About proving that maximal ideals are prime Let $R$ be a Ring with unity, and $I$ a maximal ideal in $R$. Show that $I$ is a prime ideal.
I have seen this proof in many places
If $ab\in I$ and $a\notin I$, then $I+(a)=R$ and hence there is some $r\in R$ and $i\in I$ such that $i+ra=1$. Multiplying by $b$,
$$ib+rab=b.$$
$I$ is an ideal and $i\in I$, so that $ib\in I$, and we also have $ab\in I$ by assumption. Therefore, we can conclude that  $ib+rab=b\in I$.
The 1'st question I have is why does $a\notin I$, imply $I+(a)=R$ or more specifically, why does $a\notin I$ imply that $I$ is properly contained in $I+a$? 
My 2nd question is, where do we ever use that R is commutative?
Any comments would be greatly appreciated. 
 A: In answer (at least partially) to your second question, maximal ideals are prime in non-commutative rings also (see here for example).
However, the definition of prime must be changed, since $ab \in P$ does not imply $ba \in P$ for non-commutative rings.
In non-commutative rings we define say that a (left) ideal $P$ is prime if whenever any two ideals $I$ and $J$ have a product contained in $P$ (i.e. $IJ \subseteq P$) then either $I \subset P$ or $J \subset P$.
This is equivalent to the normal definition when $R$ is commutative. There are equivalent definitions that probably mean your proof would only require minimal modification, but this is why the proof doesn't work off the bat in the non-commutative case.
Though it has been answered in the comments, I will add the answer to your first question too, for completeness.
If $a \notin I$ then $I + (a)$ properly contains $I$ since 


*

*Clearly if $x \in I$ then $x+0 \in I + (a)$, so $I$ is contained

*Clearly $a = 0 + a \in I + (a)$, but we have already said that $a \notin I$, hence the containment is proper.

A: As explained in comments, there was confusion between ideal sums and cosets. Below I explain how the proof is simply an ideal-theoretic generalization of one classical proof of Euclid's Lemma, which may lend further insight. To clarify the analogy I will use list notation for ideals, so e.g. $\,(I,J,r) = I+J+ rR.\,$ The first line below presents the classical elementwise proof, where the notation $\,(m,n)\,$ can be read as either $\,d = \gcd(m,n),\,$  or as the principal ideal $\,(m,n) = (d)\subseteq \Bbb Z.\,$ The second line gives the analogous proof using ideals, using said list notation for ideal sums.
$\qquad\qquad\,(p,a)=1,\ \ \ p\ \mid\, \ ab\ \ \Rightarrow\, \ p\ \mid\,\, pb,\ \ ab\ \ \Rightarrow\ \ p\, \mid\  \,(pb,ab)\, =\, (p,a)b\, =\, b$
$\qquad\qquad(P,a)=1,\ P\supseteq (ab)\,\Rightarrow\, P\supseteq Pb,(ab)\,\Rightarrow\,P\supseteq(Pb,ab) = (P,a)b = (b)$
Similarly, many ideal-theoretic results are generalizations of well-known results about elements, principal ideals, or gcds. To gain better intuition on such abstractions it is essential to proceed as above, going back-and-forth between the abstract generalizations and the concrete instances. Doing so helps to effectively and efficiently transfer to the abstract generalizations our well-honed intuiton from prior experience with the concrete objects. 
