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

  1. $\mathbb{Q}[X]$
  2. $\bigcup_{n=1}^{\infty}\mathbb{Q}[x^\frac{1}{n}]$
  3. $\mathbb{Q}[X,Y,Z]$
  4. $\mathbb{Z}[\frac{1}{2}]$

Here are my thoughts thus far:

  1. $\mathbb{Q}$ 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.

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up vote 4 down vote accepted
  1. $\,$ A univariate polynomial ring $\,F[x]\,$ over a field is $\rm ED$ (Euclidean) $\,\Rightarrow\rm PID \Rightarrow UFD$

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

  3. $\,$ 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.

  4. $\,$ 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 $\,r.$

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$2$. 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 $S=\{1,2,2^2,\dots\}$).

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Quick tip: in English, "a UFD" and "a unique factorization domain" are correct, but the "an" is considered incorrect. The rule about using the article "an" for vowels only applies to vowel sounds not vowel letters. So you would also say "a Euclidean domain" or "an ED". – rschwieb Jul 18 '13 at 23:16

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