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This question already has an answer here:

Let $R$ be a UFD with $1$, are the following two statements true?

  • If $S$ is a subring of $R$, then $S$ is also a UFD;

  • if $S$ is a subring of $R$, and $1\in S$, then $S$ is UFD.

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marked as duplicate by Cameron Buie, Eric Wofsey abstract-algebra Mar 24 at 15:21

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

  • $\begingroup$ Doesn’t subring automatically have $1$? $\endgroup$ – J. W. Tanner Mar 24 at 14:42
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    $\begingroup$ @J.W.Tanner: It depends on the text. Some don't require rings to have a unit, so (for example) $\{2n:n\in\Bbb Z\}$ is a ring with the usual integer operations. $\endgroup$ – Cameron Buie Mar 24 at 14:54
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Not in general. For example, note that $R=\Bbb C$ is a field, so trivially a UFD. However, consider $S=\Bbb Z[2i]:=\{x+2iy:x,y\in\Bbb Z\}.$ One can show fairly easily that $S$ is a subring of $R$ with unity. However, while $2$ is irreducible in $S,$ it is not prime in $S$. Hence, while $S$ is an integral domain (as a subring with unity of an integral domain), it is not a UFD. See here for more information on how we might go about proving this (and feel free to ask me questions if you're stuck).

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Generalizing the examples given by Bill Dubuque and Cameron Buie, we can use the fact that a UFD is normal (that is, it is integrally closed in its field of fractions) to give us a strategy for finding counterexamples.

Namely, take some UFD $R$, consider its field of fractions $K$, and then find a (unital) subring $S\subseteq R$ such that the field of fractions of $S$ is still $K$, and $S$ is not integrally closed, i.e., there is $\omega\in K\setminus S$ which satisfies a monic polynomial over $S$.

This gives us the counterexamples from both of the other examples. Cameron Buie's has $R=\Bbb{Z}[i]$, $S=\Bbb{Z}[2i]$, $\omega=i$. Bill Dubuque's has $R=\Bbb{Z}[\sqrt{n}]$, $S=\Bbb{Z}[2\sqrt{n}]$, and $\omega = \sqrt{n}$.

Both of these examples are of rings of integers, however, we can also find a subring of a polynomial ring that is not a UFD via this strategy. Consider $R=k[t]$, $S=k[t^2,t^3]$, $\omega = t$. We have $\omega^2-t^2=0$, so $\omega$ is integral over $S$. Note that $t$ is in the fraction field of $S$, since $t=\frac{t^3}{t^2}$.

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  • $\begingroup$ In case it was not obvious I chose the Rational Root form of integral closure since many beginners are not familiar with the more general ring-theoretic notion of integral closure (which is not really any more general here). $\endgroup$ – Bill Dubuque Mar 24 at 15:30
  • $\begingroup$ @Bill Dubuque, I upvoted both your answer and Cameron Buie's, I wrote this answer because I felt that it wasn't necessarily clear how those examples were arrived at. It's possible that this answer will be less useful to the OP, but I did think that it might be useful to someone. :) $\endgroup$ – jgon Mar 24 at 15:38
  • $\begingroup$ No doubt. I didn't elaborate since it's already been done in many prior answers (and the CRUDE folks will soon be deleting this thread due to the "quality" of the question) $\endgroup$ – Bill Dubuque Mar 24 at 15:41
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Pick any quadratic number ring UFD $\,R = \Bbb Z[\sqrt n]\,$ and let $\,S = \Bbb Z[2\sqrt n]\subset R.\,$ Notice that $\,w = (2\sqrt{n})/2 = \sqrt{n}\,$ is a fraction over $S$ and $\,w\not\in S\,$ but $w$ is a root of the monic polynomial $\,x^2 - n, \,$ contra the Rational Root Test (which is valid in any UFD). So $S$ is not a UFD.

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  • $\begingroup$ Did you switch $R$ and $S$? $\endgroup$ – J. W. Tanner Mar 24 at 15:23
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    $\begingroup$ @J.W.Tanner Yes, typos now fixed, thanks. $\endgroup$ – Bill Dubuque Mar 24 at 15:26

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