# Subring of a UFD [duplicate]

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.

## marked as duplicate by Cameron Buie, Eric Wofsey abstract-algebra StackExchange.ready(function() { if (StackExchange.options.isMobile) return; $('.dupe-hammer-message-hover:not(.hover-bound)').each(function() { var$hover = $(this).addClass('hover-bound'),$msg = $hover.siblings('.dupe-hammer-message');$hover.hover( function() { $hover.showInfoMessage('', { messageElement:$msg.clone().show(), transient: false, position: { my: 'bottom left', at: 'top center', offsetTop: -7 }, dismissable: false, relativeToBody: true }); }, function() { StackExchange.helpers.removeMessages(); } ); }); }); Mar 24 at 15:21

• Doesn’t subring automatically have $1$? – J. W. Tanner Mar 24 at 14:42
• @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. – Cameron Buie Mar 24 at 14:54

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).

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 $$t=\frac{t^3}{t^2}$$.

• 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). – Bill Dubuque Mar 24 at 15:30
• @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. :) – jgon Mar 24 at 15:38
• 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) – Bill Dubuque Mar 24 at 15:41

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.

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