Prove that the ring $R[X,X^{-1}]$ of Laurent polynomials over a UFD $R$ is a UFD.
I'd like someone to give a full proof.
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1234aaa
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1A more general fact is that the localization $S^{-1} D$ is a UFD if $D$ is a UFD and $S$ is a multiplicative subset of $D$> – lhf Oct 29 '13 at 02:30
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@lhf: when $0 \notin S$ (the trivial ring is not UFD because it is not an integral domain by definition). – Martin Brandenburg Oct 31 '13 at 11:18
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@MartinBrandenburg, that's by definition: http://en.wikipedia.org/wiki/Multiplicatively_closed_set – lhf Oct 31 '13 at 11:20
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See https://math.stackexchange.com/questions/140584 – Watson Nov 01 '16 at 20:19
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- If $R$ is UFD, then $R[X]$ is UFD (see any textbook).
- If $R$ is UFD and $f \in R \setminus \{0\}$, then $R[\frac{1}{f}]$ is UFD. The prime elements are those of $R$ which don't divide $f$. Proof: They are prime because of the classification of prime ideals of localizations. If $0 \neq a \in R[\frac{1}{f}]$, say $a=x/f^k$, then $x$ is a product of prime elements. Those which divide $f$ are units in $R[\frac{1}{f}]$. Thus $a$ is associated to a product of the distinguished prime elements.
Martin Brandenburg
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"Those which divide $f$ are units in $R[\frac{1}{f}]$." I don't understand why, I think it could help if you give a concrete example? Or, maybe if you want to, you can expand it to a full proof, because I find this difficult on multiple points. – 1234aaa Oct 29 '13 at 00:37
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$f$ is a unit in $R[1/f]$. Every divisor of a unit is a unit (general and trivial fact). – Martin Brandenburg Oct 31 '13 at 11:18