Adjunction of a root to a UFD

Question

Let $R$ be a unique factorization domain which is a finitely generated $\Bbbk$-algebra for an algebraically closed field $\Bbbk$. For $x\in R\setminus\{0\}$, let $y$ be an $n$-th root of $x$. My question is, is the ring

$$ A := R[y] := R[T]/(T^n - x) $$

a unique factorization domain as well?

Edit: I know the classic counterexample $\mathbb{Z}[\sqrt{5}]$, but $\mathbb{Z}$ it does not contain an algebraically closed field. I am wondering if that changes anything.

Edit: As Gerry's Answer shows, this is not true in general. What if $x$ is prime? What if it is a unit?

Answer 1

1) No, you won't get a UFD even if you take for $x$ the most unitesque of units: $1_R$ !
Namely, every $R$ will give you a counterexample:if $n$ is not divisible by $char.\mathbb k $, the Chinese remainder theorem gives
$$A=R[T]/(T^n-1)=R^n$$ which is not even a domain!

2) No, you won't get a UFD even if you take for $x$ a prime element of $R$.
Here is an example. Let $R=\mathbb C[Y,Z]$and $x=1-Y^2-Z^2$, a prime (= irreducible) element in $R=\mathbb C[Y,Z]$.
Then $$A=\mathbb C[Y,Z,T]/(T^2-x)=\mathbb C[Y,Z,T]/(T^2+Y^2+Z^2-1)$$ is not a UFD because $(t+iy)(t-iy)=(1+z)(1-z)$ in $ A$
(There is some checking to do in order to show that these are a genuinely different factorizations. If you need to quote a reference, look at page 165 of Samuel's paper here )

Answer 2

${\bf C}[x,y]$ is a UFD, but ${\bf C}[x,y,z]/(z^2-xy)$ isn't. Is that the kind of thing you had in mind?

Answer 3

I think you can get a counterexample to the unit question, even in characteristic zero, and even in an integral domain (in contrast to Georges' example), although there are a few things that need checking.

Let $R={\bf C}[x,1/(x^2-1)]$, so $1-x^2$ is a unit in $R$. Then $$(1+\sqrt{1-x^2})(1-\sqrt{1-x^2})=xx$$

It remains to check that

1. $R$ is a UFD,

2. $1\pm\sqrt{1-x^2}$ and $x$ are irreducibles in $R[\sqrt{1-x^2}]$

3. $1\pm\sqrt{1-x^2}$ and $x$ are not associates in $R[\sqrt{1-x^2}]$