Lemma. Let $R$ be a commutative ring and $S$ a multiplicative semigroup consisting of a subset of $R$ . If there exist ideals of $R$ which are disjoint with $S$ , then the set $\mathfrak{S}$ of all such ideals has a maximal element with respect to the set inclusion.

Proof. Let $C$ be an arbitrary chain in $\mathfrak{S}$ . Then the union $$\mathfrak{b} \;:=\; \bigcup_{\mathfrak{a} \in C}\mathfrak{a},$$ which belongs to $\mathfrak{S}$ , may be taken for the upper bound of $C$ , since it clearly is an ideal of $R$ and disjoint with $S$ . Because $\mathfrak{S}$ 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 ($S = \varnothing$ ). If $R$ has no zero divisors, the zero ideal $(0)$ is a prime ideal ($S = R\!\smallsetminus\!\{0\}$ ).

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

Bibliography

1

EMIL ARTIN: Theory of Algebraic Numbers. Lecture notes. Mathematisches Institut, Göttingen (1959).