About the localization of a UFD

I was wondering, is the localization of a UFD also a UFD?

How would one go about proving this? It seems like it would be kind of messy to prove if it is true.

If it is not true, what about localizing at a prime? Or what if the UFD is Noetherian?

• For a full proof using Kaplansky's criterion for UFDs see here. Sep 8, 2017 at 19:08

The localization of a UFD is a UFD provided you don't invert $0$.

Suppose $D$ is a UFD, and let $S$ be a multiplicative subset that does not contain $0$. Let $T$ be the set of all irreducibles that divide an element of $S$, and let $M$ be the set of all irreducibles not in $T$.

Claim. $p\in T$ if and only if the image of $p$ in $S^{-1}D$ is a unit.

Indeed, if $p\in T$, then there exists $s\in S$ such that $p|s$; let $x\in D$ with $px=s$. Then $$\frac{ps}{s}\cdot\frac{x}{s} = \frac{pxs}{ss} = \frac{ss}{ss} = 1_{S^{-1}D},$$ so the image of $p$ in $S^{-1}D$ is a unit. Conversely, if the image of $p$ is a unit, then there exist $t\in S$ and $x\in D$ such that $\frac{ps}{s}\cdot\frac{x}{t} = 1_{S^{-1}D}$, then $\frac{pxs}{st}=\frac{s}{s}$, hence $pxs^2 = s^2t$, hence $px=t \in S$, so $p\in T$, since $p$ divides an element of $S$. $\Box$

Claim. If $p\in M$, then the image of $p$ in $S^{-1}D$ is irreducible.

Indeed, suppose that $\frac{ps}{s} = \frac{x}{t}\frac{y}{t'} = \frac{xy}{tt'}$. Then $pstt' = xys$. Since $p$ does not divide any element of $S$, and $D$ is a UFD, an associate of $p$ appears exactly once in the factorization of $xys$ into irreducibles, and it must divide either $x$ or $y$, but not both, and all other irreducibles that appear in the factorization of $xys$ must lie in $T$. Therefore, either $\frac{x}{t}$ is a unit, or $\frac{y}{t'}$ is a unit. So $\frac{ps}{s}$ is irreducible (it is not $0$ or a unit by the previous claim).

Since $S^{-1}D$ can be embedded in the field of fraction of $D$, it follows that $S^{-1}D$ is a domain. Let $\frac{a}{s}\in S^{-1}D$; let $$a = up_1^{b_1}\cdots p_r^{b_r}q_1^{c_1}\cdots q_t^{c_t}$$ be a factorization of $a$ in $D$, where $u$ is a unit of $D$, $p_1,\ldots,p_r\in T$, and $q_1,\ldots,q_t\in M$. For a fixed $s\in S$, we have $$\frac{a}{s} = \frac{u}{s}\left(\frac{p_1s}{s}\right)^{b_1}\cdots\left(\frac{p_rs}{s}\right)^{b_r}\left(\frac{q_1s}{s}\right)^{c_1}\cdots\left(\frac{q_ts}{s}\right)^{c_t}$$ is a factorization of $\frac{a}{s}$ into a unit times a product of irreducibles (namely, the images of $q_i$ in $S^{-1}D$).

To verify uniqueness up to units, cross multiply and use unique factorization in $D$ and the claims above.

• In the second claim, "Since p does not divide any element of S", that does not seem to be the case. You only seem to have asked that $p$ is not in $S$. For instance, if you take $S \subseteq \mathbb Z$ defined as $S = \{1,p^2,p^3,\cdots,p^n,\cdots \}$ for some prime $p$, then $S$ contains no irreducibles but yet $p$ is irreducible. This implies that $M = \{ all \, primes \}$ and $T = \{p\}$ in your case. I believe you meant to define $M$ as "the set of irreducibles not in $T$". In fact, in my example, $p$ becomes a unit in the localization by your first claim. Feb 10, 2014 at 5:41
• @PatrickDaSilva: Your example is correct: although probably Arturo was assuming that $S$ was a saturated multiplicative set (e.g. if one is localizing at a prime), in which case the set of irreducibles not in $T$ is the same as the set of irreducibles not in $S$
– zcn
Feb 10, 2014 at 7:31
• @user115654 : I don't think so. I think it's really just a typo ; change that and everything follows through. We'll wait to see him confirm. Feb 10, 2014 at 14:10
• @PatrickDaSilva: It very well may be a typo - I agree that that would correct the proof. I wouldn't necessarily wait for a confirmation from him anytime soon though (although I expect no small fanfare if it appears)
– zcn
Feb 10, 2014 at 16:41
• For the sake of clarity, I think the first claim can be better written as: "Let $p \in D$ be irreducible. $p \in T$ if and only if...". Jun 7, 2021 at 17:10

