Finding $I(V)$ for $V = V(x^2 + y^2 - z^2, 2z^2 - y)$

Question

I'm trying to find the ideal of the affine variety $V(x^2 + y^2 - z^2, 2z^2 - y) \subseteq \mathbb{A}_{\mathbb{C}}^3$ in order to calculate its singularities. Is this some well-known variety or is there any common trick to prove the ideal $(x^2 + y^2 - z^2, 2z^2 - y)$ is prime (in case it is)? This came up during a problem in which I was supposed to calculate the singularities of $V$. I know the height of $I(V)$ must be $2$, since $V(x^2 + y^2 - z^2)$ is an affine variety of dimension $2$, and $V(x^2 + y^2 - z^2, 2z^2 - y)$ is stated to be an irreducible variety. Directly calculating the Jacobian of $(x^2 + y^2 - z^2, 2z^2 - y)$ gives $\dim V \leq 1$ too, assuming these are the correct generators. But this is not necessarily enough, since if $x^2 + y^2 - z^2, 2z^2 - y$ are not the generators of $I(V)$, I could be getting more (false) singularities. Maybe I'm missing something terribly obvious and the justification is easy to find.

Answer 1

I'll show you a way to rigorously prove that $\mathbb C[x,y,z] / (x^2 + y^2 - z^2, 2z^2 - y)$ is an integral domain (i.e. $(x^2 + y^2 - z^2, 2z^2 - y)$ is a prime ideal in $\mathbb C[x,y,z]$).

Step 1: $\mathbb C[x,y,z] / (x^2 + y^2 - z^2, 2z^2 - y) \cong \mathbb C[x, z] / (x^2 - z^2 + 4z^4)$.

Consider the ring homomorphisms

• $\phi: \mathbb C[x,y,z] / (x^2 + y^2 - z^2, 2z^2 - y) \to \mathbb C[x, z] / (x^2 - z^2 + 4z^4)$, defined by $$ \phi\left( \left[ f(x, y,z)\right]_{\text{mod } (x^2 + y^2 - z^2, 2z^2 - y)} \right) = \left[ f(x, 2z^2, z) \right]_{\text{mod } (x^2 - z^2 + 4z^4)},$$
• $\psi: \mathbb C[x, z] / (x^2 - z^2 + 4z^4) \to \mathbb C[x,y,z] / (x^2 + y^2 - z^2, 2z^2 - y) $, defined by $$ \psi\left( \left[ g(x, z)\right]_{\text{mod } (x^2 - z^2 + 4z^4)} \right) = \left[ g(x, z) \right]_{\text{mod } (x^2 + y^2 - z^2, 2z^2 - y)}.$$

The work you have to do is to verify that:

• $\phi$ and $\psi$ are well-defined, which is to say that the expressions we've used to define these maps are independent of our choices of representatives of residue classes.
• $\phi$ and $\psi$ really are ring homomorphisms. (In fact, they are $\mathbb C$-algebra homomorphisms too.)
• $\phi$ and $\psi$ are inverses of each other.

None of this is particularly difficult.

Step 2: $\mathbb C[x,z ] / (x^2 - z^2 + 4z^4)$ is an integral domain (i.e. $(x^2 - z^2 + 4z^4)$ is a prime ideal in $\mathbb C[x, z]$.

We can start by observing that $\mathbb C[x, z]$ is a unique factorisation domain.

To show that $(x^2 - z^2 + 4z^4)$ is a prime ideal in $\mathbb C [x, z]$, we must show that $x^2 - z^2 + 4z^4$ is a prime element in $\mathbb C [x, z]$. But $\mathbb C[x, z]$ is a unique factorization domain, so it is sufficient to show that $x^2 - z^2 + 4z^4$ is an irreducible element in $\mathbb C[x, z]$.

So suppose for the sake of argument that $x^2 - z^2 + 4z^4$ factorises as $$ x^2 - z^2 + 4z^4 = p(x, z) q(x, z),$$ where $p(x, z)$ nor $q(x, z)$ are non-units (i.e. non-constants) in $\mathbb C[x, z]$.

If you think about this for a moment, you'll see that, up to a trivial rescaling by constants, $p(x, z)$ and $q(x, z)$ must take the forms

$$ p(x, z) = x + a(z), \qquad q(x, z) = x + b(z)$$ for some polynomials $a(z), b(z) \in \mathbb C[z]$ such that $$ a(z)b(z) = -(z^2 - 4z^4), \qquad a(z) + b(z) = 0.$$

The $a(z) + b(z) = 0$ equation tells us that $a(z) = -b(z)$. The $a(z)b(z) = -(z^2 - 4z^4)$ equation then tells us that $a(z)^2 = z^2 - 4z^4$, i.e. $z^2 - 4z^4$ must be a perfect square in $\mathbb C[z]$.

Now $z^2 - 4z^4$ factorises in the UFD $\mathbb C[z]$ as $$ z^2 - 4z^4 = z \cdot z \cdot (1 - 2z)\cdot (1 + 2z).$$ So $z^2 - 4z^4$ is not a perfect square in $\mathbb C[z]$, giving a contradiction.