How to prove that $k[x,x^{-1},y]$ is a UFD

Question

Let $k$ be a field. $k[x,y]$ is a UFD by the following known argument taken from wikipedia: "If $R$ is a UFD, then so is $R[X]$, the ring of polynomials with coefficients in $R$. Unless $R$ is a field, $R[X]$ is not a principal ideal domain. By induction, a polynomial ring in any number of variables over any UFD (and in particular over a field) is a UFD".

(1) It seems that $k[x,x^{-1},y]$ is a UFD, isn't it? Is the proof based on the result that $k[x,y]$ is a UFD? (probably yes?)

(2) Can one find all irreducible=prime elements of $k[x,x^{-1},y]$? ("In UFDs, every irreducible element is prime").

Thank you very much!

Answer 1

1. The ring of Laurent polynomials $R=k[x,x^{-1}]=\bigl(k[x]\bigr)_{x}$, and a ring of fractions of a U.F.D. is a U.F.D., so $k[x,x^{-1},y]=R[y]$ is a U.F.D.
2. You can find the irreducible elements of $k[x,x^{-1},y]$ inasmuch as you can find those of $k[x,y]$. They're the same as the irreducible elements in $k[x,y]$ except $x$, which becomes a unit in $k[x,x^{-1},y]$. Which elements in $k[x,y]$ are irreducible depends on the base field.

Answer 2

1) Localization of a UFD is again a UFD (at least if you are not inverting $0$). Maybe you can try to prove that.

2) Is there any condition on $k$?