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.

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