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?

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    $\begingroup$ You need the field to be infinite. $\endgroup$ Oct 31, 2020 at 16:32

1 Answer 1


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 $f \mid g$.


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