fundamental theorem of ideal theory
Theorem. Every nonzero ideal of the ring of integers of an algebraic number field can be written as product (http://planetmath.org/ProductOfIdeals) of prime ideals of the ring. The prime ideal of the factors (http://planetmath.org/Product).
In this entry we consider the ring of the integers of a number field . We use as starting the fact that the ideals of are finitely generated submodules of (cf. basis of ideal in algebraic number field) and that its prime ideals are maximal ideals, i.e. the only ideal factors of are itself and the unit ideal .
For proving the above fundamental theorem of ideal theory, we present and prove some lemmata.
Lemma 1. The equation between the ideals of implies that .
Proof. Let and . If
then there are the elements of such that
But the of in the parentheses are elements of the ring , whence the last sum form of shows that . Consequently, .
Lemma 2. Any nonzero element of belongs only to a finite number of ideals of .
where the numbers belong to . Since we have
there can be different ideals only a finite number, at most
Lemma 3. Each ideal of has only a finite number of ideal factors.
Proof. If and , then by Lemma 1,
, whence Lemma 2 implies that there is only a finite number of such factors .
Lemma 4. All nonzero ideals of are cancellative (http://planetmath.org/CancellationIdeal), i.e. if then .
Proof. The theorem of http://planetmath.org/node/3154Steinitz (1911) guarantees an ideal of such that the product is a principal ideal . Then we may write
If and , we thus have the equation
by which there must exist the elements of such that
Consequently, the http://planetmath.org/node/7040generators of belong to the ideal , and therefore . Similarly one gets the reverse containment.
Lemma 5. If and , then has less ideal factors than .
Proof. Evidently, any factor of is a factor of . But
and , since otherwise we had
whence which would, by Lemma 4, imply .
Lemma 6. Any proper ideal of has a prime ideal factor.
Proof. Let be such a factor of that has as few factors as possible. Then
must be a prime ideal, because otherwise we had where
and are proper ideals of and, by Lemma 5, the ideal would have less factors than ; this however contradicts the fact .
Lemma 7. Every nonzero proper ideal of can be written as a product where and the factors are prime ideals.
Proof. If has only one factor distinct from , then is a prime ideal.
Induction hypothesis: Lemma 7 is in always when has at most factors. Let now have factors. Lemma 6 implies that there is a prime ideal such that where and has, by Lemma 5, at most factors. Hence, and therefore, where all ’s are prime ideals.
Lemma 8. Any two prime factor
of a nonzero ideal of are identical, i.e. and each prime factor is equal to a prime factor and vice versa.
Proof. Any prime ideal has the property that if it divides a product of ideals, it divides one of the factors of the product; now these factors are prime ideals and therefore the prime ideal coincides with one of the factors. Similarly as in the proof of the fundamental theorem of arithmetics, one sees the uniqueness of the prime factorisation of .
|Title||fundamental theorem of ideal theory|
|Date of creation||2013-03-22 19:12:40|
|Last modified on||2013-03-22 19:12:40|
|Last modified by||pahio (2872)|
|Synonym||principal theorem of ideal theory|