Multivariable polynomial roots / division .

Question

Let $R=F[x_1,...,x_n]$ be a ring of polynomials over a field $F$. let $Z(A)=\{\vec x \in F^n | \forall f \in A, f(\vec x) = 0\}$. I have already proved that $Z(A)=Z(<A>)$ where $<A>$ is the ideal generated by the set $A$. I am trying to prove that if $f$ is a polynomial such that $f(\vec x)=0$ for every $\vec x\in Z(A)$ then $f \in <A>$.

Answer 1

I think it may not be true.

Consider $n=1$ and $A=\{x^2\}$. Then $Z(A)=\{0\}$.

Then if $f(x)=x$, $f\not\in <A>$ but $f(x)=0$ $\forall$ $x\in Z(A)$.