# Verifying Hilberts Nullstellensatz on a particular example

Let $k$ be an algebraically closed field of characteristic $2$ and consider the following equations:
$$xy + z^2 = 0$$
$$uv + w^2 = 0$$
$$uy + vx = 0$$
It’s not hard to parameterize solutions to these equations, there are $4$ cases depending on whether the values of $u$ and $x$ are zero or not. In each case you can verify that the solutions also satisfy
$$vz + wy = 0$$
so according to Hilberts Nullstellensatz there should be an integer $n$ such that $(vz + wy)^n$ is contained in the ideal $I = (xy + z^2, uv + w^2, uy + vx) \subseteq k[u, v, w, x, y, z]$.

This is all fine and well so far, here’s the catch and my question. If $k = \mathbb F_2$ then the ideal $I$ is prime and does not contain $vz + wy$ (obviously the Nullstellensatz doesn’t apply), so I can’t use Macaulay2 to find $n$ and tell me how to write $(vz + wy)^n$ as a linear combination of the generators of $I$. Is there some other way to compute this? I’d like to actually see what the expression looks like so that I can get a better idea of how the larger field changes whether the ideal $I$ is prime or not.

#### Solutions Collecting From Web of "Verifying Hilberts Nullstellensatz on a particular example"

Modulo the ideal we have

$$(vz)^2=v^2 z^2 =v^2 xy = v uy y = w^2 y^2 = (wy)^2$$

Hence $(vz+wy)^2$ lies in the ideal. This works over every $\mathbb{F}_2$-algebra.