If the Quotient Ring is a Field, then the Ideal is Maximal

Problems and solutions of ring theory in abstract algebra

Problem 197

Let $R$ be a ring with unit $1\neq 0$.

Prove that if $M$ is an ideal of $R$ such that $R/M$ is a field, then $M$ is a maximal ideal of $R$.
(Do not assume that the ring $R$ is commutative.)

 
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Proof.

Let $I$ be an ideal of $R$ such that
\[M \subset I \subset R.\] Then $I/M$ is an ideal of $R/M$.

Since $R/M$ is a field by assumption, the only ideals of the field $R/M$ is $\bar{0}=M/M$ or $R/M$ itself.
So the ideal $I/M$ is either $\bar{0}$ or $R/M$.

By the fourth (or lattice) isomorphism theorem for rings, there is a one-to-one correspondence between the set of ideals of $R$ containing $M$ and the set of ideals of $R/M$. Hence $I$ must be either $M$ or $R$.

(Since the fourth isomorphism theorem gives the correspondence $M \leftrightarrow M/M=\bar{0}$ and $R \leftrightarrow R/M$, and there is no ideal of $R/M$ between $\bar{0}$ and $R/M$.)

Therefore the ideal $M$ is maximal.


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