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Let $k$ be a field with char(k) $\neq 2$. How can we decompose into irreducible components the following set: $Z(x^{2}+y^{2}+z^{2},x^{2}-y^{2}-z^{2}+1) \subseteq \mathbb{A}^{3}$?

Doing the algebra leads to $x^{2}+\frac{1}{2}=0$ and $y^{2}+z^{2}=\frac{1}{2}$. And then?

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up vote 2 down vote accepted

The ideal $I=(x^2+y^2+z^2,x^2-y^2-z^2+1)$ can also be written $I=(x^2+1/2,y^2+z^2)$, as you have computed yourself.

Now you must study the scheme $S=Spec(k[x,y,z]/I)$.
The trick is to write $k[x,y,z]/I=k[x]/(x^2+1/2)\otimes_k k[y,z]/(y^2+z^2)$, so that the scheme $S$ is the product $S=T\times U$ where $T=Spec(k[x]/(x^2+1/2))$ and $U=Spec( k[y,z]/(y^2+z^2))$.

You will then have to discuss cases , according as each of $-2$ and $-1$ is or not a square in $k$.
For example, if $k$ is algebraically closed then $S$ has four irreducible components.

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The answer depends on $k$. If you can solve for $x$ and $y^2 + z^2 = 1/2$ is irreducible, the solution is given by the two varieties $Z(x-x_i, y^2 + z^2 = 1/2)$, where $x_i$ is a solution of $x^2 + 1/2=0$.

If $y^2 + z^2 = 1/2$ is not irreducible, its factors, together with $x-x_i$, will generate the ideals for the irreducible components.

Finally, if $x^2 + 1/2=0$ has no solution, the variety is empty.

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