Theorem. Let $R$ be a commutative ring with unity. The ideal $A$ of $R$ is a maximal ideal if and only if the quotient ring $R/A$ is a field.

Proof. Suppose that $A$ is a maximal ideal of $R$. Let $b\in R\setminus A$. It suffices to show that $b+A$ has a multiplicative inverse. Consider the set $B=\{br+a : r\in R, a\in A\}$. I claim that $B$ is an ideal of $R$ that properly contains $A$. Indeed, we check the following:

Nonempty. Since $B$ contains the element $b$, we have that $B$ is nonempty.

Subtraction. Let $br_1+a_1, br_2+a_2\in B$. Then we have that

$(br_1+a_1)-(br_2+a_2) = b(r_1-r_2)+(a_1-a_2)$.

Since $R$ is a ring and $A$ is an ideal, $r_1-r_2\in R$ and $a_1-a_2\in A$. Thus, $(br_1+a_1)-(br_2+a_2)\in B$.

Multiplication. Let $x\in R$ and $br+a\in B$. Since $R$ is commutative, it suffices to show that $x(br+a)\in B$. We have that

$x(br+a)=bxr+xa$.

Since $R$ is a ring and $A$ is an ideal, $xr\in R$ and $xa\in A$. Thus, $x(br+a)\in B.$

Properly contains $A$. By definition of $B$, $A\subseteq B$. Moreover, notice that $b\in B$ and by assumption, $b\notin A$. Thus, $A$ is a proper subset of $B.$

Therefore, $B$ is as claimed. Now, since $A$ is maximal, we must have that $B=R$. Thus $1\in B$, say, $1=bc+a'$. Then,

$1+A=(bc+a')+A=bc+A=(b+A)(c+A).$

We conclude that $c+A$ is the multiplicative inverse of $b+A$; hence, $R/A$ is a field.

Conversely, suppose that $R/A$ is a field. Let $B$ be an ideal of $R$ that properly contains $A$ (i.e. $A\subset B\subseteq R$) and let $b\in B\setminus A$. Then $b+A$ is a nonzero element of $R/A$ and so there exists an element $c+A\in R/A$ such that $(b+A)(c+A)=1+A$, the multiplicative identity of $R/A$. Since $b\in B$, $c\in R$, and $B$ is an ideal, we have that $bc\in B$. Moreover, since

$1+A=(b+A)(c+A)=bc+A$,

we have that $1-bc\in A$. Thus, $1=(1-bc)+bc\in B$ and since $B$ contains $1$ it must be that $B=R$. We conclude that $A$ is a maximal ideal of $R$. $\Box$