$I$ is maximal $\implies I$ is prime 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.
 A: There is an assumption of commutativity lurking around, namely in the equivalences given.  E.g., the trivial ideal is maximal in the ring of $n\times n$ matrices with coefficients in a field, but the ring is not a field if $n\gt 1$. 
For non-commutative rings with identity, recall that an ideal $P$ is a prime ideal if and only if for any two ideals $A$ and $B$, if $AB\subseteq P$ then $A\subseteq P$ or $B\subseteq P$. (This is equivalent to the element-wise condition in commutative rings, but for noncommutative rings, the element-wise condition is stronger; such ideals are called "completely prime").
To show that a maximal ideal must be prime, let $\mathfrak{M}$ be a maximal ideal. We show that if $A$ and $B$ are ideals and neither is contained in $\mathfrak{M}$, then their product is not contained in $\mathfrak{M}$. 
If $A$ is not contained in $\mathfrak{M}$, then by maximality we have $A+\mathfrak{M}=R$. Similarly, $B+\mathfrak{M}=R$. Therefore,
$$R = RR = (A+\mathfrak{M})(B+\mathfrak{M})=AB + A\mathfrak{M} + \mathfrak{M}B + \mathfrak{M}^2 \subseteq AB+\mathfrak{M}\subseteq R.$$
Therefore, $AB+\mathfrak{M}=R$, so $AB$ is not contained in $\mathfrak{M}$, as claimed. Thus, if $\mathfrak{M}$ is maximal, then it is prime.
Of course, this proof works equally well for commutative rings, and does not require the equivalence via structure of the quotients.

For completelyness:
Proposition. Let $R$ be a ring with identity, and let $\mathfrak{P}$ be an ideal.


*

*If for every $a$ and $b$ in $R$, $ab\in \mathfrak{P}\Rightarrow a\in\mathfrak{P}\text{ or }b\in\mathfrak{P}$, then $\mathfrak{P}$ is a prime ideal (if $A,B$ are ideals and $AB\subseteq \mathfrak{P}$, then $A\subseteq \mathfrak{P}$ or $B\subseteq \mathfrak{P}$).

*If $R$ is commutative, then the converse of (1) also holds. 
Proof. 


*

*Let $A$, $B$ be ideals such that $AB\subseteq \mathfrak{P}$; if $A\subseteq\mathfrak{P}$ we are done. Otherwise, let $a\in A$ such that $a\notin \mathfrak{P}$. Then for every $b\in B$ we have $ab\in AB\subseteq\mathfrak{P}$, hence $ab\in\mathfrak{P}$. By our assumption, $b\in\mathfrak{P}$. Thus, $B\subseteq \mathfrak{P}$, as claimed. 

*Assume $R$ is commutative and that for any ideals $A$, $B$, if $AB\subseteq \mathfrak{P}$, then $A\subseteq \mathfrak{P}$ or $B\subseteq \mathfrak{P}$. Let $a,b\in R$ be such that $ab\in \mathfrak{P}$. Then $(a)(b) = (ab)\subseteq \mathfrak{P}$, so either $(a)\subseteq \mathfrak{P}$ or $(b)\subseteq\mathfrak{P}$. Hence either $a\in \mathfrak{P}$ or $b\in\mathfrak{P}$, as claimed. $\Box$
The assumption of commutativity in 2 is behind the equality $(a)(b)=(ab)$. In a noncommutative ring, $(a)$ consists of all sums of elements of the form $ras$ with $r,s\in R$, and similarly for $(b)$. The product will be generated by all elements of the form $rasbt$ with $r,s,t\in R$, and without commutativity we may be unable to rewrite them as sums of elements of the form $u(ab)v$.
For an example of a noncommutative ring and an ideal $\mathfrak{P}$ that is prime but not completely prime, let $S$ be the ring of $n\times n$ matrices over a field $F$. It's known that if $R$ is a ring with identity, then the ideals of $M_n(R)$, the ring of $n\times n$ matrices over $R$ are exactly of the form $M_n(\mathfrak{J})$, where $\mathfrak{J}$ is an ideal of $R$. So if $R=F$ is a field, then the only ideals of $R$ are the trivial ideal and the whole field, so the only ideals of $M_n(R)$ are the trivial ideal and the whole ideal. Therefore, the zero ideal is prime. However, you can find nonzero zero divisors in $S$, so you can find products of two elements which lie in the zero ideal with  neither element being the zero element. So the zero ideal is prime but not completely prime.

For rings without identity things get more annoying. For example, consider the ring $2\mathbb{Z}$ of even integers. An ideal is maximal if and only if it is of the form $2p\mathbb{Z}$ for some prime $p$. In particular, $4\mathbb{Z}$ is maximal; but $4\mathbb{Z}$ is not a prime ideal, since $(2\mathbb{Z})(2\mathbb{Z}) \subseteq 4\mathbb{Z}$, but $2\mathbb{Z}$ is not contained in $4\mathbb{Z}$. Or take a ring with zero multiplication; the ideals correspond exactly to the subgroups of its additive structure, so maximal ideals correspond to maximal subgroups. However, no proper ideal is prime (under either definition). 
A: 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.
A: Everything's fine in your answer ; there is absolutely no problem in your proof. +1!
A: Yes, your proofs seem fine.  You can make the proof of $R/I$ a field $\Leftrightarrow$ $I$ maximal a bit more compact by using the following propositions:
Proposition 1:  If $R$ is commutative with identity then $R$ is a field if and only if the only ideals of $R$ are $0$ and $R$.
Proposition 2:  Let $R$ be commutative with identity, and let $I$ be an ideal of $R$.  Then every ideal of $R/I$ is of the form $J/I$ where $J$ is an ideal of $R$ and $I\subseteq J$.  Conversely, if $J$ is an ideal of $R$ and $I\subseteq J$, then $J/I$ is an ideal of $R/I$.
The proofs are fairly basic (and canonical exercises), and I suggest giving them a try.
With the propositions above in hand, suppose $R/I$ is a field.  Then the only ideals of $R/I$ are $I$ (ie $0+I$) and $R/I$.  Therefore the only ideals of $R$ that contain $I$ are $I$ and $R$.  Therefore $I$ is maximal.
On the other hand, suppose $I$ is maximal, and consider an ideal of $R/I$.  This ideal must be of the form $J/I$ for $I\subseteq J \subseteq R$.  However, as $I$ is maximal, we have either $J=I$ or $J=R$.  Thus, the only ideals of $R/I$ are $0+I$ and $R/I$, hence $R/I$ is a field.