One slick way is via Kaplansky's characterization: a domain is a UFD iff every nonzero prime ideal contains a nonzero prime. This is easily seen to be preserved by localization, hence the proof. Alternatively, proceed directly by showing that primes stay prime in the localization if they survive (don't become units). Thus UFD localizations are characterized by the set of primes that survive.

The converse is also true for atomic domains, i.e. domains where nonzero nonunits factor into atoms (irreducibles). Namely, if $\rm\:D\:$ is an atomic domain and $\rm\:S\:$ is a saturated submonoid of $\rm\:D^*$ generated by primes, then $\rm\: D_S$ UFD $\rm\:\Rightarrow\:D$ UFD $\:\!$ (a.k.a. Nagata's Lemma). This yields a slick proof of $\rm\:D$ UFD $\rm\Rightarrow D[x]$ UFD, viz. $\rm\:S = D^*\:$ is generated by primes, so localizing yields the UFD $\rm\:F[x],\:$ $\rm\:F =\,$ fraction field of $\rm\:D.\:$ Thus $\rm\:D[x]\:$ is a UFD, by Nagata's Lemma. This gives a more conceptual, more structural view of the essence of the matter (vs. traditional argument by Gauss' Lemma).

• I like the way this answer relates conceptually to a "bigger world" by invoking more general concepts. It wouldn't be obviously my first encounter with the idea of localisation, but it extends my understanding of how powerful it is, and why. May 3, 2012 at 22:14

As far as I can tell, the localization of a UFD is always a UFD. Let $R$ be a UFD and $S \subseteq R$ multiplicatively closed.

A ring is a UFD if and only if every height 1 prime ideal is principal. So, let $P$ be a height 1 prime ideal of $S^{-1}R$. Then there is a prime ideal $I \lhd R$ such that $P = S^{-1}I$. Now, localization does not change height, so $I$ has height 1, hence it is principal as $R$ is a UFD, say $I = \langle a \rangle$. Then $P = S^{-1}I = S^{-1}\langle a \rangle = \langle \frac{a}{1} \rangle$, so $P$ is principal. Hence, all height 1 prime ideals of $S^{-1}R$ are principal, hence $S^{-1}R$ is a UFD.

Edit: As Arturo mentions, this requires $0 \not\in S$. Also, it requires that $R$ is noetherian.

• Dear Johannes, the criterion for factoriality you quote requires $R$ to be noetherian (which is admittedly a very mild condition: I'm nitpicking!) May 3, 2012 at 21:32
• I knew I had missed something along the line... May 3, 2012 at 21:33
• @GeorgesElencwajg Why is the Noetherian hypothesis considered 'mild'? How does this argument fail if $R$ is not Noetherian? Jun 3, 2021 at 17:45
• @user46372819 Because the criterion mentioned in line 3 requires noetherianity. "Mildness" is an empirical observation: most results in commutative algebra or algebraic geometry require some noetherianity hypothesis. Jun 4, 2021 at 8:11

I'll provide a solution based on hints in Aluffi (V.4.15). Suppose $$R$$ is UFD and $$S$$ is a multiplicatively closed subset. $$S$$ cannot contain $$0$$ (otherwise $$S^{-1}R=0$$), and since $$R$$ is integral domain, $$\frac{r_1}{s_1}=\frac{r_2}{s_2}$$ in $$S^{-1}R$$ implies $$r_1s_2=r_2s_1$$.

Claim 1: if $$q\in R$$ is irreducible, then $$q/1$$ is irreducible in $$S^{-1}R$$.

