Let R be a commutative ring with identity 1 \neq 0. The ideal M of R is maximal iff the quotient ring R/M is a field.
Proof. This proof can be easily done by using the Lattice Isomorphism Theorem of Rings.
“\Rightarrow“: Given maximal ideal M of R. By the Lattice Isomorphism Theorem of Rings, the ideals of R containing M correspond bijectively with the ideals of R/M. The only ideals of R that consist M are M and R itself. So we have that R/R \cong 0 and R/M are ideals of R/M. We use a handy proposition which we have proven here which implies that R/M is a field.
“\Leftarrow“: Given R/M a field. Then by this proposition here, we have that the only ideals of R/M are 0 and R/M. We use the Lattice Isomorphism Theorem of Rings again, which means that there are two ideals consisting M. We know that at least it is M and R, which is already two. This implies there is no ideal I of R such that M \subset I \subset R. Therefore, M is a maximal ideal of R.
