I'm learning abstract algebra and there's an exercise which states

Let $R$ be a commutative ring with unit. Call an ideal $M$ to be maximal if the only ideal containing $M$ is $R$ and $M$ itself. Prove for any ideal $I$ of $R$, $R/I$ is a field iff $I$ is maximal.

I've a proof for this, but I'm not sure if it's correct or whether the entire proof doesn't makes sense. Can you check my proof ?

We use the following notation: we denote $I \leq R$ if $I$ is an ideal of $R$ and we denote $I < R$ if $I \leq R$ and there's an element which is in $R$ but not in $I$ (i.e $R$ is "bigger" than $I$ and "contains" it).

Lemma: Let $S$ be a commutative ring with unit. Then $S$ is a field if and only if the only ideals of $S$ are $0$ and $S$ itself.


  • For the first part, i.e ($S$ is a field) $\Rightarrow$ (Only ideals of $S$ are $0$ and $S$ itself), consider an ideal $I$ of $S$ such that $0 < I \leq S$. I will prove that $I = S$. As $0 < I$, pick $a$ such that $a \in I$ and $a \neq 0$. Now for any nonzero element $c \in S$, $c \in I$ too since for any $a \in I$, $a (a^{-1}c) \in I \Rightarrow c \in I$, as desired.

  • For the other direction, i.e (Only ideals of $S$ are $0$ and $S$ itself) $\Rightarrow$ ($S$ is a field), we prove the contrapositive, i.e if $S$ is not a field, then there's a $M$ such that $0 < M < S$. Since $S$ is not a field, we can find a nonzero noinvertible element $a \in S$. Now it's easy to see that $M = aS$ is an ideal, hence $M \leq S$. To prove it's strict inequality, i.e $M < S$, note that $1 \not \in M$ since $1 \in M \Leftrightarrow \exists b \in R \text{ such that } ba = 1$, but it contradicts the invertibility of $a$.

This completes the proof of this lemma $\blacksquare$.

Now returning to the main problem, let $I$ be an ideal of $R$.

  • Suppose $R/I$ be a field. I will show that there doesn't exist $J$ such that $I < J < R$. For the sake of contradiction, assume such $J$ exists. But then some routine calculation reveals $0 < J/I < R/I$, a contradiction to our Lemma.

  • Suppose $M$ is a maximal ideal of $R$. I will prove that $R/M$ is a field. Actually I prove the contrapositive, i.e if $R/M$ is not a field, then $M$ is not a maximal ideal of $R$.

    • By contrapositive to Lemma, we can find $J$ such that $0 < J < R/M$. Since the elements of $J$ are of the form $j + M$ for $j \in R$, define $S = \{j \in R| j+M \in J \}$. It's easy to see that $S < R$ and $M \subset S \subset R$ (strict subset)
    • Clearly $M \leq S \leq R$. I now claim that $M < S < R$. Indeed, pick $s \in S - M, b \in R - S$ (they exists since $ M \subset S \subset R$). Now since $s \not \in M$, but $s \in S$, we have $M < R$. Similarly, since $b \not \in S$ but $b \in R$, we get $S < R$. Hence $M$ is not maximal, as desired.

This finishes the proof of this theorem $\blacksquare$.

  • $\begingroup$ You don't have a question, you have a solution. Why don't you try posting it as a solution for math.stackexchange.com/q/106909/29335 say? Among all the duplicates of that question, that ones seems pretty well attended. $\endgroup$ – rschwieb Jul 24 '18 at 18:14
  • $\begingroup$ @rschwieb I'm not too much familiar with how proof-verification questoins are dealt but should this be a stand alone question since this is asking to verify a possibly wrong proof ? $\endgroup$ – cdt Jul 24 '18 at 18:15
  • $\begingroup$ I don't speak for everybody, but IMO, unless you have a specific question about a place you do think is wrong, I think solutions should be posted as solutions. Probably someone is going to answer your question anyway, since your proof seems to be a lot more involved than is necessary, and there is "a lot to say" about what you wrote. $\endgroup$ – rschwieb Jul 24 '18 at 18:16
  • $\begingroup$ And when you post a new solution to a question that is duplicated so many times, people are bound to think you aren't looking for content on the site before asking. Nobody wants to help people who don't help themselves to the content we've been trying to maintain. Keep that in mind for future reference, and have a good day. $\endgroup$ – rschwieb Jul 24 '18 at 18:18
  • 1
    $\begingroup$ @rschwieb Is that really so? When I post something in the proof-verification tag, I am most of the time sure, that the question was asked before, or could be looked up. But I want to learn from my own mistakes and want to find my solution to solve the problem, where I might need some help to do so. And I thought this is appreciated. $\endgroup$ – Cornman Jul 24 '18 at 18:27

We want to show, that $\mathcal{m}\subset R$ is a maximal ideal iff $R/\mathcal{m}$ is a field.

Note that $(0)\subseteq R$ is a maximal ideal iff $R$ is a field.


$(0)\subseteq R$ is a maximal ideal iff $(0)$ and $R$ are the only ideals in $R$ iff $R$ is a field.

Now for what we initialy wanted to show:

The only ideals $\mathcal{m}\subseteq\mathcal{A}\subseteq R$ are $\mathcal{m}$ and $R$ iff the only ideals $(0)\subseteq\overline{\mathcal{A}}\subseteq R/\mathcal{m}$ are $(0)$ and $R/\mathcal{m}$.

  • $\begingroup$ Thanks for your reply, but is my proof correct ? Thanks. $\endgroup$ – cdt Jul 24 '18 at 18:29
  • $\begingroup$ I have not checked it yet. I just wanted to point out, that a proof for this statement can be quite small. $\endgroup$ – Cornman Jul 24 '18 at 18:30
  • $\begingroup$ Yes, I've also checked the thread rschwieb had previously pointed out before posting this. I thought my proof is very wrong (since the proofs given there was very short and mine is very long and clumpsy) that's why I posted. $\endgroup$ – cdt Jul 24 '18 at 18:31
  • $\begingroup$ @AlexKChen I comment here, because there are already alot of comments under your initial questions. Do you confuse $S$ with $R$ in your starting post? This is the first that strikes me off guard. $\endgroup$ – Cornman Jul 24 '18 at 18:33
  • $\begingroup$ Ooops yes sorry. I've corrected the typo. $\endgroup$ – cdt Jul 24 '18 at 18:34

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