proof that a domain is Dedekind if its ideals are products of primes
For the equivalence of 1 and 2 see proof that a domain is Dedekind if its ideals are products of maximals. Also, as every maximal ideal is prime, it is immediate that 2 implies 3. So, we just need to consider the case where 3 is satisfied and show that 2 follows, for which it enough to show that every nonzero prime ideal is maximal.
We first suppose that is an invertible (http://planetmath.org/FractionalIdeal) prime ideal and show that it is maximal. To do this it is enough to show that any gives . First, we have the following inclusions,
Then, consider the prime factorizations
As is strictly contained in and , it must also be strictly contained in and . So, are nonzero prime ideals, and by uniqueness of prime factorization (see, prime ideal factorization is unique) Equation (3) gives and , after reordering of the factors. So and,
But was assumed to be invertible, and can be cancelled giving , showing that is maximal.
Now let be any prime ideal and . Factoring into a product of primes
|Title||proof that a domain is Dedekind if its ideals are products of primes|
|Date of creation||2013-03-22 18:35:07|
|Last modified on||2013-03-22 18:35:07|
|Last modified by||gel (22282)|