implications of having divisor theory
Proof. Let be an element of the quotient field of which is integral over . Then satisfies an equation
where . Now, we can write with , whence (1) may be written
Let us make the antithesis that does not belong to itself. Then and therefore we have for the corresponding principal divisors . We infer that there is a prime divisor factor of and an integer such that
By the condition 2 of the definition of divisor theory (http://planetmath.org/DivisorTheory), the right hand side of the equation (2) is divisible by
On the other side, the highest power of , by which the divisor is divisible, is . Accordingly, the different sides of (2) show different divisibility by powers of . This contradictory situation means that the antithesis was wrong and thus the proposition has been proven.
Proposition 2. When an integral domain has a divisor theory , then each element of has only a finite number of non-associated (http://planetmath.org/Associates) factors (http://planetmath.org/DivisibilityInRings).
This is unique up to the ordering of the factors; . Then we form of the prime divisors all products having factors () and choose from the products those which are principal divisors. Thus we obtain a set of factors of containing at most elements. All different principal divisor factors of are gotten, as runs all integers from 0 to , and their number is at most equal to
(see 5. in the binomial coefficients). To every principal divisor , there corresponds a class (http://planetmath.org/EquivalenceClass) of associate factors of , and the elements of distinct classes (http://planetmath.org/EquivalenceClass) are non-associates. Since has not other factors, the number of its non-associated factors is at most .
|Title||implications of having divisor theory|
|Date of creation||2013-03-22 17:59:13|
|Last modified on||2013-03-22 17:59:13|
|Last modified by||pahio (2872)|
|Synonym||properties of rings having a divisor theory|