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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  • 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.


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