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.
Proof:
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 $\blacksquare$.
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.
Proof:
$(0)\subseteq R$ 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 $R/\mathcal{m}$.