TITLE: $I$ is maximal $\implies I$ is prime
QUESTION [6 upvotes]: Been asked to show this is true with hints $R/I$ field $\Longleftrightarrow$  $I$ is maximal and $R/I$ integral domain $\Longleftrightarrow$  $I$ prime.
Can you check this please, I have had a ten month break from university mathematics and am returning shortly so need to know I'm on the right track). How do you format TeX on this site as well?
Let $R/I$ be a field (so neccessarily $I \neq R$) and $J$ be an ideal of $R$ where $I$ is a subset of $J$. If $I$ is not equal to $J$ then there exists an $x \in J\backslash I$ ( $J$ minus the set $I$) so the coset of $x$, $I + x$, is non-zero hence there is an element $a \in R$ such that $(I+a)(I+x) = (I+1)$ so $I + (ax - 1) = I$ hence $ax - 1$ is in $I$. But $x \in J$ and $J$ is an ideal so $ax \in J$ and as $I \subseteq J$, $ax - (ax-1) = 1 \in J$; this implies $J = R$.
Now, consider $I$ maximal and $(I+a) \neq I$. This means $a \notin I$ so consider the ideal $I + <a>$ (the principal ideal generated by $a$) which obviously contains $I$. It follows that the ideal must be $R$ so there exist $i \in I$, $r \in R$ such that the following holds;
$$
i + ra = 1
$$ 
as $1 \in R$. Thus $I + (i+ra) = I + 1$, so $I + ra = I + 1 = (I+a)(I+r)$ hence $I+a \in R/I$ does indeed have an inverse if $a$ is non zero.
Let $R/I$ be an integral domain. Neccesarily $I \neq R$. Let $xy \in I$. Then $(I+xy) = I = (I+x)(I+y)$ so as $R/I$ is an integral domain this implies $x$ or $y$ is in $I$, hence $I$ is prime. 
Now, let $I$ be prime. Suppose $(I+x)(I+y) = I$. Then $I+xy = I$ hence $xy$ is in $I$ so $x \in I$ or $y \in I$ so $I+x$ or $I+y$ is $I$ (which is zero).
A field is an integral domain thus
$I$ maximal implies $I$ is prime.

REPLY [5 votes]: This is a prototypical example of modular reduction - which serves to simplify and/or reduce to canonical form. Here, reducing mod $\rm\:I\:,\:$ the problem simplifies to $(0)$ maximal $\implies (0)$ prime, i.e. field $\implies$ domain, a simple "canonical" form of this implication - packaged in a form for optimal reuse. It's important to bring this structure to the fore, and master these techniques in simple contexts such as here, since it will be much more difficult to learn in more complex complex contexts, where additional complexity obfuscates matters. You can find further discussion, including examples by the master algebraist Irving  Kaplansky in my post here.