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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.

Jim
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1 Answers1

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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.

Martin Brandenburg
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  • This is really weird. I've just double checked and Macaulay2 insists that over $\mathbb F_2$ the ideal $I$ is prime and $vz + wy$ is not contained in $I$, but it also agrees with you that $(vz + wy)^2$ is contained in $I$. This is a contradiction right? Am I missing something completely obvious? – Jim Apr 09 '13 at 17:19
  • Hm. Can you show by hand that $vz+wz \notin I$? Because then we know that $I$ is not prime. – Martin Brandenburg Apr 09 '13 at 18:02
  • Well, by degree considerations it would have to be a $k$-linear combination of the generators, which it is not. – Jim Apr 09 '13 at 18:05
  • I agree. So perhaps there is a bug in the software? – Martin Brandenburg Apr 09 '13 at 21:40
  • Yea, I guess I'll go report it. That's a little nerve-racking as I've used Macaulay2 calculations in my research.... – Jim Apr 09 '13 at 23:10