Let $\mathbb R$ be the field of real numbers. From Nagata's criterion for unique factorization domains, it follows that $\frac{\mathbb R[X_1,\ldots,X_n]}{(X_1^2+\ldots+X_n^2)}$ is a unique factorization domain for all $n\ge 5$. But if $n=2$, then it is an integral domain which is not even normal. What happens from $n=3$ and $n=4$? Are they unique factorization domains? Of course, if we take $\mathbb C$ instead of $\mathbb R$, then they are not UFDs. I suspect that the presence of $\sqrt{-1}$ in the base field will perhaps dictate the answer. Dietrich Burde pointed out the following link see here which answers my question in affirmative for $n=3$. But for $n=4$, we need to answer the related question - Is $\frac{\mathbb R[X,Y,Z]}{(X^2+Y^2+Z^2+1)}$ a UFD?
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Let me check. Thanks. – sagnik chakraborty Jul 01 '22 at 13:30
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How to prove that $\mathbb R[X,Y,Z]/(X^2+Y^2+Z^2+1)$ is a UFD? – sagnik chakraborty Jul 01 '22 at 13:46
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Please don't change the title. You can make an edit, and add more to it. – Dietrich Burde Jul 01 '22 at 14:09
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Okay, thank you. – sagnik chakraborty Jul 01 '22 at 14:11
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By the way, with $-1$ it is always a UFD. See here, how to use Nagata's criterion. – Dietrich Burde Jul 01 '22 at 14:13
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Yes, this one i know. – sagnik chakraborty Jul 01 '22 at 14:19
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So I'm unclear: the question isn't the title question anymore? I think if it's not then the title really SHOULD be changed. – rschwieb Jul 01 '22 at 15:46
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@rschwieb The title question question was a duplicate , but then the OP changed it after I had said this. This felt a little bit strange. – Dietrich Burde Jul 01 '22 at 16:32
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@sagnikchakraborty Is the question in the final paragraph now the main question? – rschwieb Jul 01 '22 at 17:07
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I thought Dietrich requested me not to change the title. So i did not change it. Yes, now the question is whether $\frac{\mathbb R[X,Y,Z]}{(X^2+Y^2+Z^2+1)}$ is a UFD or not. Please let me know if i should change the title accordingly. – sagnik chakraborty Jul 01 '22 at 18:49
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@sagnikchakraborty In general you should not edit questions in such a way as to invalidate answers. That said, you have no answers at the moment, and probably have time to update your question. It is definitely not good right now that your title question doesn't match your real question. – rschwieb Jul 01 '22 at 20:53
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Why Nagata's criterion only works for $n\geq5$? – Bromelain Jul 01 '22 at 22:08
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@rschwieb Ok, got the point. So i'm changing the title. – sagnik chakraborty Jul 02 '22 at 04:38
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@Bromelain Because the assumptions required to apply Nagata's criterion forces us to take $n\ge 5$. For a proof of it, you may look at the original paper titled `A remark on unique factorization theorem'. – sagnik chakraborty Jul 02 '22 at 04:44