# prime ideals by Artin are prime ideals

Due to Artin, a prime ideal of a commutative ring $R$ is the maximal element among the ideals not intersecting a multiplicative subset $S$ of $R$.  This is equivalent (http://planetmath.org/Equivalent3) to the usual criterion

 $\displaystyle ab\in\mathfrak{p}\quad\Rightarrow\quad a\in\mathfrak{p}\;\lor\;b% \in\mathfrak{p}$ (1)

of prime ideal (see the entry prime ideal (http://planetmath.org/PrimeIdeal)).

Proof.$1^{\underline{o}}$.  Let $\mathfrak{p}$ be a prime ideal by Artin, corresponding the semigroup $S$, and let the ring product $ab$ belong to $\mathfrak{p}$.  Assume, contrary to the assertion, that  neither of $a$ and $b$ lies in $\mathfrak{p}$.  When  $(\mathfrak{p},\,x)$  generally means the least ideal containing $\mathfrak{p}$ and an element $x$, the antithesis implies that

 $\mathfrak{p}\subset(\mathfrak{p},\,a)\;\;\land\;\;\mathfrak{p}\subset(% \mathfrak{p},\,a),$

whence by the maximality of $\mathfrak{p}$ we have

 $(\mathfrak{p},\,a)\cap S\neq\varnothing\;\;\land\;\;(\mathfrak{p},\,b)\cap S% \neq\varnothing.$

Therefore we can chose such elements  $s_{i}=p_{i}+r_{i}a+n_{i}a$  of $S$ (N.B. the multiples) that

 $p_{i}\in\mathfrak{p},\;\,r_{i}\in R,\;\,n_{i}\in\mathbb{Z}\quad(i\,=\,1,\,2).$

But then

 $s_{1}s_{2}\;=\;(p_{2}+r_{2}b+n_{2}b)p_{1}+(r_{1}a+n_{1}a)p_{2}+(r_{1}r_{2}+n_{% 2}r_{1}+n_{1}r_{2})ab+(n_{1}n_{2})ab\in\mathfrak{p}.$

This is however impossible, since the product $s_{1}s_{2}$ belongs to the semigroup $S$ and  $\mathfrak{p}\cap S=\varnothing$.  Because the antithesis thus is wrong, we must have  $a\in\mathfrak{p}$  or  $b\in\mathfrak{p}$.

$2^{\underline{o}}$.  Let us then suppose that an ideal $\mathfrak{p}$ satisfies the condition (1) for all  $a,\,b\in R$.  It means that the set  $S=R\!\smallsetminus\!\mathfrak{p}$  is a multiplicative semigroup.  Accordingly, the $\mathfrak{p}$ is the greatest ideal not intersecting the semigroup $S$, Q.E.D.

Remark.  It follows easily from the theorem, that if $\mathfrak{p}$ is a prime ideal of the commutative ring $\mathfrak{O}$ and $\mathfrak{o}$ is a subring of $\mathfrak{O}$, then $\mathfrak{p\cap o}$ is a prime ideal of $\mathfrak{o}$.

Title prime ideals by Artin are prime ideals PrimeIdealsByArtinArePrimeIdeals 2013-03-22 18:44:55 2013-03-22 18:44:55 pahio (2872) pahio (2872) 10 pahio (2872) Theorem msc 13C99 msc 06A06 IdealGeneratedByASet PrimeIdeal