Application of the Nullstellensatz The question is as follows 
Let $k\subset K$ be algebraically closed fields. And $I \leq k[x_1,...x_n]$ an ideal.
Show that if $f \in K[x_1 ,...x_n]$ vanishes on $Z(I)$  it vanishes on $Z_K(I)$.
Where $Z(I)$  is the set of zeros of $I$  in $k^n$, $Z_K(I)$   is the set of zeros in $K^n$.
I've tried proving this following the proof of the nullstellensatz but I'm stuck, The only connection I can think of  between $Z(I)$  and $Z_K(I)$  is of inclusion which does not seem to help.
Hints will be most welcome.
 A: The idea is the following: 
We consider the ideal $J$ generated from te set $I$ in $K[x_1, \dots , x_n]$.
What is the structure of this ideal? 
$J=\{\sum_{i=1}^kh_if_i : f_i\in I ,h_i\in K[x_1, \dots , x_n]\}$
We observe that 
$Z(I)=Z_K(J)$
In fact: 
Let $x\in k^n$ such that $x\in Z(I)$. Then for each $r=\sum_{i=1}^kh_if_i\in J$ we have that 
$r(x)=0$ so 
$x\in Z_K(J)$
Conversely, if $x\in K^n$ such that $x\in Z_K(J)$ then, if we choose an $r\in I\subseteq J$, we get that 
$r(x)=0$
but $r$ is a polynomial with coefficients in $k$ and $k$ is algebraically closed, so 
$x\in k^n$ and this means 
$x\in Z(I)$ 
that is what we wanted to prove. 
By hypothesis, 
$Z_K(I)=Z_K(J)=Z(I)\subseteq Z_K(f)$ 
that is what we wanted to prove.
There is a problem with this proof. If holds surely for $n=1$ but when we have $n>1$ it is not clear that if 
$r(x)=0$ then $x\in k^n$ 
A: I'm not sure about this, but if you say that $k$ is algebraically closed and $I\vartriangleleft k[x_1,...,x_n],$ all the zeros of the polynomials in $I$ will be in $k^n,$ so $Z(I)=Z_K(I),$ and obviously if $f(P)=0\ \forall\ P\in Z(I)\in k^n$ we have that $f(P)=0\ \forall\ P\in Z_K(I)\in K^n.$
