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Give an example of a UFD having a subring which is not a UFD.

I thought of $\mathbb{Z}[\sqrt{2},\sqrt{3}]$. Could you please explain my question. I am trying grasp the concepts, need help.

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    $\begingroup$ $\mathbb{Z}[\sqrt{-3}] \subset \mathbb{Z}[e^{2\pi i/3}]$. $\endgroup$ – Daniel Fischer Dec 20 '13 at 12:55
  • $\begingroup$ @DanielFischer I don't know if $\mathbb Z[e^{2\pi i / 3}]$ is a UFD ? $\endgroup$ – Complex analysis Dec 20 '13 at 13:04
  • $\begingroup$ @Complexanalysis It's a Euclidean ring. $\endgroup$ – Daniel Fischer Dec 20 '13 at 13:12
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Take any integral domain which is not a UFD and consider it as a subring of its field of fractions. (Fields are UFD for trivial reasons and if you don't accept this, take the polynomial ring over it)

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    $\begingroup$ how can you be so fast... :D ...Simple. elegant.. $\endgroup$ – user87543 Dec 20 '13 at 12:57
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Consider $\mathbb C$ , it is a field hence obviously $UFD $ but if u consider a subring $\mathbb Z[\sqrt {-5}]$ is not a UFD . $9=3.3$ and also $9= (2+\sqrt{5} ) . (2-\sqrt5)$ , Hence the factorization is not unique.

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One question to ask after reviewing the two answers already posted is whether the existence of an extension $S\subset R$, with $R$ a UFD and the nonunits of $S$ being nonunits in $R$, makes $S$ a UFD.

While plausible, this is also false. For instance $k[x^2,x^3]\subset k[x]$ is a counterexample for any UFD $k$. This is because $x^2$ and $x^3$ are irreducible and yet $x^6=(x^2)^3=(x^3)^2$.

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$\mathbb{Z}[2i]$ as a subset of $\mathbb{Z}[i]$. First one is not integrally closed, so can't be UFD, second is Euclidean.

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Maybe I am beating the dead horse, but...

$\mathbb{Q} + X \mathbb{R}[X] \subset \mathbb{R}[X]$

is one of many similar examples.

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