What is the big picture behind the quotient of a ring by its maximal ideal being a field? A ring $(X, +, \times, 0, 1)$ is an algebraic structure such that $(X, +, 0)$ is a commutative group, $(X, \times, 1)$ is a monoid, and $\times$ distributes over $+$. 
A sub-ring is a subset $Y  \subset X$ such that $(Y, +)$ is a subgroup, and $Y$ is closed under $\times$.
An ideal of a ring $X$ is a subset $Z \subset X$ such that $Z$ is closed under $+$, and for any $x \in X$ and $z \in Z$, we have $z \times x \in Z$ and $x \times z \in Z$.
$X$ is itself an ideal of $X$, and it is called the unit ideal. An ideal of $X$ which is not the unit ideal is called a proper ideal.
A proper ideal $M$ is a maximal ideal of $X$ if for any ideal $I$ in $X$ such that $M \subseteq I \subseteq X$, then $I = M$ or $I = X$. 
Let $-$ be the inverse of $+$. Define an equivalence relation $\sim$ where $x \sim y$ if and only if $x - y \in I$, where $I$ is an ideal of $X$. Let $X/I$ be the set of equivalence classes of $X$ under $\sim$.
Fact: Let $[0], [1]$ be the equivalence classes containing $0$ and $1$ respectively, and $\oplus, \otimes$ the generalizations of $+, \times$ over the equivalence classes in $X/I$. Then, $(X/I, \oplus, \otimes, [0], [1])$ is a ring. 
Fact: Let $M$ be the maximal ideal of $X$, and let $\sim$ be defined so that $x \sim y$ if and only if $x - y \in M$. Then $X/M$ is a field. That is $(X/M, \oplus, \otimes, [0], [1])$ is such that $(X/M, \oplus, [0])$ and $(X/M \setminus [0], \otimes, [1])$ are groups, and $\otimes$ distributes over $\oplus$.
Question: from a big picture/intuition perspective, why might one expect the quotient group of a ring by its maximal ideal to produce a field? What is special about the maximality of the maximal ideal, that $X/M$ is no longer just a ring, but also a field?
 A: (Caveat: Groups and rings are somewhat special, in that we can “code” quotients via “substructures with something extra”; for more general algebraic objects, such as monoids or semigroups, you need congruences. That said, I will cast my answer in terms of groups and rings.)
Consider quotients in general; the idea of a quotient/homomorphic image is that you can try to study aspects of your object via a quotient, which is (hopefully) simpler than your initial object. Homomorphic images are also important ways of studying an algebraic structure.
The objects for which this doesn’t work are the “simple objects”: objects whose only homomorphic images are the trivial object (everything is identified) and the object itself (the quotient doesn’t do anything). Understanding simple objects is important and interesting, in its own right, of course.
Now look at groups: you have abelian groups and you have nonabelian groups. For abelian groups, every subgroup yields a quotient, so to be an abelian simple group you must have no proper nontrivial subgroups. Those groups have a very basic structure: they must be cyclic of prime order. By contrast, the nonabelian simple groups are much more complicated, because while they can have no proper nontrivial normal subgroup, they can have lots of subgroups. The study of nonabelian simple groups turns out to be much more complicated, as witnessed by the length of time and the length of paper needed to classify them.
Now let’s look at rings; the role of normal subgroups is played by two-sided ideals. Again, we have two types of rings: commutative and noncommutative. 
A commutative ring is going to be simple if and only if it has no proper nontrivial ideals. This forces no proper one-sided ideals. What does that tell us about the structure? 
Given any ring $R$ and any element $a$, the set $aR=\{ar\mid r\in R\}$ forms a right ideal. If $a\neq 0$, and $R$ has a unity, then if $R$ has no proper one-sided ideals this means that $aR=R$, and in particular that $a$ has a right inverse, since $1\in aR$. Symmetrically, $Ra$ is a left ideal, and so $a$ has a left inverse; thus, in a simple commutative ring, every nonzero element must have an inverse. That is, we must be in a field. 
This observation is akin to the observation that in a simple abelian group, we must be in a cyclic group of prime order.
And just as in groups, if you drop commutativity then you end up with a lot more simple rings: for example, the ring of $n\times n$ matrices of a field is simple, even though it has one-sided ideals. While a commutative ring with no proper ideals must be a field, a ring with no proper nontrivial ideals need not be a division ring. (However, a ring with no proper nontrivial one-sided ideals must be a division ring, by the same argument as given above for commutativity).
So, for abelian groups, maximal subgroups must have prime index, by the isomorphism theorem: $A/M$ has no proper nontrivial subgroups, so it must be cyclic of prime order, hence $[A:M]$ is prime. For commutative rings, maximal ideals satisfy that $R/M$ has no proper nontrivial ideals, and therefore that $R/M$ is a field (because those are the only simple commutative rings with unity). 
