§ Handy characterization of adding an element into an ideal, proof that maximal ideal is prime

§ The characterization

Let $I$ be an ideal. The ideal generated by adding $(a \in R)$ to $I$ is 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, $b \in I$.