1

Let $X = \lbrace x_{\alpha}\rbrace_{\alpha \in I}$ be a set of variables and $F[X]$ the ring of polynomials in the variables from $X$. This ring can be thought of as the union of all the rings $F[x_{\alpha_{1}}, ..., x_{\alpha_{n}}]$, as the union ranges over all finite subsets $\lbrace x_{\alpha_{1}}, ..., x_{\alpha_{n}} \rbrace$ of $X$. Show that $F[X]$ is a UFD if $F$ is a field.

I know that:

"$F$ field implies $F[x]$ UFD" and "$R$ UFD implies $R[x]$ UFD". Thus, I can show by induction that $F[x_{\alpha_{1}}, ..., x_{\alpha_{n}}]$ is a UFD.

If it's right so far, how do I proceed with the union?

user26857
  • 53,190
Lucas
  • 4,260
  • 1
  • 21
  • 57
  • Perhaps this question could help you : https://math.stackexchange.com/questions/635185/if-p-in-rx-1-dots-x-n-is-irreducible-is-it-still-irreducible-in-rx-1-do?rq=1 – Arnaud D. Mar 21 '18 at 16:57

1 Answers1

4

Obviously an element of the union has at least one factorization, because it is a member of one of the UFD's in the union.

Suppose now there are two complete factorizations of your single element in the union. Only finitely many variables can be mentioned in both factorizations, so in fact your factorizations (both of them) are contained in a finite set of variables $X'\subseteq X$, and you know for sure that $F[X']$ is a UFD. So in fact those two factorizations must be equivalent in that ring. So, the factorization is unique up to equivalence, and the union is a UFD.

rschwieb
  • 160,592