UFDs and Localization

This post will be a little random. I plan on talking about a couple sufficient criteria for a ring $R$ to be a UFD, and then use (one of) them to show that the group ring $R[\Z^n]$ is a UFD when $R$ is 1. These criteria will involve the concept of localizing a ring, which is something I have wanted to talk about on this blog for a while now, so let’s start with that.


Localization 2 is just about the nicest algebraic operation 3 one can apply; although this is not apparent from its definition. In essense, localization gives us a way to invert a subset of elements of a ring $R$. In this way, it is a generalization of the field of fractions of a ring. Before constructing the localization of a ring, we need to know which subsets of elements we can invert. Throughout this post, all rings are commutative with unity.

Let $R$ be a ring. A subset $S\subset R$ is called multiplicative if $1\in S$ and $a,b\in S\implies ab\in S$.
Given a multiplicative set $S\subset R$, we can form the localization $\sinv R$ which is the ring $$\sinv R=\brackets{\left.\frac ab\right| a\in R,b\in S}$$ where addition and multiplication are defined as expected, and we say $$\frac ab=\frac cd\iff\exists u\in S:0=u(ad-bc)$$
Prove that addition and multiplication on $\sinv R$ are well-defined and actually do give it a ring structure.

Note that when $R$ is a domain (and $0\not\in S$), the condition on equality of fractions becomes the usual $a/b=c/d\iff ad=bc$. Note furthermore that $\sinv R\subset\Frac(R)$ and $\Frac(\sinv R)=\Frac(R)$. Finally, $\sinv R=0\iff0\in S$. In order to prevent me from saying something technically false, for the rest of the post, assume that $0\not\in S$ for any multiplicative set we consider.

There is a natural map $R\to\sinv R$ given by $r\mapsto r/1$, and this map is injective when $S$ contains no zero divisors (e.g. if $R$ is a domain) as $r/1=0/1\iff\exists u\in S:0=ur$.
There are two standard classes of multiplicative sets. These are
  • $S=\{1,x,x^2,\dots\}$ for some $x\in R$. In this case, we write $\sinv R=R\sqbracks{\frac1x}\simeq R[y]/(1-xy)$
  • $S=R\sm\mfp$ where $\mfp\subset R$ is a prime ideal. In this case we write $\sinv R=R_\mfp$

Note that when $R$ is a domain and $\mfp=(0)$, we get $R_{(0)}=\Frac(R)$. In general, the second example above is particularly prevalent because it let’s you study rings “one prime at a time”. In particular,…

Given an ideal $I\subset R$, the set $\sinv I=I\cdot\sinv R=\brackets{\left.\frac as\right|a\in I,b\in S}$ is an ideal of $\sinv R$.
$R_\mfp$ has exactly 1 maximal ideal.
We claim that $\mfp R_\mfp$ is the only maximal ideal of $R_\mfp$. This will follow from showing that every $\mfp R_\mfp$ is literally the set of all non-units in $R_\mfp$. Pick som non-unit $a/s\in R_\mfp$. Then, $a\not\in S$ since otherwise $s/a\in R_\mfp$ would be an inverse. Hence, $a\in R\sm S=\mfp$ so $a/s\in\mfp R_\mfp$. Conversely, if $a/s\in\mfp R_\mfp$, then we can assume that $a\in\mfp$. Suppose that $b/t$ was an inverse to $a/s$; then, $$\frac{ab}{st}=\frac11\iff\exists u\not\in\mfp:0=u(ab-st)\iff ab-st\in\mfp\iff st\in\mfp$$ which is impossible since $st\in S$, so $a/s$ must not be a unit. Thus, $\mfp R_\mfp$ is the set of all non units, and hence the unique maximal ideal.
A ring $R$ is called local if it only has 1 maximal ideal.

In addition to this, localization also respects many properties of the original ring. For example, we have the following theorems.

If $R$ is a domain, then so is $\sinv R$.
Let $\mfp\subset R$ be a prime ideal in a domain $R$. Then $\sinv\mfp$ is prime as well.
If $\mfp\cap S\neq\emptyset$, then $\sinv\mfp=\sinv R$ as it contains units, and so is prime. Hence, suppose that $\mfp\cap S=\emptyset$. Let $x=a/s,y=b/t\in\sinv A$ ($s,t\in S$ and $a,b\in R$) with $xy\in\sinv\mfp$. Then, $\frac{ab}{st}=\frac c{s'}$ for some $c\in\mfp$ and $s'\in S$ which means that $abs'=cst\in\mfp$. Since $s'\not\in\mfp$, either $a\in\mfp$ or $b\in\mfp$ so $x$ or $y$ lies in $\mfp$, making $\mfp$ primes.
Let $\mfq\subset\sinv R$ be a prime ideal and let $f:R\to\sinv R$ denote the map $f(r)=r/1$. Then, $\mfp:=\inv f(\mfq)$ is a prime ideal of $R$. When $R$ is a domain, we have $\mfp=\mfq\cap R$.
In general, given two rings $R,R'$ with a map $g:R\to R'$ and a prime ideal $\mfq\subset R'$, $\inv g(\mfq)$ is a prime ideal of $R$ since $$ab\in\inv g(\mfq)\iff g(a)g(b)\in\mfq\iff g(a)\in\mfq\iff a\in\inv g(\mfq)$$ where I should really say $g(a)$ or $g(b)$ is in $\mfq$ above, but meh.
$\sinv R$ is a PID when $R$ is a PID.

