Prove fact about polynomial in uncountable fields

Question

$F$-uncountable field.
$I_{i}$-ideal in $F[x_{1},...,x_{n}]$
$F^{n}=\cup_{i=1}^{\infty}V(I_{i})$  
$V(I_{i})\subseteq V(I_{i+1})$
Prove that $\exists k, V(I_{k})=F^{n}$
All that I've find is that using Hilbert basis theorem we can prove that all ideal are finite generated.

Answer 1

You must note that the statement is not true for countable field $\mathbb Q$. Write $\mathbb Q$ as $\{a_1, a_2, \ldots, a_n, \ldots \}$. Then in $\mathbb Q[X]$ define $I_k = (X-a_1)(X-a_2) \ldots (X-a_n)$. Here no $V(I_k)$ is $\mathbb Q$.

Now consider the field is uncountable and $n=1$. Then each $V(I)$ is either empty or finite set or the full space $F$. So the result is clear in this case. So for some $I_k$ we have $V(I_k) = F$ and $I_k =0$.