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We know that if $R/M$ is a field, then it only has two ideals: $R/M$ and $0$. Thus by the Correspondence Theorem, then since $M$ is an ideal, there cannot be another ideal $J$ in $R$ where $M\subset J$. Is commutativity in $R$ used anywhere in this proof?

I was wondering this because in Abstract Algebra e3, Dummit and Foote Chapter 7 Section 4, they give the same proof assuming that $R$ is commutative, but one of the exercises asks to prove the same thing without the commutativity assumption.

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  • $\begingroup$ In the proof of Proposition 12, Dummit and Foote uses the commutativity of $R$ in the last part where they utilize Proposition 9(2). $\endgroup$
    – Decaf-Math
    Commented May 14, 2017 at 5:23
  • $\begingroup$ @Decaf-Math In that proof, they utilize Proposition 9(2) that assumes commutativity. But in the proof for Proposition 9(2), it does not look like it uses commutativity. If R is a field then every non-zero ideal has to have a unit i.e. R is the only ideal. If 0 and R are the only ideals, then the principal ideal of $x\in R$ has $(x)=R$. So then it must contain 1, i.e. $x$ must be a unit. Can't you make these conclusions without commutativity? $\endgroup$
    – kdavid2
    Commented May 14, 2017 at 5:33
  • $\begingroup$ This may or may not answer your question, but recall that a field is a commutative division ring. (top of page 224) $\endgroup$
    – Decaf-Math
    Commented May 14, 2017 at 5:39
  • $\begingroup$ @Decaf-Math Oh, it does! In the converse for Prop 9(2), even if every element is a unit, that only shows $R$ is a division ring. So for $R$ to be a field, then it must be commutative. Thank you! $\endgroup$
    – kdavid2
    Commented May 14, 2017 at 5:43
  • $\begingroup$ In general if $R$ is non-commutative then $R/I$ where $I$ is a maximal right ideal is not a ring but a simple $R$-module $\endgroup$
    – JJR
    Commented May 15, 2017 at 18:28

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