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Ring of functions that are polynomials in $\cos t$ and $\sin t$, with real coefficients

Let $S=\mathbb R[X,Y]/(X^2+Y^2-1)$. Is the ring $S$ a UFD?

We obviously have a factorization $x^2=(1-y)(1+y)$ here but I have trouble showing that $x,1-y$ are irreducible, or that they are not equivalent (i.e. one is multiple of another by a unit in this ring). How would one go about proving these?

Is there an easier proof using the equivalence of UFD with Krull domain in which every height 1 prime ideal is principal?

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Bernard
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    Using SEARCH will find this http://math.stackexchange.com/questions/244460/ring-of-functions-that-are-polynomials-in-cos-t-and-sin-t-with-real-coeff/244487#244487 –  Dec 16 '12 at 18:31
  • I hadn't seen that, true, but my question is not answered there anyway (namely why x, 1-y are irreducible and if you have a second proof using the equivalent definition above). So I don't see why the topic was closed. – Bernard Dec 17 '12 at 08:53
  • @Bernard: It might help you to complete the missing parts, if you first map $S$ into the ring of complex Laurent polynomials as described in the linked question. – Jyrki Lahtonen Dec 17 '12 at 10:19
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    I've added some more details to my answer there. If $S$ is an UFD, then (without any reference to Krull domains) every height $1$ prime ideal is principal. If I understand well you want to prove that this is not the case here. I suggest you to try to see if the prime ideal $(x,y-1)$ is principal or not (its height is $1$). –  Dec 17 '12 at 11:05
  • @YACP Thank you. Yes I tried that but I showed it IS principal. I'm probably wrong but here is how it goes: S is normal so localizations at maximal ideals are normal. Let $m$ any codimension 1 prime ideal in S; then $S_m$ is a Noetherian local domain of dimension 1 which is normal, hence a DVR, hence principal. What am I missing here? – Bernard Dec 17 '12 at 18:08
  • If $mS_m$ is principal this doesn't mean that $m$ is principal. –  Dec 17 '12 at 18:17
  • Dear Bernard, I have added another answer to the previous question on this which is more geometric, and possibly of interest (depending on your background). Regards, – Matt E Dec 17 '12 at 21:53

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