equivalent definitions for UFD
Let be an integral domain. Define
Of course and is a multiplicative subset (recall that a prime element multiplied by an invertible element is again prime). Furthermore is a UFD if and only if (see the parent object for more details).
Lemma. If are such that , then both .
for some prime elements . We can group these prime elements in such way that divides and divides . Thus and for some . Since is an integral domain we conclude that , which means that both are invertible in . Therefore (for example) is prime and thus . Analogously , which completes the proof.
Proof. Without loss of generality we may assume that is not a field, because the thesis trivialy holds for fields. In this case always contains nonzero prime ideal (just take a maximal ideal).
,,” Let be a nonzero prime ideal. In particular is proper, thus there is nonzero which is not invertible. By assumption and since is not invertible, then there are prime elements such that . But is prime, therefore there is such that , which completes this part.
,,” Assume that is not a UFD. Thus there is a nonzero such that . Consider an ideal . We will show, that . Assume that there is such that . It follows that (by lemma). Contradiction.
Since and is a multiplicative subset, then there is a prime ideal in such that and (please, see this entry (http://planetmath.org/MultiplicativeSetsInRingsAndPrimeIdeals) for more details). But we assumed that every nonzero prime ideal contains prime element (and is nonzero, since ). Obtained contradiction completes the proof.
|Title||equivalent definitions for UFD|
|Date of creation||2013-03-22 19:04:04|
|Last modified on||2013-03-22 19:04:04|
|Last modified by||joking (16130)|