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

*Proof. *Suppose that is a maximal ideal of . Let . It suffices to show that has a multiplicative inverse. Consider the set . I claim that is an ideal of that properly contains . Indeed, we check the following:

*Nonempty. *Since contains the element , we have that is nonempty.

*Subtraction.* Let . Then we have that

.

Since is a ring and is an ideal, and . Thus, .

*Multiplication.* Let and . Since is commutative, it suffices to show that . We have that

.

Since is a ring and is an ideal, and . Thus,

*Properly contains .* By definition of , . Moreover, notice that and by assumption, . Thus, is a proper subset of

Therefore, is as claimed. Now, since is maximal, we must have that . Thus , say, . Then,

We conclude that is the multiplicative inverse of ; hence, is a field.

Conversely, suppose that is a field. Let be an ideal of that properly contains (i.e. ) and let . Then is a nonzero element of and so there exists an element such that , the multiplicative identity of . Since , , and is an ideal, we have that . Moreover, since

,

we have that . Thus, and since contains it must be that . We conclude that is a maximal ideal of .

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