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Let $I$ be an ideal in a commutative, unital ring $R.$ Prove that if $R/I$ is a field, then $I$ is maximal.

Here's my take on this problem.

Suppose $R/I$ is a field. Then every nonzero element is invertible. We first show that $I$ is proper. If $1\in I,$ then $I=R\Rightarrow R/I$ is not a field (as it is the trivial ring, which is assumed to be nonunital for the purpose of this problem). We now show that $I$ is maximal. Suppose there exists an ideal $J$ such that $I\subsetneq J\subsetneq R.$ Take $0\neq a\in R/I.$ Then $\exists a^{-1}\in R/I$ such that $aa^{-1}=a^{-1}a=1\in R/I.$ By assumption, $J$ contains an element other than $1$ that is not in $I$ (otherwise $J$ would not be a proper ideal of $R$). So there exists $b\in J$ such that $b\neq 0\in R/I.$ However, since $R/I$ is a field $\exists b^{-1}\in R/I$ such that $bb^{-1}=b^{-1}b=1.$ Since $b^{-1}\neq 0, b^{-1}\in R\setminus I.$ Hence $bb^{-1}=1\in J,$ which is a contradiction. Hence $J=R$ and $I$ is maximal.

I know I did a lot of unnecessary steps in my proof (like recalling definitions), but other than that, are there any major flaws in my logic or ways to shorten the proof?

Michael Hardy
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  • You can compare your proof with the standard one (it is a popular post topic), see for example here. – Dietrich Burde Nov 10 '19 at 17:43
  • There is a typographical difference between $R\backslash I$ (with \backslash) and $R\setminus I$ (with \setminus), the former lacking the spacing appropriate to binary operation symbols. I edited the question accordingly. (I would have writte $R\smallsetminus I,$ but maybe that's just my own tastes.) $\qquad$ – Michael Hardy Nov 10 '19 at 17:50
  • @MichaelHardy thanks. –  Nov 10 '19 at 17:51

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