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# prime ideals by Artin are prime ideals

Theorem. 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 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).

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 $\mathfrak{o}$.

## Mathematics Subject Classification

13C99*no label found*06A06

*no label found*

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