Greatest Common Divisor Domains and Polynomial Extensions

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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