On the existence of a polynomial in several variables with only one zero

Question

Given an ordered field $\mathbb{K}$, then for all $n>1$ there exists $f\in\mathbb{K}[X_1,\ldots,X_n]\ s.t.\ \mathcal{V}^{\mathbb{K}^n}(f):=\{p\in\mathbb{K}^n:f(p)=0\}=\{(0,\ldots,0)\}$

For instance, the unforgettable $\sum_{k=1}^nX_n^2$.

On the other hand, Hilbert's Nullstellensatz shows that if $\mathbb{K}$ is algebraically closed a polynomial as such doesn't exist.

So I started wondering for which fields $\mathbb{K}$ and $n>1$ $\ \exists f\in\mathbb{K}[X_1,\ldots,X_n]\ \mathcal{V}^{\mathbb{K}^n}(f)=\{(0,\ldots,0)\}$

I do not hope for a complete solution (but I wouldn't dislike it either): a little bestiary of examples and/or necessary conditions would be more than appreciated.

For instance: what happens in finite fields?

Chevalley-Warning theorem, paired with the observation that $\forall \alpha \in \mathbb{F}_q\ \alpha^q=\alpha$ made me think that such polynomials did not exist, but I soon proved myself wrong by realizing that $xy+x+y$ works for $n=2$ in $\mathbb{F}_2$ and $\mathbb{F}_4$.

Thanks for any help.

Answer 1

Examples of such polynomials are (obviously the field won't be algebraically closed):

1. If $\Bbb{K}$ is a finite field of $q$ elements, then $f(x)=x^{q-1}$ satisfies $f(0)=0$ and $f(z)=1$ for all $z\in\Bbb{K}, z\neq0$. Therefore we easily get several variable versions such as $$F(x_1,x_2,\ldots,x_n)=1-\prod_{j=1}^n(1-f(x_i)).$$ If any $x_i\in\Bbb{K}^*$, then the product is zero, and $F(x_1,x_2,\ldots,x_n)=1$.
2. If $\Bbb{L}/\Bbb{K}$ is a degree $n$ extension of fields, then the norm map $N^L_K:\Bbb{K}\to\Bbb{L}$ vanishes only at $0\in\Bbb{L}$. But elements of $\Bbb{L}$ can be viewed as vectors of $n$ elements of $\Bbb{K}$ turning the norm map into a polynomial in $n$ variables. For example, when $\Bbb{K}=\Bbb{Q}$, $n=3$, we can select $\Bbb{L}=\Bbb{Q}(\root3\of2)$. If $z=x_1+x_2\root3\of2+x_3\root3\of4$ is an arbitrary element of $\Bbb{L}$, then we can conclude that the cubic polynomial $$N(x_1,x_2,x_3):=N(z)=x_1^3+2x_2^3+4x_3^3-6x_1x_2x_3$$ is non-zero whenever $(x_1,x_2,x_3)\in\Bbb{Q}^3\setminus\{(0,0,0)\}.$