Zeros of multivariables polynomials over algebraically closed fields extension

Question

I am trying to prove that if two algebraically closed fields, $F$ and $K$ such that $K$ contains $F$, J is ideal in $F[x_1,...,x_n]$, and $f$ is a polynomial in $K[x_1,...,x_n]$ that vanishes on Z(J) with respect to $F$ (i.e, if $(a_1,..,a_n)$ in $F^n$ vanishes every $g$ in $J$ than it vanishes $f$), than $f$ vanishes on $Z(J)$ with respect to $K$. Everything I tried got me to a dead end. I'll be glad for some help with this one

Answer 1

If $f$ vanishes on $Z(J)$ with respect to $F$, then by Nullstellensatz, since $F$ is algebraically closed, $f^n\in J$ for some $n$. Hence $f$ vanishes on $Z(J)$ with respect to any field $K$ containing $F$ (algebraically closed or not).

Edit: It was pointed out to me that $f$ is in $K[x_1,\ldots,x_n]$. In that case, I believe the solution goes as follows: (editing this on my phone at a layover, will edit to format properly later)

Let $I$ be the radical of $J$. By Nullstellensatz, we have the exact sequence $$ 0\to I\to F[x_1,\ldots,x_n]\overset{\mathbf{ev}}{\to} F^{V(J)}.$$ Tensoring with $K$ over $F$ (noting that $F$ is a field, so $K$ is flat), we get $$0\to KI\to K[x_1,\ldots, x_n]\overset{\mathbf{ev}\otimes K}{\to} K^{V(J)}$$ Then if $f$ vanishes on $V(J)$ then by definition, it lies in the kernel of the evaluation map, and hence $f\in KI$. $KI$ is the radical of $KJ$, so $f^n$ is in $KJ$ for some $n$, and thus $f$ vanishes on $V(J)$ with respect to $K$. This solution doesn't require $K$ to be algebraically closed.