definition of prime ideal by Artin

Lemma.  Let $R$ be a commutative ring and $S$ a multiplicative semigroup consisting of a subset of $R$.  If there exist http://planetmath.org/node/371ideals of $R$ which are disjoint with $S$, then the set $𝔖$ of all such ideals has a maximal element with respect to the set inclusion.

Proof.  Let $C$ be an arbitrary chain in $𝔖$.  Then the union

$$𝔟:=\bigcup _{𝔞\in C}𝔞,$$

which belongs to $𝔖$, may be taken for the upper bound of $C$, since it clearly is an ideal of $R$ and disjoint with $S$.  Because $𝔖$ thus is inductively ordered with respect to “$\subseteq$”, our assertion follows from Zorn’s lemma.

Definition.  The maximal elements in the Lemma are prime ideals of the commutative ring.

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

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