mathbb{Z}\times 3\mathbb{Z}\bigr) = \phi\bigl( (c,d)+10\mathbb{Z}\times 3\mathbb{Z}\bigr),$$
which shows that $\phi$ is well defined.
It is also straightforward to check that $\phi$ is a ring homomorphism, and that $f(a,b) = \phi(\pi(a,b))$ for all $(a,b)\in R$. To show that $\phi$ is an isomorphism, note that it must be a surjection (because $\phi\circ\pi = f$ is a surjection, and if a composition is surjective, then the second map is surjective), and if 
$$(0\bmod 10,0\bmod 3) = \phi\bigl( (a,b) + 10\mathbb{Z}\times 3\mathbb{Z}\bigr) = (a\bmod 10, b\bmod 3),$$
then $a\equiv 0\pmod{10}$ and $b\equiv 0 \pmod{3}$, so $(a,b)\in 10\mathbb{Z}\times 3\mathbb{Z}$; this means that
$$(a,b)+ 10\mathbb{Z}\times 3\mathbb{Z} = (0,0)+10\mathbb{Z}\times3\mathbb{Z},$$
so the map is one-to-one. Thus, $\phi$ is an isomorphism, as desired.
Also, $\phi$ is the only homomorphism that "fits" into the diagram: because if $\psi$ also "fits", then given 
$(a,b)+10\mathbb{Z}\times 3\mathbb{Z}$, we can write this as $\pi(a,b)$, so
$$\psi\bigl((a,b)+10\mathbb{Z}\times 3\mathbb{Z}\bigr) = \psi(\pi(a,b)) = f(a,b) = \phi(\pi(a,b)) = \phi\bigl( (a,b)+10\mathbb{Z}\times 3\mathbb{Z}\bigr),$$
so $\psi=\phi$. 
"Up to isomorphism." What about your final question? Consider the integers "in English" and the integers "in Spanish" ("los enteros"). The names of the elements are different ("one", "two", etc., vs. "uno", "dos", etc.) But "the integers" and "los enteros" are essentially the same ring: it's just a question of what we call the elements, not how we add them or multiply them.
Explicitly, we have a "translation" betwen "the integers" and "los enteros." This translation is such that if you take two integers, add them, and then translate the answer, you get the same thing as if you first translate the integer and then add them in Spanish. And the same thing with multiplication. Even though "the integers" and "los enteros" are technically different things (as sets), if all we are concerned about is the ring-theoretic properties of these rings, then they are essentially "the same ring." We don't really care whether we call the unit element "one" or "uno", the important point is that it is a unit, that if you add it to itself you will never get the zero element ("zero" or "cero", depending on which ring you are in) etc. The translation from English to Spanish and the translation from Spanish to English lets us go back and forth between the two at will, and none of the ring-theoretic properties will depend on which language we are using, but only on the ring-theoretic properties of the elements.
So, for all ring theoretic mathematical purposes, "the integers" are the same as "los enteros": because we have a perfect translation between the two rings, namely, an isomorphism. So, the two are really "equivalent" (for any ring-theoretic investigation you might want to undertake).
This means that there is really no point in distinguishing between the integers and los enteros. They are "equivalent rings", because they is an isomorphism between them. We say this by saying that they are "equivalent up to isomorphism."
(More formally: if you imagine the collection of all rings (okay, there are foundational problems with that, but ignore them; they are easy to get around), then the relation $R\cong S$ given by "there is a ring isomorphism between $R$ and $S$" is an equivalence relation: it is reflexive (the identity map shows $R\cong R$); symmetric (if $R\cong S$, then there is an isomorphism $f\colon R\to S$, so $f^{-1}\colon S\to R$ is an isomorphism from $S$ to $R$, proving $S\cong R$); and transitive: if $R\cong S$ and $S\cong T$, then there exist $f\colon R\to S$ and $g\colon S\to T$ that are isomorphisms, and $g\circ f\colon R\to T$ is an isomorphism, showing $R\cong T$. So we have an equivalence relation between rings, which partitions the collection of all rings into equivalence classes. When two rings are in the same equivalence class, we say they are "equivalent up to isomorphism", because the equivalence relation is "there is an isomorphism between".)
A: The rank nullity theorem in linear algebra is an example of this theorem. In other words, two vector spaces are isomorphic if and only if they have the same dimension. So we have $$\text{dim}(R/ \text{ker} \ f) = \text{dim}(R) - \text{dim}(\text{ker} \ f) = \text{dim}(\text{Im} \ f)$$ or $$\text{dim} \ R = \text{dim}(\text{ker} \ f)+ \text{dim}(\text{Im} \ f)$$
The phrase "up to isomorphism" just means that we consider isomorphic structures to be equivalent. So $\mathbb{Z}_3$ and $\mathbb{Z}/3 \mathbb{Z}$ are equivalent up to isomorphism. 
