# Nullstellensatz for hyperplanes over a general field?

Let $$\mathbb{F}$$ be an arbitrary field and consider a polynomial of degree one: $$f(x_1,\ldots,x_n)=a_1x_1+\cdots+a_nx_n+b\in \mathbb{F}[x_1,\ldots,x_n].$$ Let $$H:f=0$$ be the corresponding affine hyperplane and let $$g\in\mathbb{F}[x_1,\ldots,x_n]$$ be any polynomial that vanishes on $$H$$. If $$\mathbb{F}$$ is algebraically closed then the classical Nullstellensatz tells us that $$f$$ divides $$g$$.

Question: Do we really need algebraic closure in this case?

• You need the field to be infinite. Oct 31, 2020 at 16:32

Applying a linear change of coordinates to the variables, we can assume without loss of generality that $$f = x_1$$. Let $$h(x_2, \dots, x_n) = g(0, x_2, \dots, x_n)$$, considered as an element of $$\mathbb{F}[x_2, \dots, x_n]$$. Then $$h(a_2, \dots, a_n) = 0$$ for all $$a_2, \dots, a_n \in \mathbb{F}$$. This shows how we can construct counterexamples: if $$\mathbb{F}$$ is finite, then there are polynomials that vanish on all inputs, e.g., $$\prod_{a \in \mathbb{F}} (x - a)$$. However, if $$\mathbb{F}$$ is infinite, then polynomial interpolation shows that the polynomial $$h$$ must be identically zero, which means $$f \mid g$$.