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An ideal $M$ of a commutative ring $A$ (with unity) is maximal iff $A/M$ is a field.

This is easy with the correspondence of ideals of $A/I$ with ideals of $A$ containing $I$, but how can you prove it directly? Take $x + M \in A/M$. How can you construct $y + M \in A/M$ such that $xy - 1 \in M$? All I can deduce from the maximality of $M$ is that $(M,x) = A$.

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What do the elements of $(M, x)$ look like? – Dylan Moreland Feb 8 '12 at 1:58
Also take a look at a closely related question (it's asking about a different approach, so I don't think we should close as a duplicate). – Zev Chonoles Feb 8 '12 at 2:02
up vote 4 down vote accepted

From $(M,x)=A$ you can infer that there are $m\in M, y\in A$ so that $m+xy=1$. Thus, $xy+M=1+M$.

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Since $(M,x)=A$, you have that $1\in (M,x)$. The elements of $(M,x)$ are expressions of the form $$am+bx$$ where $a,b\in A$ and $m\in M$ (of course, since $M$ is an ideal, we in fact have $am\in M$ as well.)

Thus, if $1\in (M,x)$, there exists a $m\in M$ and an $y\in A$ such that $$1=m+xy$$ and modding out by $M$, we get that $$1+M = (x+M)(y+M).$$

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I don't think anyone has mentioned the converse yet so I'll post it here.

Converse: If $A/M$ is a field then $M$ is a maximal ideal of $A$.

Proof: Suppose there exists an ideal $I$ of $A$ such that $M \subsetneqq I \subsetneqq A$. Then this means that there is an $x \in I$ such that $x \notin M$. Now because $A/M$ is a field, this means $\exists y \in A \backslash M$ such that

$$xy \equiv 1 \operatorname{mod} M.$$

Equivalently this is saying that $xy - 1 = m$ for some $m \in M$. Rearranging the equation we get that $xy - m = 1$. But then as $M \subsetneqq I$ this means that $m$ is necessarily contained in $I$ too. Since $I$ is an ideal this means that $1 \in I$ which is a contradiction since $I$ was assumed not to be the whole ring.

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Hint: $\ \exists\: y:\ xy-1\in M\: \iff\: (x) + M \:=\: (1),\: $ which is true since $M$ is maximal and $x\not\in M$

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