As far as integral domains go, the UFD or **unique factorisation domain** is certainly one of the nicest. In this post we recall a few facts about UFDs and look at a wider class of rings called GCD-domains, which are domains in which every two elements have a greatest common divisor, and explain a proof following Gilmer and Parker that a polynomial extension of a GCD domain is a GCD domain. Along the way, we will see many cool algebraic facts!

Let's recall that an integral domain is a UFD if every nonzero nonunit can be written uniquely (up to order and multiplication by a unit) as a product of irreducible (equivalently, prime) elements. The ultimate example of such a docile creature is the integral domain $ \mathbb{Z}$, the integers. Fields are UFDs as well, and a basic theorem of algebra is: if $ R$ is a UFD then $ R[x]$ is a UFD, where $ R[x]$ is the polynomial ring in one variable.

Thus we get that $ F[x_1,\dots,x_n]$ for $ F$ a field and $ \mathbb{Z}[x_1,\dots,x_n]$ are UFDs. given a UFD $ R$, however, it is not always true that $ R[[x]]$, the power series ring in one variable, is a UFD. On the bright side, if $ R$ is a principal ideal domain (hence a UFD), then $ R[[x]]$ is a UFD. The proof of this boils down to examining the evaluation map $ R[[x]]\to R$ at zero, which would tell us after a bit of thought that if $ P$ is a prime ideal in $ R[[x]]$ and $ P'$ is its image, then $ P$ is principal if $ P'$ is principal and $ x\not\in P$. We could then finish the proof by using the very useful fact that an integral domain is a UFD if and only if every prime ideal contains a principal prime ideal.

To reiterate, if $ R$ is a principal ideal domain then $ R[[x]]$ is a unique factorisation domain. **I wonder if the converse is true?** But let us move on.

### GCD Domains

A particular property of a UFD is that any two elements have a greatest common divisor, which can be found just by writing out the unique factorisation of two elements and finding the common factorisation. However, UFDs are a bit strange in that they satisfy the ascending chain conidition on principal ideals: i.e., there is no strictly increasing infinite chain of principal ideals. Sometimes, we might need to work with rings without this property, and yet still crave that every two elements have a greatest common divisor (GCD). So let us call an integral domain (after Kaplansky) a GCD-domain if every two elements have a GCD. These are sometimes also called HCF rings or pseudo-Bezout domains.

Why would we do this? Aside from the above-mentioned reason that the ascending chain condition on prinicpal ideals might be annoying at times, another reason is that GCD-domains share many of the same properties as UFDs, and so it is instructive to see how much the ascending chain condition actually comes into play.

For instance, every GCD-domain is integrally closed, just like a UFD. Moreover, if $ R$ is a GCD-domain, then $ R[x]$ is as well (here is a fun consequence: if $ R$ is integrally closed and $ K$ its quotient field then $ R = \cap V_i$ where $ V_i$ are all the valuation domains between $ R$ and $ K$. Every valuation domain in a GCD so $ V_i[x]$ is a GCD-domain, and hence $ R[x]$ is integrally closed. See [1]).

### Polynomial Extensions of GCD-Domains

In fact, much more is true. If $ G$ is a torsion-free abelian group and $ R$ is a GCD-domain then $ R[G]$, the group ring of $ G$ with coefficients in $ R$, is a GCD-domain. This is a result of Gilmer and Parker [2], and in that paper they also prove a similar result for cancellative semigroups with whose principal ideals are closed under finite intersection, which also implies the result for polynomial rings.

For some intuition, we shall go over their proof in [2] in the special case of a polynomial ring. To be precise:

**Theorem**. If $ R$ is a GCD domain then $ R[x]$ is a GCD-domain.

*Proof.*We call a polynomial $ f\in R[x]$ primitive if the greatest common divisor of its coefficients is $ 1$. We leave it to the reader to prove the following straightforward facts (or, to accept them as black boxes):

- If $ f\in R[x]$ is primitive and $ r\in R$ then in $ R[x]$, we have the equality of ideals $ (f)\cap (d) = (fd)$.
- Let $ K$ be the fraction field of $ R$. If $ f\in R[x]$ is primitive then $ fK[x]\cap R[x] = fR[x]$. Hint: use #1.

