# 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 $\mathfrak{S}$ 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.

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 $R$.

## References

• 1 Emil Artin: .  Lecture notes.  Mathematisches Institut, Göttingen (1959).
Title definition of prime ideal by Artin DefinitionOfPrimeIdealByArtin 2013-03-22 18:44:31 2013-03-22 18:44:31 pahio (2872) pahio (2872) 9 pahio (2872) Definition msc 13C99 msc 06A06 EveryRingHasAMaximalIdeal prime ideal