Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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?

share|improve this question
add comment

2 Answers

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.

share|improve this answer
add comment

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.

share|improve this answer
add comment

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.