Now that we have these technical pieces out of the way, let's proceed to the proof. We let $ \gamma(f)$ denote a greatest common divisor of the coefficients of $ f$, which exists because $ R$ is a GCD-domain (this is sometimes called the *content* of $ f$). Let $ f,g\in R[x]$. The name of the game is to prove that they have a greatest common divisor. Write $ f = \gamma(f)f_1$ and $ g = \gamma(g)g_1$ so that $ f_1$ and $ g_1$ are primitive polynomials.

Let $ d = \mathrm{lcm}{\gamma(f),\gamma(g)}$. Since greatest common divisors exist in $ R$, so do least common multiples (prove this!). Since $ K[x]$ is a GCD-domain (since it is a UFD), $ f$ and $ g$ have a GCD $ t\in K[x]$. Of course, since we are working over $ K$, we may assume that $ t$ has coefficients in $ R$ and $ t$ is primitive.

Our claim is that in $ R[x]$, we have the equality of ideals $ fR[x]\cap gR[x] = (dt)R[x]$, so that in other words, $ dt$ is a GCD of $ f$ and $ g$ in $ R[x]$. Now, $ t\in f_1K[x]\cap R[x]$ so $ t\in f_1R[x]$ via Fact #2, and so $ dt\in fR[x]$. Similarly, $ dt\in gR[x]$. Thus $ (dt)\subseteq fR[x]\cap gR[x]$. We just need to show the reverse inclusion. Let $ h\in fR[x]\cap gR[x]$.

Write $ h = \gamma(h)h_1$ where $ h_1$ is primitive. Then $ \gamma(f) | \gamma(h)$ and $ \gamma(g) | \gamma(h)$ so $ d | \gamma(h)$ (characterising property of least common multiple). Now observe that $ hK[x]\cap R[x] = h_1K[x]\cap R[x] = h_1R[x]$.

However, also note that $ h_1R[x]\subseteq fK[x]\cap gK[x]\cap R[x] = tK[x]\cap R[x] = tR[x]$ and thus $ t | h_1$ and so $ dt | h$ and thus $ h\in (dt)$, which shows the reverse inclusion.

Note that this gives an easy algorithm to compute greatest common divisors in the polynomial ring $ R[x]$ where $ R$ is a GCD-domain, assuming you know how to compute greatest common divisors in $ R$. If $ f,g\in R[x]$, use the Euclidean algorithm to compute $ t$, the greatest common divisor in $ K[x]$, and put $ t$ into the form of a primitive polynomial with coefficients in $ R$. Then $ dt$ is the greatest common divisor of $ f$ and $ g$ in $ R[x]$, where $ d$ is the lowest common multiple of $ \gamma(f)$ and $ \gamma(g)$. So in particular, although $ \mathbb{Z}[x]$ is not a Euclidean domain, we can still use the Euclidean algorithm twice (once in $ \mathbb{Q}[x]$ and once in $ \mathbb{Z}$) to find a greatest common divisor of any two polynomials in $ \mathbb{Z}[x]$.

The above argument was abstracted by Gilmer and Parker in [2] to give a template for showing that an integral domain $ D$ is a GCD-domain whenever $ D_S$ is for particular multiplicatively closed sets $ S$, which applies in particular to $ D = R[x]$ and $ S = R – \{0\}$, since $ K[x]$ is always a GCD.

Interestingly, in their paper, they also give conditions for $ R[S]$, the semingroup ring, to be a principal ideal domain. This happens if and only if $ R$ is a field and $ R[S]$ is isomorphic to either $ R[x]$ or $ R[x,x^{-1}]$!

### References

[1] Kaplansky, Irving. Commutative rings. Chicago: University of Chicago Press, 1974.

[2] Gilmer, Robert, and Tom Parker. "Divisibility properties in semigroup rings." The Michigan Mathematical Journal 21.1 (1974): 65-86.