Alternative proof to $M$ maximal $\iff R/M$ is a field

Question

This is a long-winded way of proving the theorem. I restate the original statement as

Let $R$ be a commutative ring with unity, then $M$ is maximal $\iff R/M$ is a quotient ring containing only 2 ideals, itself $R/M$ and the zero ring $\{ M \}$.

I am proving this without the use of

1. That $R/M$ is a field

2. The correspondence theorem or any thing to do with the isomorphisms.

I do not want to use those facts.

$(\Leftarrow)$ Now assume $M$ is maximal relative to $R$ where $R$ is a commutative ring with unity. We need to show $R/M$ contains only two trivial ideals, the zero ring $\{0 \} = \{ M \}$ and itself $R/M$. Consider the principal ideal generated by the element $\{a + M\}$ with $a\not\in M$. That is, $J = R/M\{a + M\} = \{ra + M\}$ Now the ideal $ra + M \supset M$, but $M$ is maximal, so $ra + M = R$. This means $J$ cannot exist. So there cannot be any ideal in between the containment.

$(\Rightarrow)$ Now assume the quotient ring $R/M$ contains two ideals: $\{ M \}$ and $R/M$ itself, and if there is an ideal $K$ of $R$ such that $$M \subset K \subset R,$$ then there is an $k \in K$ where $k \not\in M$. Now for this choice of $k$ and using the ideal definition on $R/M$ itself, notice $$(k + M)R/M = (k + M) \{ a + M \}_{a \in R} = \{ka + M\}_{a \in R} \subset \{a + M \}_{a \in R} = R/M.$$

In fact this containment is equality because we are considering the ideal property on the $\textit{whole}$ ring, so $\{ka + M \} = \{a + M\}$. This means $ka - a \in M \subset K$. Hence if for $k' \in K$, we have $-a = k' - ka \in K$. Then $K = R$ and in fact $M$ is maximal.

Answer 1

Really long indeed!

Suppose $M$ is maximal and that $J$ is an ideal of $S=R/M$. Suppose $J$ contains a nonzero element $r+M$. Then $r\notin M$, so $rR+M=R$ and so $1=rx+y$, with $x\in R$ and $y\in M$. Therefore $$ 1+M=rx+y+M=rx+M=(r+M)(x+M)\in J $$ so $J=S$.

Suppose $S=R/M$ has only the trivial ideals. Suppose $I$ is an ideal of $R$ with $M\subsetneq I$. Let $r\in I$, $r\notin M$. Then $r+M$ is a nonzero ideal of $S$, hence the ideal it generates is $S$. Therefore $$ 1+M=(r+M)(x+M) $$ for some $x\in R$. This means $1=rx+y$, for some $y\in M$. Since $y\in M\subset I$, we conclude $1\in I$ and so $I=R$.

Answer 2

Alternate proof: $M \subset R$ is maximal iff $R/M$ is a field.

$\textbf{Proof:}$

First suppose $M \subset R$ is maximal. As $M$ is an ideal, $R/M$ is a commutative ring with $1$ and as $M \neq R$, $R/M \neq 0$ so let $a+M \in R/M$ be non-zero. Consider the ideal

$$I=\{ra+m : \text{some $r \in R,m \in M$}\}.$$

Clear to see that $I \subset R$ is an ideal and as $a \in I$ and $M$ is maximal, $I=R$. Thus $1 \in I$ and one has

$$1=ra+m ; \space \text{some $r \in R, m \in M$}.$$

This holds iff

$$1+M=ra+M=(r+M)(a+M)$$

thus $(r+M) \in R/M$ is a multiplicative inverse for $a+M$ and $R/M$ is a field.

Conversely, suppose $R/M$ is a field. And suppose there exists and ideal $I \subset R$ such that

$$M \subsetneq I \subsetneq R.$$

Let $a \in I \setminus M$ then $a+M \in R/M$ is nonzero and thus it has a multiplicative inverse. I.e,

$$(a+M)(u+M)=1+M$$

for some $u +M \in R/M$ but this holds if and only if

$$1=au-m \in I$$

Thus $I = R$ and $M \subset R$ is maximal. $\square$

Also theres a proof which uses the Fourth isomorphism theorem for Rings.