16D25-MaximalIdealIsPrime.tex

\documentclass[12pt]{article}
\usepackage{pmmeta}
\pmcanonicalname{MaximalIdealIsPrime}
\pmcreated{2013-03-22 17:37:59}
\pmmodified{2013-03-22 17:37:59}
\pmowner{pahio}{2872}
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\pmtitle{maximal ideal is prime}
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\pmtype{Theorem}
\pmcomment{trigger rebuild}
\pmclassification{msc}{16D25}
\pmclassification{msc}{13A15}
\pmrelated{SumOfIdeals}
\pmrelated{MaximumIdealIsPrimeGeneralCase}
\pmrelated{CriterionForMaximalIdeal}

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\begin{document}
\textbf{Theorem.}  In a commutative ring with non-zero unity, any maximal ideal is a prime ideal.

{\em Proof.}\, Let $\mathfrak{m}$ be a maximal ideal of such a ring $R$ and let the ring product $rs$ belong to $\mathfrak{m}$ but e.g. \,$r \notin \mathfrak{m}$.  The maximality of $\mathfrak{m}$ implies that\, 
$\mathfrak{m}\!+\!(r) = R = (1)$.\, Thus there exists an element \,$m \in \mathfrak{m}$\, and an element\, $x \in R$\, such that\, $m\!+\!xr = 1$.\, Now $m$ and $rs$ belong to $\mathfrak{m}$, whence
       $$s = 1s = (m\!+\!xr)s = sm\!+\!x(rs) \in \mathfrak{m}.$$ 
So we can say that along with $rs$, at least one of its \PMlinkname{factors}{Product} belongs to $\mathfrak{m}$, and therefore $\mathfrak{m}$ is a prime ideal of $R$.
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