$\newcommand{\Q}{\mathbb Q}$ 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...
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).