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$\newcommand{\Q}{\mathbb Q}$ I saw an argument that the ideal $I=(x^2+y^2-1, z^2+w^2-1)$ is a prime ideal in $\Q[x,y,z,w]$ but I cannot see why. I tried to find a surjective homomorphism from $\Q[x,y,z,w]$ onto some integral domain with kernel $I$ but in vain. Or should I consider to show the set $\{(x,y,z,w)\mid x^2+y^2=1,z^2+w^2=1\}\subset\Q^4$ being irreducible?

Thanks in advance...

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    $\begingroup$ Hm, if we were working over an algebraically closed field $k$, we could write $\frac{k[x,y,z,w]}{(x^2+y^2-1, z^2+w^2-1)} \cong \frac{k[x,y]}{(x^2 + y^2 - 1)} \otimes_k \frac{k[z,w]}{(z^2+w^2-1)}$ and then use the fact that the product of irreducible varieties is a variety. But since $\mathbb{Q}$ is not algebraically closed, I'm not sure that will work... $\endgroup$ – André 3000 Nov 2 '18 at 14:54
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    $\begingroup$ Maybe, there is a general result like this. "Let $A$ and $B$ be associative algebras over $K$ (an integral domain or a field, I don't know). If $A$ and $B$ are themselves integral domains, then $A\otimes_K B$ is also an integral domain." If there is something like that then we can use $$\mathbb{Q}[x,y,z,w]/(x^2+y^2-1,z^2+w^2-1)\cong \big(\mathbb{Q}[x,y]/(x^2+y^2-1)\big)\otimes_\mathbb{Q}\big(\mathbb{Q}[z,w]/(z^2+w^2-1)\big)$$ @André3000 suggested. Can anybody confirm or contradict the quote? $\endgroup$ – user593746 Nov 2 '18 at 17:15
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    $\begingroup$ This answer might be relevant. $\endgroup$ – Ennar Nov 2 '18 at 17:20
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    $\begingroup$ @Zvi I think your conjecture is false in that level of generality because of varieties that are irreducible, but not geometrically irreducible. Take $K = \mathbb{R}$, $A = \mathbb{R}[x,y]/(x^2 + y^2)$ and $B = \mathbb{C}$. Then $A$ and $B$ are domains, but since $x^2 + y^2 = (x + iy) (x-iy)$ factors over $\mathbb{C}$, then $A \otimes_\mathbb{R} B$ is not a domain. $\endgroup$ – André 3000 Nov 2 '18 at 19:23
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    $\begingroup$ @Zvi I think this answer in the same spirit as your conjecture might help, though. This question is also related, though it again is working over an algebraically closed field. $\endgroup$ – André 3000 Nov 2 '18 at 19:24
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It is well known that the product of geometrically irreducible varieties is again geometrically irreducible (this is equivalent to the statement that if $A,B$ are $k$-algebras for $k$ algebraically closed and $A,B$ have no zero-divisors, then $A\otimes_k B$ has no zero divisors). We show that $\operatorname{Spec}\Bbb Q[x,y]/(x^2+y^2-1)$ is geometrically irreducible.

Base changing to the algebraic closure $\overline{\Bbb Q}$, we get that our variety is $\operatorname{Spec}\overline{\Bbb Q}[x,y]/(x^2+y^2-1)$. So it suffices to show that $x^2+y^2-1$ is irreducible over $\overline{\Bbb Q}$. Up to the linear change of coordinates $x=x+iy,y=x-iy$, this polynomial factors as $xy-1$, which can quickly be shown to be irreducible (it could only factor as a product of linear polynomials, but you can see for yourself that it doesn't).

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