If $f \in k[x_{1},...,x_{n}]$ with $f(P) = 0$ for all $P \in k^{n}$, show that $f=0$ in $k[x_{1},...,x_{n}]$

Question

Let $k$ be an infinite field. If $f \in k[x_{1},...,x_{n}]$ with $f(P) = 0$ for all $P \in k^{n}$, show that $f=0$ in $k[x_{1},...,x_{n}]$.

I tried to get $S = \lbrace f \rbrace$. But, I don't know if $k$ is algebrically closed, so, I don't know if I can say that $Z(f) = k^{n} = V(k)$. I don't know what to do. Any hints?

Answer 1

Use induction, let $$f=a(x_1,\ldots ,x_{n-1})x_n^k+\cdots$$ now there are $a_1\ldots ,a_{n-1}\in k$ such that $$a(a_1,\ldots ,a_{n-1})\neq 0.$$ Now we see that $$f(a_1,\ldots ,a_{n-1},x_n)=0$$ has only finitely many solutions.