defined as $A \equiv (I \cup \{ a\})$. We prove that $A = I + aR$.
$\begin{aligned}
&(I \cup \{a \}) \\
&= \quad \{ \alpha i + \beta a | i \in I, \alpha, \beta \in R \} \\
&= \quad \{ i' + \beta a | i' \in I, \alpha, \beta \in R \} \qquad \text{($I$ is closed under multiplication by $R$)} \\
&= I + aR
\end{aligned}$

#### § Quotient based proof that maximal ideal is prime

An ideal $P$ is prime iff the quotient ring $R/P$ is an integral domain. An
ideal $M$ is maximal $R/M$ is a field. Every field is an integral domain,
hence:
$M \text{ is maximal } \implies R/M \text{ is a field } \implies R/M \text {is an integral domain} \implies M \text{ is prime}$.
I was dissatisfied with this proof, since it is not ideal theoretic: It argues
about the behaviour of the quotients. I then found this proof that argues
purly using ideals:
#### § Ideal theoretic proof that maximal ideal is prime

#### § Sketch

Let $I$ be a maximal ideal. Let $a, b \in R$ such that $ab \in I$. We need
to prove that $a \in I \lor b \in I$. If $a \in I$, the problem is done.
So, let $a \notin I$. Build ideal $A = (I \cup {a})$. $I \subsetneq A$. Since
$I$ is maximal, $A = R$. Hence, there are solutions for
$1_R \in A \implies 1_r \in I + aR \implies \exists i \in I, r \in R, 1_R = i + ar$.
Now, $b = b \cdot 1_R = b(i + ar) = bi + (ba)r \in I + IR = I$. ($ba \in I$ by assumption).
Hence, $b \in I$.
#### § Details

let $i$ be a maximal ideal. let $a, b \in r$ such that $ab \in i$. we need
to prove that $a \in i \lor b \in i$.
if $a \in i$, then the problem is done. so, let $a \notin i$. consider
the ideal $A$ generated by adding $a$ into $I$. $A \equiv (I \cup \{a\})$.
We have shown that $A = I + aR$. Hence, $I + a0 = I \subset A$.
Also, $0 + ac \dot 1 = a \in A$, $a \neq I$ \implies $A \neq I$. Therefore,
$I \subsetneq A$. Since $I$ is maximal, this means that $A = R$
Therefore, $I + aR = R$. Hence, there exists some $i \in I, r \in R$ such
that $i + ar = 1_R$.
Now, $b = b \cdot 1_R = b \cdot (i + ar) = bi + (ba) r \in I + IR = I$ Hence,
$b \in I$.