Application of the Nullstellensatz

Question

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.

Answer 1

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 $f(P)=0\ \forall\ P\in Z_K(I)\in K^n.$

Answer 2

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 $x\in k^n$