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Show that a proper ideal M of a commutative ring R is maximal if and only if R/M is a field.

What I know:

Because M is a proper ideal $M \neq R$. The ideal M is maximal if it is a maximal element among the proper ideals. R is a field if it is commutative and $R^{x} = R^{*} $ where $R^{x}$ is the group of units of R and $R^{*}$=$R$ \ {$0$}.

I need help working through and writing the actual proof.

Some ideas:

I know that a ring is a field iff the zero ideal is maximal.

I assume I will need to reference the Correspondence Theorem for rings.

Any advice or help would be greatly appreciated!

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    $\begingroup$ Is the ring assumed to be commutative? $\endgroup$ – Elchanan Solomon Feb 27 '14 at 19:22
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    $\begingroup$ Why did you mark it with group theory??? $\endgroup$ – tomasz Feb 27 '14 at 19:24
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If $M$ is a maximal ideal and $a\notin M$ then the ideal $(a)+M$ must equalize $R$ so there are elements $r\in R$ and $m\in M$ with $ra+m=1$ or equivalently $(r+M)(a+M) =1+M$ in ring $R/M$.

Conversely if $R/M$ is a field and $a\notin M$ then $(r+M)(a+M) =1+M$ for some $r\in R$. Then $ra-1\in M$ so that $1\in (a)+M$ and consequently $R=(a)+M$. This for every $a\notin M$ so apparently $M$ is maximal.

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You should know the ideals of $R$ containing $M$ correspond bijectively (and preserving inclusion) to the ideals of $R/M$ (under the natural projection).

The theorem should be obvious from that if you know a ring is a field iff the zero ideal is maximal. Just ask what does correspond to $M$ in $R/M$?

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By way of contradiction, suppose M is not maximal. Then $\exists P$ such that $M \subset P \subset R$ (proper containment) and $P / M$ an ideal in $R/M$. (Proof for that in my text.)

$P/M \neq \{0\}$ because $M \subset P$ proper.

$\implies$ $P/M = R/M$ since $R/M$ a field. (Note that the only ideals of fields are $\{0\}$ and the entire field.)

Now I claim that $P/M = R/M \implies P=R$.

Let $r \in R$, $p \in P$ and $P \subset R$. Then $r+M = p+M$.

$\implies r-p \in M$ by coset properties

$\implies r-p \in P$ by $P \subset M$

$\implies r+P = p+P$ by coset properties

$\implies r+P = 0+P$ by $p \in P$

$\implies R \subset P$.

Since $P \subset R$, $P=R$.

But $P=R$ contradicts our proper containment above. So $M$ is a maximal ideal of $R$.

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