A: The first isomorphism theorem relates two rings $A$ and $B$ which have a surjective homomorphism from $A$ to $B$, meaning that every element of $B$ is the image of an element of $A$. In order to understand the theorem, it is crucial to understand what an isomorphism is. 
An isomorphism is a surjective homomorphism between two rings which can be inverted to produce another surjective homomorphism. What is important about an isomorphism is that, for any algebraic purposes, if the rings $A$ and $B$ have an isomorphism $f$ from $A$ to $B$, $A$ and $B$ differ only in notation. This is because if something is true for $A$, then the same thing can be verified for $B$ by taking $f^{-1}(b)$ for each element of $B$, performing computations in $A$, and then returning to ring $B$ by taking $f(f^{-1}(b)) = b$. An example of two isomorphic rings are $\mathbb{Z}_2 = {0,1}$, the integers mod 2 with addition and multiplication defined by $0+0 = 0, 0+1 = 1+0 = 1, 1+1=0, 0*0 = 1*0 = 0*1 = 0$ and $1*1 = 1$, and $R = {a,b}$ with addition and multiplication defined by $a+a = a, a+b=b+a=b, b+b=a, a*a = b*a = a*b = a$ and $b*b = b$. Clearly, these rings are essentially the same, we've just replaced "0" with "a" and "1" with "b", and thus there is an isomorphism $f$ from $\mathbb{Z}_2$ to $R$ given by $f(0) = a$ and $f(1) = b$ (you should verify for yourself that this is an isomorphism. It is because isomorphisms preserve the important properties of a ring that we say rings are "equivalent up to isomorphism" to mean that they are algebraically identical.
Another concept we need is the quotient ring $R/ \text{ker}f$. This ring is formed by taking the ring $R$ and considering the elements $a$ and $b$ "equivalent" if $f(a) = f(b)$. For example, if the function $f$ is a homomorphism from the ring of integers to $\mathbb{Z}_2$ defined by $f(x) = 0$ if $x$ is even and $f(x) = 1$ if $x$ is odd, then it is easy to see that the ring $R/ \text{ker}f$ consists of the set of even integers (all of which are equivalent to zero) and the set of odd integers (all of which are equivalent to one) and thus this is a ring with two elements (which is isomorphic to $\mathbb{Z}_2$).
Hopefully I did a good enough job explaining these concepts. The theorem says that $S$ and $R/ \text{ker}f$ are isomorphic - that is, essentially the same in the sense I described above.
A: To say that a diagram of objects and morphisms is commutative (or that the diagram commutes) is to say that for any two objects of the diagram, the morphism obtained by following a path between the two objects in the direction of the arrows is independent of the choice of the path.
In your diagram, we have to paths from $R$ to $S$: 1) the arrow (ring morphism) $f$, and 2) the path given by the composition of the arrow (ring morphism) from $R$ to $R/\ker f$ (let's call it $h$) and the arrow (ring morphism) from $R/\ker f$ to $S$ (let's call it $g$). In this case, to say that your diagram is commutative means that $f=gh$.
Basically, the theorem says that any ring epimorphism $f:r\rightarrow S$ can be factored as $gh$, where $h:R\rightarrow R/\ker f$ is epi and $g:R/\ker f\rightarrow S$ is iso (in many textbooks this last part is refereed to as the first isomorphism theorem).
Let's take a look at (a sketch of) the proof: since $f:R\rightarrow S$ is an (epi)morphism, its kernel $\ker f$ is a two-sided ideal of $R$, so we immediately have the quotient ring $R/\ker f$ and the canonical epimorphism from a ring to one of its quotients (in the general case, if $I$ is a two-sided ideal of a ring $R$ then the canonical epimorphism $\pi:R\rightarrow R/I$ is naturally defined by $\pi(r):=r+I$, for all $t\in R$). In our specific case, we have that $h:R\rightarrow R/\ker f$ is defined by $h(r)=r+\ker f$, for all $r\in R$.
Now we must construct an isomorphism $g:R/\ker f\rightarrow S$. For this, we simply define (again, this definition is natural) $g(r+\ker f):=f(r)$, for all $r+\ker f\in R/\ker f$. It's easy to show that $g$, as we defined it, is in fact a ring isomorphism.
Finally, from our definitions, it follows that for all $r\in R$ we have $gh(r)=g\big(h(r)\big)=g(r+\ker f)=f(r)$ so $f=gh$.
