Which of these Rings are Euclidean Domains and which are UFDs? I want to determine which of the following cases are the rings Euclidean Domains and in which they are UFDs. 


*

*$\mathbb{Q}[X]$

*$\bigcup_{n=1}^{\infty}\mathbb{Q}[x^\frac{1}{n}]$

*$\mathbb{Q}[X,Y,Z]$

*$\mathbb{Z}[\frac{1}{2}]$


Here are my thoughts thus far:


*

*$\mathbb{Q}$ is a field and so $\mathbb{Q}[X]$ is a Euclidean Domain. It is also a UFD (I think I have shown correctly why this is the case, so I only need comments on this if it is in fact not a UFD).


I'm really struggling for ideas for the other cases. Guidance would be very appreciated.
 A: *

*$\,$ A univariate polynomial ring $\,F[x]\,$ over a field is $\rm ED$ (Euclidean) $\,\Rightarrow\rm PID \Rightarrow UFD$

*$\ \bigcup_{n=1}^{\infty}\mathbb{Q}[x^\frac{1}{n}]\,$ fails ACCP by
$(x) \subsetneq (x^{1/2}) \subsetneq (x^{1/4}) \subsetneq\,\ldots\ $ so it's $\ \lnot \rm UFD\,\Rightarrow\,\lnot ED$

*$\,$ Polynomial rings over UFDs are UFDs, thus so is $\,\Bbb Q[x,y,z],\,$ but it's $\,\lnot\rm PID\,\Rightarrow\,\lnot ED,\,$ since $\,(x,y)\,$ is not principal.

*$\,$ Localization preserves Euclidean domains.  Hint:  lift the Euclidean function from $\,\Bbb Z\to \Bbb Z[1/2]\,$ by ignoring all factors of $\,2,\ $ i.e. for odd  $\,a,b\in\Bbb Z,\,$ $\ b 2^i\mid a 2^j \in \Bbb Z[1/2]\iff b\mid a\in \Bbb Z.\,$ If $\,b\nmid a\,$  then $\, 0 < r = a-qb < b,\,$ so $\,0 < v(r) \le r < b = v(b 2^j),\,$ where $\,v(r)\,$ is the odd part of $\,r.$ 
A: $2$. This ring is not an UFD (since $x=(x^{\frac{1}{n}})^n$ has many different decompositions).
$3$. $\mathbb{Q}[X,Y,Z]$ is an UFD, but is not Euclidean (since it is not a PID).
$4$. Every ring of fractions of an Euclidean domain is also Euclidean, so in this case your ring is Euclidean (since $\mathbb Z[\frac{1}{2}]=S^{-1}\mathbb Z$, where $S=\{1,2,2^2,\dots\}$). 
