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

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