equivalent definitions for UFD

Let R be an integral domainMathworldPlanetmath. Define

T={uR|u is invertible}{p1pnR|pi is prime}.

Of course 0T and T is a multiplicative subset (recall that a prime elementMathworldPlanetmath multiplied by an invertible element is again prime). Furthermore R is a UFD if and only if T=R\{0} (see the parent object for more details).

Lemma. If a,bR are such that abT, then both a,bT.

Proof. If ab is invertiblePlanetmathPlanetmath, then (since R is commutativePlanetmathPlanetmathPlanetmath) both a,b are invertible and thus they belong to T. Therefore assume that ab is not invertible. Then


for some prime elements piR. We can group these prime elements in such way that p1pn divides a and pn+1pk divides b. Thus a=αp1pn and b=βpn+1pk for some α,βR. Since R is an integral domain we conclude that αβ=1, which means that both α,β are invertible in R. Therefore (for example) αp1 is prime and thus aT. Analogously bT, which completesPlanetmathPlanetmathPlanetmathPlanetmathPlanetmath the proof.

Theorem. (Kaplansky) An integral domain R is a UFD if and only if every nonzero prime idealMathworldPlanetmathPlanetmath in R contains prime element.

Proof. Without loss of generality we may assume that R is not a field, because the thesis trivialy holds for fields. In this case R always contains nonzero prime ideal (just take a maximal idealMathworldPlanetmath).

,,” Let P be a nonzero prime ideal. In particular P is proper, thus there is nonzero xP which is not invertible. By assumptionPlanetmathPlanetmath xT and since x is not invertible, then there are prime elements p1,,pkR such that x=p1pkP. But P is prime, therefore there is i{1,,k} such that piP, which completes this part.

,,” Assume that R is not a UFD. Thus there is a nonzero xR such that xT. Consider an ideal (x). We will show, that (x)T=. Assume that there is rR such that rxT. It follows that xT (by lemma). ContradictionMathworldPlanetmathPlanetmath.

Since (x)T= and T is a multiplicative subset, then there is a prime ideal P in R such that (x)P and PT= (please, see this entry (http://planetmath.org/MultiplicativeSetsInRingsAndPrimeIdeals) for more details). But we assumed that every nonzero prime ideal contains prime element (and P is nonzero, since xP). Obtained contradiction completes the proof.

Title equivalent definitions for UFD
Canonical name EquivalentDefinitionsForUFD
Date of creation 2013-03-22 19:04:04
Last modified on 2013-03-22 19:04:04
Owner joking (16130)
Last modified by joking (16130)
Numerical id 4
Author joking (16130)
Entry type Theorem
Classification msc 13G05
Related topic UniqueFactorizationAndIdealsInRingOfIntegers