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