Let $$\frac{q}{1}=\frac{r_1}{s_1}\frac{r_2}{s_2}$$, then $$s_1s_2q=r_1r_2.$$ Since the multiset of irreducibles of $$r_1r_2$$ contains $$q$$, either $$r_1=rq$$ for some $$r\in R$$, or $$r_2=rq$$ for some $$r\in R$$. In the first case the multiset of irrecudibles of $$r_2$$ is contained in the multiset of irreducibles of $$s_1s_2$$, so $$\exists r, r_2r=s_1s_2$$, so $$\frac{r_2}{s_2}\frac{r}{s_1}=\frac{1}{1}$$, so $$\frac{r_2}{s_2}$$ is a unit and $$q/1$$ is irreducible. The second case is similar.$$\Box$$

Claim 2: if $$a/s\in S^{-1}R$$ is irreducible, then $$a/s$$ is an associate of $$q/1$$ for some irreducible $$q\in R$$.

$$a/s$$ is an associate of $$a/1$$ since $$s/1$$ is a unit in $$S^{-1}R$$. Since associate of irreducible is still irreducible, $$a/1$$ is irreducible. Since $$R$$ is UFD, we can write $$a=q_1q_2...q_n$$ where all $$q_i$$ are irreducible. Then $$\frac{a}{1}=\frac{q_1}{1}\frac{q_2}{1}...\frac{q_n}{1}$$. Since $$\frac{a}{1}$$ is irreducible, and since associate of irreducible is irreducible, we can argue inductively to get that $$\frac{a}{1}$$ is associate of $$\frac{q_i}{1}$$ for some $$i$$.$$\Box$$

To see $$S^{-1}R$$ is a UFD, we first show factorization exists. For any $$a/s\in S^{-1}R$$ and $$a/s$$ not a unit and not zero, it suffices to show $$a/1$$ has factorization into irreducibles since $$1/s$$ is a unit. Note $$a$$ is nonzero, nonunit in $$R$$. (If it is unit, then $$\frac{a}{s}\frac{a^{-1}s}{1}=\frac{1}{1}$$ and $$\frac{a}{s}$$ is a unit, contradiction). Thus $$a$$ can be written as product of irreducibles, then using Claim 1 we see that $$a/1$$ can be written as a product of irreducibles.

Finally, to show factorization is unique. suppose $$\frac{r_1}{s_1}...\frac{r_n}{s_n}=\frac{r'_1}{s'_1}...\frac{r'_m}{s'_m}$$ where the fractions are irreducible. Using Claim 2, we can write $$\frac{r}{s}\frac{q_1}{1}...\frac{q_n}{1}=\frac{q'_1}{1}...\frac{q'_m}{1}$$ where $$\frac{r}{s}$$ is a unit and $$\frac{r_i}{s_i}$$ is an associate of $$\frac{q_i}{1}$$ and $$\frac{r'_i}{s'_i}$$ is an associate of $$\frac{q'_i}{1}$$, and $$q_i$$ and $$q'_i$$ are all irreducble in $$R$$. Thus $$rq_1...q_n=sq'_1...q'_m \quad (*)$$ Each $$q_i$$ cannot divide any element in $$S$$ because otherwise $$\frac{q_i}{1}$$ is a unit, contradiction with the fact that it is irreducible. Thus the multiset of irreducibles of $$q_1...q_n$$ is contained in the multiset of irreducibles of $$q'_1...q'_m$$.

Since $$\frac{r}{s}$$ is a unit $$\exists \frac{r'}{s'}, \frac{r}{s}\frac{r'}{s'}=\frac{1}{1}$$, that is, $$rr'=ss'$$. Multiply both sides of $$(*)$$ by $$r'$$ we get $$ss'q_1...q_n=sr'q'_1...q'_m$$. Note $$ss'\in S$$ as $$S$$ is multiplicative, then by similar argument as in the previous paragraph we get that the multiset of irreducibles of $$q_1'...q'_m$$ is contained in the multiset of irreducibles of $$q_1...q_n$$. Thus $$m=n$$, and $$q_i=q'_i$$ after relabeling. Thus $$(\frac{r_i}{s_i})=(\frac{q_i}{1})=(\frac{q'_i}{1})=(\frac{r'_i}{s'_i})$$, so factorization is unique. Thus $$S^{-1}R$$ is UFD.