Let $F$ be an algebraically closed (hence infinite) field and $n$ a positive integer. For a subset $S$ of $F^n$, let us use the following notation:
$J(S):=\{f \in F[x_1,...,x_n] : f(a_1,...,a_n)=0$ for all $(a_1,...,a_n) \in S \}$
There is a theorem in Hungerford's Algebra, Section VIII.7, which asserts
$J(F^n)=\emptyset$,
with proof omitted. Is this an typo? I think, at least, the zero polynomial $0$ belongs to $J(F^n)$. On the other hand, how do I have to prove the inclusion
$J(F^n) \subset 0$?
It seems obvious, but I cannot write down rigorously.
Here's an inductive proof that if $f \neq 0$ (in the sense that the sum has a non-zero coefficient), then there exists some $(a_1,\dots,a_n) \in F^n$ such that $f(a_1,\dots,a_n) \neq 0$.
The case of $n=1$ is straightforward (but uses the fact that $F$ is algebraically complete).
For the inductive step: suppose that $f \neq 0$. Write $f$ as a polynomial in $x_n$. That is, regard $f(x_1,\dots,x_{n-1},x_n)$ as a function of $x_n$ with coefficients that are expressions in $F[x_1,\dots,x_{n-1}]$. One of these coefficients must be non-zero. By our inductive hypothesis, there exists some $v = (a_1,\dots,a_{n-1})$ such that this coefficient evaluated at $v$ is non-zero. Now, consider the function $f(a_1,\dots,a_{n-1},x)$, which is a non-zero element of $F[x]$.