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I was reading Arturo Magidin's answer here, which states that the localization of a UFD over a multiplicative subset not containing $0$ is still a UFD. It makes sense that a factorization into units and irreducibles exists, but I don't see uniqueness.

He says it follows by cross multiplying and using the claims. Ok, I let $a/s\in S^{-1}D$. I can factor it into irreducibles $$ \frac{a}{s}=\frac{p_1}{s_1}\cdots\frac{p_n}{s_n}=\frac{q_1}{t_1}\cdots\frac{q_m}{t_m}. $$ If I crossmultiply I get different types of terms all over the place and don't know what to compare?

Community
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Kally
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2 Answers2

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Remember that the factorization in an UFD is only unique up to multiplication by units.

If you have $a=s\cdot \prod \frac{p_i}{s_i}$, where $\frac{p_i}{s_i}$ are irreducible, then ${p_i}$ must also be irreducible in the localisation, we can assume without loss of generality that they are irreducible in the original ring as well (if necessary putting away their factors from $S$ into $s$), and you have $a\cdot \prod s_i=s\prod p_i$.

Then $p_i$, along with irreducible factors of $s$, factor $a\cdot \prod s_i$ into irreducible terms. Some of them will come from the factorization of $\prod s_i$, and those are units in the localisation anyway (because their product is a unit) and as such don't really matter, similarly the factors coming from $s$, and others come from the factorization of $a$ and those are uniquely determined up to equivalence in the original ring, and even more so in the localisation (as it's easier to be equivalent with more units).

tomasz
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  • Thanks tomasz, I follow most of this, but why does $p_i/s_i$ irreducible imply $p_i$ irreducible in the ring? – Kally Jan 11 '14 at 01:40
  • @Kally: good point. They may be reducible. But if $p_i=a\cdot b$ where $a,b$ are not units in the original ring, then one of $a,b$ must be in $S$. – tomasz Jan 11 '14 at 01:47
  • @Kally: I fixed it now, should be correct. – tomasz Jan 11 '14 at 01:55
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It suffices to show primes $\,p\,$ in $D\,$ that do not map to units in $\,E = S^{-1}D\,$ remain prime in $E.$

But that is easy to prove. $ $ Suppose $\ \dfrac{p}u {\biggm|} \dfrac{a}s\dfrac{b}t.\ $ Then $\ \dfrac{p}u \dfrac{c}v\, =\, \dfrac{a}s\dfrac{b}t\ \,$ so $\,\ pcst\, =\, abuv$.

Since $\,p\,$ is prime in $D,\,$ we infer $\,p\,$ divides one of $\,a,b,u,v.\,$ But if $\,p\mid u\,$ or $\,p\mid v\,$ in $D\,$ then that remains true in $E\,$ so, being a factor of a unit in $E,\,$ we infer $\,p\,$ is a unit in $E,\,$ contra hypothesis. Thus $\,p\mid a\,$ or $\,p\mid b\,$ in $D,\,$ so also in $E,\,$ so $\,p/u\mid a/s\,$ or $\,p/u\mid b/t,\,$ hence $\,p\,$ remains prime in $E.$

Bill Dubuque
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  • Thanks Bill, I follow your proof. But why does that suffice to show uniqueness? – Kally Jan 11 '14 at 03:18
  • @Kally Products of primes always factor uniquely in any domain (same proof as in $\Bbb Z$). – Bill Dubuque Jan 11 '14 at 03:28
  • Sorry Bill, but what is the map you're talking about from $D$ to $E$? Is it $p\mapsto pu/u$? The proof seems like it sent $p$ to $p/u$, for arbitrary $u$. OK, so consider arbitrary $a/u\in E$. Factor $a$ in $R$ as $vp_1\dots p_n$ for a unit $v$, and primes/irreducibles $p_i$. So $a/u=(vu/u)(p_1u/u)\cdots (p_nu/u)$. We can collect all the $p_iu/u$ which are units, and wouldn't we have to know that remaining $p_ju/u$ are primes? – Kally Jan 11 '14 at 03:37
  • @Kally The irreducibles in $E$ are precisely (the associates of) those irreducibles of $D$ that do not map to units in $E$. If we show these remain prime in $E$ then every irreducible in $E$ is prime, which implies uniqueness of factorizations into irreducibles (existence of such is clearly inherited from $D$). Informally, the factorizations in $E$ arise from those in $D$ by simply ignoring the primes that becomes units. As for $p/u$ vs $p/1$ etc, it doesn't matter which associate of $p$ we choose in $E$ since it does not affect divisibility matters (e.g. primality). – Bill Dubuque Jan 11 '14 at 04:03
  • What is justification that the irreducibles of $E$ are the irreducibles of $D$ which do not map to units? I had the same trouble with that fact in tomasz' answer. Expliclity, if $d/s$ is irreducible in $E$, what is the actual irreducible preimage in $D$? – Kally Jan 11 '14 at 04:07
  • @Kally The preimage is the lone prime factor of $d$ that is not a unit in $E$. – Bill Dubuque Jan 11 '14 at 04:23
  • Thanks, I really liked seeing this approach. – Kally Jan 11 '14 at 04:31