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
Canonical name	CriterionForMaximalIdeal
Date of creation	2013-03-22 19:10:40
Last modified on	2013-03-22 19:10:40
Owner	pahio (2872)
Last modified by	pahio (2872)
Numerical id	6
Author	pahio (2872)
Entry type	Theorem
Classification	msc 16D25
Classification	msc 13A15
Related topic	MaximalIdealIsPrime