Rings of functions. I'm trying to compile a list of examples of rings (not the usual $\mathbb{R}, \mathbb{Z}, \mathbb{Z/nZ}$ and so on) in order to develop my intuition.
Here are some examples and some questions about them. I'm using slightly nonstandard notation by denoting $C(R)$ as the set of all functions (continuous as well as discontinuous ones).

($a_1$) Let $C(\mathbb{R})$ denote the ring of real valued functions defined on $\mathbb{R}$, and Let $I_S \subset C(\mathbb{R})$ be the subring of all function which vanish on the subset $S \subset \mathbb{R}$. Clearly $I_S$ is an ideal since the operations on the functions are defined to be pointwise.
If $S_0 \subseteq S_1 \subseteq \cdots$ are a sequence of nested subsets of $\mathbb{R}$ then $I_{S_0} \subseteq I_{S_1} \subseteq \cdots$ is a sequence of nested ideals of $C(\mathbb{R})$. 
($a_1$?) Can i say safely that $C(\mathbb{R})$ is not noetherian? Is it right to identify the quotient ring $C(\mathbb{R})/I_S$ with the ring $C(S)$ of real valued defined on $S$?. ANSWERED
($a_2$) Let $C(\mathbb{R})$ be as before and let $I$ be denote the subring of functions which are eventually zero. Clearly $I$ is an ideal of $C(\mathbb{R})$. 
(We can think of the coset $f + I$ as the set of all functions that eventually coincide with $f$ and then by multiplying $(f+I)(g+I)= fg + I$ we get functions which eventually coincide with $fg$.)
($a_2?$) Is $I$ maximal in this case? What is a natural way of thinking about the quotient $C(\mathbb{R})/I$? ANSWERED

[$b$] Let $C(\mathbb{R})_B$ denote the ring of real valued bounded functions defined on $\mathbb{R}$. 
That is for every $f(x) \in C(\mathbb{R})_B$ there exists $M>0$ such that $|f(x)| \leq M$ for all $x$. $C(\mathbb{R})_B$ is a ring under pointwise addition and multiplication of functions. 
Let $I \subset C(\mathbb{R})_B$ denote the the subset of all functions $f(x)$ with $\lim_{x \to \pm \infty} f(x)=0$. 
$I$ is an ideal (any bounded function times a function that vanishes at infinity vanishes too). 
[$b?$] Is $I$ maximal in this case? How should i think about the quotient ring $C(\mathbb{R})_B/I$? 
My thought is that if $C(\mathbb{R})_B$ contained only function with well defined limits at $\pm \infty$ then $C(\mathbb{R})_B/I$ was isomorphic to $\mathbb{R}^2$ yet this is not the case so I'm not sure at all what to make of this.

I have some more examples which i would like to verify but it's getting to be a long question so I'll end it here.
 A: Please see $\S$ 5.2 of my commutative algebra notes for some material on rings of continuous functions.  (For a much more serious treatment, see the text Rings of Continuous Functions by Gillman and Jerison.)
In particular, Exercise 5.16 outlines a proof of a result that implies that $C(\mathbb{R})$ is not Noetherian, and the following exercise asks you to show that if for a normal topological space $X$ the ring $C(X)$ is Noetherian, then $X$ is finite.
As for your particular questions: well, you've asked rather a lot of them.  In order to get more complete answers, you may want to break them up into several math.SE questions and say a little about your thoughts on each.  But here are some comments:
Regarding a1): you can say "$C(R)$ is not Noetherian" and you'll have said something true, but I don't think that what you've suggested establishes that fact.  First of all the correspondence $S \mapsto I_S$ is inclusion-reversing [by a weird coincidence I am currently typing up a paper on another inclusion-reversing correspondence between zero sets and functions, a generalization of Alon's Combinatorial Nullstellenatz!], so you want a descending chain of subsets.  Moreover you need your chain of ideals to strictly ascending -- i.e., no equality -- and for that it is not enough to choose your chain of sets to be strictly descending, because if $S' \subset S \subset \overline{S'}$ then $I_S = I_{S'}$.  But you can fix this...
Can you identify $C(\mathbb{R})/I_S$ with $C(S)$?  You are asking whether every continuous function on $S$ extends to a continuous function on $\mathbb{R}$.  And the answer is "not necessarily", but there is a simple necessary and sufficient condition on $S$ for this.
Regarding a2): yes, the set $I$ of functions with compact support is an ideal of $\mathbb{R}$.  One can think of $C(\mathbb{R})/I$ as equivalence classes of continuous functions which are eventually equal.  This is clearly still quite a large ring.  In particular I claim that the image of the sine function in this ring is neither zero nor a unit, so the ring is not a field and $I$ is not maximal.
Added: It has been pointed out that the OP neither said nor intended that the functions be continuous (I was misled by the notation $C(\mathbb{R})$, which until now in my experience has always meant "continuous functions".)  This simplifies things and makes what the OP wrote much more correct: to show that $C(\mathbb{R})$ is not Noetherian one can use any strictly descending chain of subsets, and it is then always true that $C(\mathbb{R})/I_S = C(S)$.  However, from the perspective of my answer I would say that this amounts to putting the discrete topology on $\mathbb{R}$, which is a case to consider, but there is another even more interesting topology to consider as well...
A: I wouldn't use $C(\mathbb{R})$ to denote the set of all functions $\mathbb{R}\to\mathbb{R}$, so I'll denote the ring by $F(\mathbb{R})$.
This ring is just $\mathbb{R}^{\mathbb{R}}$ which is clearly not noetherian; consider the sets
$$
\mathbb{N}\supset\mathbb{N}\setminus\{0\}
\supset\mathbb{N}\setminus\{0,1\}
\supset\dots\supset\mathbb{N}\setminus\{0,\dots,n\}\supset\dotsb
$$
and clearly you have
$$
I_{\mathbb{N}}\subset I_{\mathbb{N}\setminus\{0\}}
\subset I_{\mathbb{N}\setminus\{0,1\}}\subset\dotsb
$$
(all proper inclusions). Yes, the correspondence $S\mapsto I_S$ is order reversing, as others have pointed out. Note that this easily extends to any function ring: if $R$ is a (commutative) ring and $X$ is an infinite set, then the ring $R^X$ is not noetherian.
Can you identify $F(S)$ with $F(\mathbb{R})/I_S$? Certainly so, because any map from $S$ to $\mathbb{R}$ can be extended to $\mathbb{R}$ by zero. This is nothing more than the obvious isomorphism
$$
R^X\cong R^S\times R^{X\setminus S}.
$$
The ring $F_b(\mathbb{R})$ of bounded functions is a bit more complicated, but not too much. If $I$ denotes the set of functions in $F_b(\mathbb{R})$ that have limit $0$ at $\pm\infty$, then it is not true that this ideal is maximal. If $f(x)=\sin x$, then $f$ is not invertible modulo $I$. This would mean there exists a bounded function $g$ such that $fg-\mathbf{1}\in I$ ($\mathbf{1}$ is the constant function taking only the value $1$), that is,
$$
\lim_{x\to\infty}(g(x)\sin x-1)=0
$$
or
$$
\lim_{x\to\infty}g(x)\sin x=1
$$
which is impossible.
