## In a unital ring, if the quotient corresponding to an ideal is a field, then the ideal is maximal

Let $R$ be a ring with $1 \neq 0$. Prove that if $M \subseteq R$ is an ideal such that $R/M$ is a field, then $M$ is maximal in $R$. (Do not assume that $R$ is commutative.)

Suppose we have a two-sided ideal $I$ with $M \subseteq I \subseteq R$. By the Lattice Isomorphism Theorem for rings, $I/M$ is a two-sided ideal of the field $R/M$. In particular, $R/M$ is commutative, and so by Proposition 9, the only ideals of $R/M$ are $M/M$ and $R/M$. Again using the Lattice Isomorphism Theorem, we have $I = M$ or $I = R$. So $M$ is maximal in $R$.

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### Comments

• Ken  On October 31, 2011 at 12:33 pm

I think you accidentally assumed that \$R\$ is commutative. The use of Proposition 9 sort of depends on it. Any way to solve this without assuming such?

• nbloomf  On November 2, 2011 at 2:43 pm

I think we’re only assuming that $R/M$ is commutative, which is true since we assumed $R/M$ is a field (and all fields are commutative in D&F). I rearranged the proof to (hopefully) make this more clear. Of course it is possible that I am completely confused.

Thanks!