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 <title>definition of prime ideal by Artin</title>
 <name>DefinitionOfPrimeIdealByKrull</name>
 <created>2009-01-17 10:10:02</created>
 <modified>2009-04-19 20:50:32</modified>
 <type>Definition</type>
<parent id="409">prime ideal</parent>
 <creator id="2872" name="pahio"/>
 <author id="2872" name="pahio"/>
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	<category scheme="msc" code="06A06"/>
	<category scheme="msc" code="13C99"/>
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	<concept>prime ideal</concept>
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	<object name="EveryRingHasAMaximalIdeal"/>
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 <content>\PMlinkescapeword{ideals} \PMlinkescapeword{ideal}

\textbf{Lemma.}\, Let $R$ be a commutative ring and $S$ a multiplicative semigroup consisting of a subset of $R$.\, If there exist \PMlinkid{ideals}{371} of $R$ which are disjoint with $S$, then the set $\mathfrak{S}$ of all such ideals has a maximal element with respect to the set inclusion.

{\em Proof.}\, Let $C$ be an arbitrary chain in $\mathfrak{S}$.\, Then the union
$$\mathfrak{b} \;:=\; \bigcup_{\mathfrak{a} \in C}\mathfrak{a},$$
which belongs to $\mathfrak{S}$, may be taken for the upper bound of $C$, since it clearly is an ideal of $R$ and disjoint with $S$.\, Because $\mathfrak{S}$ thus is inductively ordered with respect to ``$\subseteq$'', our assertion follows from Zorn's lemma.\\

\textbf{Definition.}\, The maximal elements in the Lemma are {\em prime ideals} of the commutative ring.\\

The ring $R$ itself is always a prime ideal ($S = \varnothing$).\, If $R$ has no zero divisors, the zero ideal $(0)$ is a prime ideal ($S = R\!\smallsetminus\!\{0\}$).

If the ring $R$ has a non-zero unity element 1, the prime ideals corresponding the semigroup \,$S = \{1\}$\, are the maximal ideals of $R$.


\begin{thebibliography}{9}
\bibitem{Artin} {\sc Emil Artin}: {\em Theory of Algebraic Numbers}.\, Lecture notes.\, Mathematisches Institut, G\"ottingen (1959).
\end{thebibliography}
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