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?



Nullstellensatz for hyperplanes over a general field?
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 $f \mid g$.