I think this should be everything we need for this post. With that done, when is a ring a UFD?

UFD Criteria

I always hate trying to prove rings are UFDs because my go-to tactics are to prove something stronger (e.g. its a PID), but this clearly isn’t always possible. The definition of a UFD is kind of messy, so it’s nice to have alternative, sufficient conditions for the property. One4 such condition is

Let $R$ be an integral domain. If every element of $R$ is a product of primes, then $R$ is a UFD.
First, let $\pi$ be an irreducible element, and write $\pi=\tau_1\dots\tau_n$ as a product of primes. Then, since $\pi$ is irreducible, we really have (possibly after rearranging the $\tau_i$) that $\pi=u\tau_1$ for some unit $u$, so $\pi$ is a prime. Thus, every element of $R$ factors into a product of irreducibles. Suppose that $$u\pi_1\dots\pi_n=U\Pi_1\dots\Pi_N$$ where $u,U$ are units and $\pi_i,\Pi_i$ are irreducibles ($\iff$ primes). We need to show that (possibly after rearrangement) $\pi_i,\Pi_i$ are associates and $n=N$. Well, $\pi_1$ is prime so $\pi_1\mid\Pi_j$ for some $j$. Rearrange to assume that $j=1$, and divide by $\pi_1$ to get that $u\pi_2\dots\pi_n=U'\Pi_2\dots\Pi_N$ for some unite $U'$. The claim then follows by induction, so $R$ is a UFD.
Let $R$ be an integral domain. Then, $R$ is a UFD iff every nonzero prime ideal of $R$ contains a nonzero principal prime ideal (i.e. contains a prime element).
$(\implies)$ This direction is easy. If $R$ is a UFD and $\mfp$ is a prime ideal, then consider any $x\in\mfp$. We can write $x=\pi_1\dots\pi_n$ as a product of prime elements, and primality of $\mfp$ implies that $(\pi_i)\subset\mfp$ for some $i$, giving us the result.
$(\impliedby)$ For this direction, let $S$ be the set of all (finite) products or prime elements of $R$ (including the empty product so $1\in S$). Then, $0\not\in S$ and $S$ is clearly multiplicative, so we can form the (nonzero) ring $A=\sinv R$. We claim that $A$ is a field. Pick any nonzero nonunit $r\in R$. Suppose that $1/r\not\in A$, so $(r)\subsetneq A$ meaning that it is contained in some maximal ideal $\mfm\subset A$. Then, $\mfp=\mfm\cap R$ is prime, and hence contains a principal prime ideal $(\pi)$. But this means that $\sinv(\pi)\subset\sinv\mfm\subsetneq A$ which is impossible as $\pi\in S$ so it's definitely a unit in $A$. Thus, $1/r\in A$, so we can write $$\frac1r=\frac b{\pi_1\dots\pi_n}\in A$$ where $b\in R$ and the $\pi_i\in R$ are all prime. We now prove by induction on $n$ that $r$ is the product of prime elements. If $n=1$, then $rb=\pi_1$ so since primes are irreducible and $r$ is not a unit, $r$ must itself be prime. For $n>1$, we have $rb=\pi_1\dots\pi_n$ so each $\pi_i$ divides $r$ or $b$. If some $\pi_j$ divides $b$, then $b=c\pi_j$, so $rc$ is a product of $n-1$ primes, meaning $r$ is a product of primes by induction. Otherwise, each $\pi_i$ divides $r$ so $\parens{\frac r{\pi_1\dots\pi_n}}b=1$, making $b$ a unit and $r$ a product of primes. Hence, we have that every element of $R$ is a product of primes, and so $R$ is a UFD.
Every PID is a UFD.
Let $R$ be a Dedekind domain. Then, $R$ is a UFD iff $R$ is a PID.

It is maybe worth mentioning that the above 5 can be proven without localizing. You replace the supposition that $(r)\subsetneq\sinv R$ with the supposition that $(r)\cap S=\emptyset$ to eventually show that $rb=\pi_1\dots\pi_n$ for some $b$. When doing this, you have to use Zorn’s lemma to show that there’s some prime ideal of $R$ containing $(r)$ that is disjoint from $S$ whereas we get the analagous fact more easily by considering an appropriate maximal ideal of $\sinv R$6.

So UFDs are integral domains where every element is a product of primes or are integral domains where each prime ideal contains a principle prime ideal. To me, these conditions 7 sound more reasonable to prove than the original unique factorization into irreducibles formulation. As an application of the second of them, we also prove the following.

