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

(1)

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 $s_i = p_i+r_ia+n_ia$ 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 $$s_1s_2 \;=\; (p_2+r_2b+n_2b)p_1+(r_1a+n_1a)p_2+(r_1r_2+n_2r_1+n_1r_2)ab+(n_1n_2)ab \in \mathfrak{p}.$$ This is however impossible, since the product $s_1s_2$ 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 $\mathfrak{o}$ .