Results 1 to 1 of 1

Thread: Localization of a UFD is again a UFD

  1. #1
    Member
    Joined
    Nov 2010
    Posts
    193

    Localization of a UFD is again a UFD

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

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

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

    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 $\displaystyle r\in R$ is irreducible, then $\displaystyle r/1$ is either irreducible or a unit.
    2. This means that a factorization of $\displaystyle r$ into irreducibles in $\displaystyle R$ will give a corresponding factorization of $\displaystyle r/1$ into irreducibles (and units). Since an arbitrary element of $\displaystyle W^{-1}R$ looks like $\displaystyle \frac{r}{w}=\frac{1}{w}\cdot \frac{r}{1}$, and $\displaystyle \frac{1}{w}\in W^{-1}R^*$, this implies that every nonzero nonunit of the localization has a factorization into irreducibles (existance).
    3. Prove that if $\displaystyle r/w$ is irreducible, then $\displaystyle r/w$ is an associate of an element of the form $\displaystyle \frac{x^e}{1}$, where $\displaystyle x\in R$ 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; Nov 9th 2010 at 10:52 AM. Reason: Solved
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. localization of modules
    Posted in the Advanced Algebra Forum
    Replies: 1
    Last Post: Oct 19th 2011, 11:45 PM
  2. localization of a ring
    Posted in the Advanced Algebra Forum
    Replies: 1
    Last Post: Apr 13th 2011, 01:20 AM
  3. localization of a UFD
    Posted in the Advanced Algebra Forum
    Replies: 0
    Last Post: Aug 21st 2010, 02:45 AM
  4. Localization - isomorphism
    Posted in the Advanced Algebra Forum
    Replies: 1
    Last Post: May 30th 2010, 07:05 PM
  5. Localization
    Posted in the Advanced Algebra Forum
    Replies: 0
    Last Post: Mar 7th 2009, 09:31 PM

Search tags for this page

Click on a term to search for related topics.

Search Tags


/mathhelpforum @mathhelpforum