Let $R$ be a UFD. Then, $\sinv R$ is a UFD.
Let $\mfq\subset\sinv R$ be a prime ideal, so $\mfp=\mfq\cap R$ is a nonzero prime ideal in $R$. This means that it contains a principal prime $(\pi)\subset\mfp$, so $\pi\in\mfq$. Thus, it suffices to show that $\pi$ is a prime of $\sinv R$. Pick $a/s,b/t\in\sinv R$ and suppose that $\pi\mid\frac{ab}{st}$. Then, $\pi\mid ab$ (multiply by $st$) so $\pi\mid a$ or $\pi\mid b$ ($\pi$ is prime in $R$), but this means that $\pi\mid a/s$ or $\pi\mid b/t$ (since $s,t$ are units). Thus, $\pi$ is prime in $\sinv R$ and $\sinv R$ is a UFD.

If that’s not a clean proof, then I don’t know what is. If you think about it, the above really proves the following more general theorem. 8

Let $\phi:A\to R$ be a ring map with $A$ a UFD. Then if
  • $\phi(a)$ is irreducible whenver $a$ is irreducible; and
  • $\inv\phi(\mfp)\neq(0)$ for all nonzero primes $\mfp\subset R$
we must have that $R$ is a UFD.

If you read the first footnote, then you may be wondering if I’ll prove that $R$ UFD $\implies R[x]$ UFD. I won’t 9.

An Interesting Conjecture

First, a crash course on group rings.

Let $R$ be a ring and $G$ be a group. The group ring $R[G]$ is the ring of formal sums $$R[G]=\brackets{\left.\sum_{g\in G}r_g\cdot e_g\right|r_g\in R}$$ where $r_g=0$ for all but finitely many $g\in G$. Multiplication is given by $e_g\cdot e_h=e_{gh}$ for $g,h\in G$, and more generally: $$\parens{\sum_{g\in G}r_g\cdot e_g}\parens{\sum_{h\in G}s_h\cdot e_h}=\sum_{g\in G}\sum_{h\in H}r_gs_h\cdot e_{gh}$$ The additive identity is the sum 0, and the multiplicative identity is the identity element of the group.
Group rings satisfy the following:
  • $R[G]$ is commutative iff $G$ is abelian.
  • $R[G]$ has zero divisors if $G$ has a torsion element. Indeed, if $x^n=1$ in $G$, then $(1-e_x)\in R[G]$ is a zero divisor (where by $1$ we technically mean $1e_1$).
  • For $G=\Z$, we have $R[G]\simeq R[t,\inv t]\simeq R[t,s]/(1-st)$

Now, I came across Kaplansky’s Criterion while a friend and I were trying to resolve the following:

Let $k$ be a field. Is the group ring $k[\Z^n]$ a UFD (or even a domain)?

This turns out to be true with a simple proof once you know that localization preserves UFD-ness.

Let $R$ be a UFD. Then, $R[\Z^n]$ is a UFD.
It is not hard to see that $$R[\Z^n]\simeq R[x_1,\inv x_1,\dots,x_n,\inv x_n]$$ We now proceed by induction. The claim is true when $n=0$ since $R[\Z^0]\simeq R$, so suppose that $n>0$ and that $A:=R[x_1,\inv x_1,\dots,x_{n-1},\inv x_{n-1}]$ is a UFD. Then, $A[x_n]$ is a UFD as is its localization $A[x_n]\sqbracks{\frac1{x_n}}$ but clearly, $R[\Z^n]\simeq A[x_n]\sqbracks{\frac1{x_n}}$ so $R[\Z^n]$ is a UFD as claimed.

At this point, the only remaining question is why would I care about that. Well, one of my friends mentioned the following problem, and I was surprised that this is not already known to be true, so we set about convincing ourselves it held in the simplest possible case: finitely generated abelian groups.

Let $K$ be a field, and let $G$ be a torsion-free group. Then, the group ring $K[G]$ does not contain nontrivial zero divisors.

I should maybe mention that this is only one of Kaplansky’s conjectures, so the name isn’t well-defined.

  1. Depending on how I’m feeling when I get to that part, I might also say why I care about this fact and/or “plug a hole” in this blog by proving that R[x] is a UFD when R is a UFD since I didn’t last time I had a chance to. 

  2. I won’t get into why it’s called this, but it’s related to studying functions defined near a point of a space by looking at the functions that do not vanish at that point. 

  3. “functor” 

  4. Two? 

  5. theorem, not lemma or corollary 

  6. The kicker is that proving that any ideal is contained in a maximal one involves using Zorn’s lemma, so we still appeal to Zorn’s lemma; we just do so more implicitly. 

  7. Or at least the second one 

  8. I’m not gonna lie: Kaplansky’s Criterion has way more applications than I realized when I started writing this post. Why isn’t it more popular? 

  9. I mistakingly thought that Kaplansky’s Criterion gave a simple proof of this along the lines of the localization proof: “Pick a prime ideal I of R[x] and intersect it with R to get a prime ideal J of R. Then J contains a prime element p which remains prime in R[x] so I contains the prime element p. Apply Kaplansky and you win.” The issue with this is that J may be the zero ideal of R, in which case you have to go through all the annoyingness of the standard proof that I’d rather avoid writing up. 

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