Let $k$ be a field which is not algebraically closed and $n>1$ be an integer . Then does there exist $f\in k[X_1,...,X_n]$ such that $Z(f) (:=\{(a_1,...,a_n)\in k^n : f(a_1,...,a_n)=0\}) =\{0\}$ ?

My Work : I was thinking like this : Considering $k$ as a sub-field of its algebraic closure $\bar k$, we have $k \subsetneq \bar k$ . Also, there is an irreducible polynomial $g(X) \in k[X]$ such that deg $g(X)>1$ . May be we could somehow apply Hilbert Nullstelensatz to $\bar k$, but I am not sure .

I can easily do it for a finite field $k$; in that case just take

$f(X_1,...,X_n)=\sum_{(y_1,...,y_n)\in k^n \setminus \{0\} } \prod_{i=1}^n (1-(X_i-y_i)^{q-1})$ , where $|k|=q$ .

I can also do it if $[\bar k : k] $ is finite and $n=2$ . By Artin-Schreier theorem (http://www.math.uconn.edu/~kconrad/blurbs/galoistheory/artinschreier.pdf) , we will have $[\bar k : k]=2$ . Let $u\in \bar k \setminus k$, then the minimal polynomial of $u$ over $k$ has degree 2, and let $\bar u$ be another root of the minimal polynomial of $u$ over $k$ . Then $f(X_1,X_2)=(X_1+uX_2)(X_1+\bar u X_2)$ satisfies our required condition . Even in this case when $\bar k$ is a finite extension over $k$, I don't know what happens if $n>2$ .

Please help . Thanks in advance


marked as duplicate by Eric Wofsey, Community Nov 20 '17 at 19:46

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

  • $\begingroup$ If $k$ is ordered, one can take $x_1^2+ \dots+ x_n^2$ $\endgroup$ – MatheinBoulomenos Nov 20 '17 at 19:10

Similar to Eric Wofsey's linked answer, the case $n=2$ solves the general case :

Take any $\alpha \in \overline{k}/k$, let $\alpha_1, \ldots,\alpha_l$ be its conjugates, set $f_2(X_1,X_2) = \prod_{j=1}^l (X_1+\alpha_j X_2)$ having only one root at $(0,0)$, thus $$f_{m+1}(X_1,\ldots,X_{m+1}) = f_2(f_m(X_1,\ldots,X_m),X_{m+1}) = \prod_{j=1}^l (f_m(X_1,\ldots,X_m)+\alpha_j X_{m+1})$$ has only one root at $(0,\ldots,0)$.