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Math Help - Localization of a UFD is again a UFD

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    Localization of a UFD is again a UFD

    Ilm been trying to do this problem for a while now.

    Suppose our UFD is R and we have the multiplicative set W\subseteq R.

    The problem I'm having is that irreducible elements in one ring (either R or W^{-1}R) don't seem to always carry over. As a simple example, consider the ring \mathbb{Q}[x,y]. Then x is irreducible here, but is no longer irreducible in \mathbb{Q}[x,y]_{(y)} (it becomes a unit). Conversely, the element \frac{xy}{1}\in \mathbb{Q}[x,y]_{(y)} is irreducible (it is an associate of the irreducible element \frac{y}{1}). However, xy is certainly not irreducible in \mathbb{Q}[x,y]_{(y)}.

    So I'm basically stuck on BOTH conditions of having a UFD (showing there always EXISTS a factorization into irreducibles, and then proving that it is unique). Does anyone have any ideas?

    EDIT: I believe I figured it out. A quick sketch of the steps I used:

    1. Prove that if r\in R is irreducible, then r/1 is either irreducible or a unit.
    2. This means that a factorization of r into irreducibles in R will give a corresponding factorization of r/1 into irreducibles (and units). Since an arbitrary element of W^{-1}R looks like \frac{r}{w}=\frac{1}{w}\cdot \frac{r}{1}, and \frac{1}{w}\in W^{-1}R^*, this implies that every nonzero nonunit of the localization has a factorization into irreducibles (existance).
    3. Prove that if r/w is irreducible, then r/w is an associate of an element of the form \frac{x^e}{1}, where x\in R is irreducible.
    4. Use (3) to prove uniqueness, writing any irreducible in this form.

    So, it works (I believe), but it's really just a lot of playing with elements and factorizations. I'm sure there's a better proof.
    Last edited by topspin1617; November 9th 2010 at 11:52 AM. Reason: Solved
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