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I would like to prove that $A = \mathbb{C}[x,y,z]/(x^2+z^2-1, x^2-y^2-z^2)$ is an integral domain.

I feel that it would be easy enough to prove the denominator is prime using the techniques of Grobner bases, but my professor suggests another way. He says to consider the inclusions $\mathbb{C}[x]$ $\subset$ $B = \mathbb{C}[x,z]/(z^2-(1-x^2))$ $\subset$ $A$ and use a result which says that $k[x,y]/(y^2-f(x))$ is an integral domain if and only if $f(x)$ is not a square in $k[x]$. This shows $B$ is a domain, but here I get stuck. I consider $A=B[y]/(y^2 - (x^2-z^2))$ and I can see that the denominator is irreducible, but $B$ is not a UFD, so irreducible is not necessarily equivalent to prime. How should I proceed?

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Hi, I haven't tried Grobner basis, so I don't have an answer. Doesn't the relation defining B make it not a UFD, since z*z = (1-x)(1+x)? – Cass Nov 24 '12 at 13:36
Ah, I see. I will have to refresh myself on the algorithm for Grobner basis. I believe that the condition you derived in your latest post is correct, since when I assume there are zero divisors, I derive the condition that $a^2(x^2-z^2)=b^2$ for some $a,b$ in $B$. Is it obvious that this cannot happen in this case? – Cass Nov 24 '12 at 13:47
Those elements are prime. $C[x,z]$ is a UFD by Gauss's lemma, and they are clearly irreducible. – Cass Nov 24 '12 at 13:49
up vote 5 down vote accepted

Since the question contains a mistake, let me fix it.

The ring $R=\mathbb{C}[X,Y]/(X^2+Y^2-1)$ is a UFD because it is isomorphic to $\mathbb{C}[U,V]/(UV-1)$ (via the substitutions $U\mapsto X+iY$ and $V\mapsto X-iY$) and the last one is a ring of fractions of $\mathbb{C}[U]$ (with respect to the multiplicative system $\{1,U,U^2,\dots\}$).

Remark. If $R$ is an integral domain and $\alpha\in R$, then $R[X]/(X^2−\alpha)$ is an integral domain if and only if there are no non-zero elements $a,b\in R$ such that $b^2=\alpha a^2$.

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Thank you. I also see my error regarding $1-x$ and $z$ being prime. – Cass Nov 24 '12 at 14:25

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