For which fields $k$ does the following hold for all $n \geq 1$? Let $(a\_1,\ldots,a\_n) \in k^n$. Then there exists a polynomial $f(x\_1,\ldots,x\_n) \in k[x\_1,\ldots,x\_n]$ such that $f(a\_1,\ldots,a\_n) = 0$ but $f(b\_1,\ldots,b\_n) \neq 0$ for all $(b\_1,\ldots,b\_n) \neq (a\_1,\ldots,a\_n)$?


This is clearly impossible for $k$ algebraically closed.


It does hold for $k$ a subfield of $\mathbb{R}$; indeed, in that case we can take 


$$f(x\_1,\ldots,x\_n) = (x\_1-a\_1)^2 + (x\_2-a\_2)^2 + \cdots + (x\_n-a\_n)^2.$$



Fields for which there exist multivariable polynomials vanishing at single specified point
If $k$ is not algebraically closed, such a polynomial always exists (the opposite is also true and is mentioned in the post). 


We may assume that $a\_i=0$ for all $i$. Take an irreducible polynomial $g(x)$ of degree $d>1$, then for the homogeneous form $G(x,y)=y^dg(x/y)$ we have $G(x,y)=0$ if only if $x=y=0$. This solves the case $n=2$, for $n=3$ consider the polynomial $G(G(x,y),z)$, it takes zero value only when $x=y=z=0$, and so on. 


