# criterion for maximal ideal

Theorem.  In a commutative ring $R$ with non-zero unity, an ideal $\mathfrak{m}$ is maximal if and only if

 $\displaystyle\forall a\in R\!\smallsetminus\!\mathfrak{m}\;\;\exists r\in R% \quad\mbox{such that}\;\;1\!+\!ar\;\in\;\mathfrak{m}.$ (1)

Proof.$1^{\circ}$.  Let first $\mathfrak{m}$ be a maximal ideal of $R$ and  $a\in R\!\smallsetminus\!\mathfrak{m}$.  Because  $\mathfrak{m}\!+\!(a)=R$,  there exist some elements  $m\in\mathfrak{m}$  and  $-r\in R$  such that  $m\!-\!ar=1$.  Consequently,  $1\!+\!ar=m\in\mathfrak{m}$.
$2^{\circ}$.  Assume secondly that the ideal $\mathfrak{m}$ satisfies the condition (1).  Now there must be a maximal ideal $\mathfrak{m}^{\prime}$ of $R$ such that

 $\mathfrak{m}\;\subseteq\;\mathfrak{m}^{\prime}\;\subset\;R.$

Let us make the antithesis that $\mathfrak{m}^{\prime}\!\smallsetminus\!\mathfrak{m}$ is non-empty.  Choose an element

 $a\;\in\;\mathfrak{m}^{\prime}\!\smallsetminus\!\mathfrak{m}\;\subset\;R\!% \smallsetminus\!\mathfrak{m}.$

By our assumption, we can choose another element $r$ of $R$ such that

 $s\;=\;1\!+\!ar\;\in\;\mathfrak{m}\;\subset\;\mathfrak{m}^{\prime}.$

Then we have

 $1\;=\;s\!-\!ar\;\in\;\mathfrak{m}^{\prime}\!+\!\mathfrak{m}^{\prime}\;=\;% \mathfrak{m}^{\prime}$

which is impossible since with 1 the ideal $\mathfrak{m}^{\prime}$ would contain the whole $R$.  Thus the antithesis is wrong and  $\mathfrak{m}=\mathfrak{m}^{\prime}$  is maximal.

Title criterion for maximal ideal CriterionForMaximalIdeal 2013-03-22 19:10:40 2013-03-22 19:10:40 pahio (2872) pahio (2872) 6 pahio (2872) Theorem msc 16D25 msc 13A15 MaximalIdealIsPrime