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.)

 
LoadingAdd to solve later

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.


LoadingAdd to solve later

Sponsored Links

More from my site

You may also like...

Leave a Reply

Your email address will not be published. Required fields are marked *

This site uses Akismet to reduce spam. Learn how your comment data is processed.

More in Ring theory
Problems and solutions of ring theory in abstract algebra
Finite Integral Domain is a Field

Show that any finite integral domain $R$ is a field.  

Close