Let $I ⊆ K[x_1 ,...,x_n ]$ be an ideal in a polynomial ring. Show that $\mathcal{I}_{K[x_1,...,x_n]}(\mathcal{V}_{\overline {K^n}}(I))=\sqrt{I} $ (A generalization of Hilbert’s Nullstellensatz) Let $K$ be a field
and $\overline K$ its algebraic closure. Let $I ⊆ K[x_1 ,...,x_n ]$ be an ideal in a polynomial
ring. Show that
\begin{equation}
\mathcal{I}_{K[x_1,...,x_n]}(\mathcal{V}_{\overline {K^n}}(I))=\sqrt{I}
\end{equation}
\begin{equation}
\mathcal{V}_{\overline {K^n}}(I)=\{(\xi_1,...,\xi_n)\in \overline {K^n}| f(\xi_1,...\xi_2)=0,\ for\ all\ f\in I\}
\\
\mathcal{I}_{K[x_1,...,x_n]}(\mathcal{V}_{\overline {K^n}}(I))=\{f\in K[x_1,...x_n]|f(\xi_1,...,\xi_n)=0\ for\ all\ (\xi_1,...,\xi_n)\in \mathcal{V}_{\overline {K^n}}(I)\}
\end{equation}
The $\supset$ is easy. For the converse, need to show that $f^k\in I$ for some $k$. And always have
$\sqrt{I}=\bigcap_{I\subset P}P$, $P$ is prime ideal in $K[x_1,...,x_n]$.
If $f\notin \sqrt{I}.$ Then there are some $P\supset I$ such that $f\notin P.$ $K[x_1,...,x_n]/P$ is a domain for this prime ideal. Then $K[x_1,...,x_n]/P$ is a finitely generated ring over K. Is there exists a maximal ideal $m$ such that $(K[x_1,...,x_n]/P)/m$ is $\overline K$. And let $x_i$ be $\xi_i.$ Then its a contradiction.
How can I get this. Or this ideal is right? What's the right way.
 A: As you note, the inclusion $\supset$ is straightforwards. To prove the other inclusion, we need to show that if $f$ is an element of the LHS, then some power $f^k$ is in $I$.
To do this, let $J$ be the ideal of $\overline{K}[x_1,\cdots,x_n]$ generated by $I$ and let $f\in \mathcal{I}_{K[x_1,\cdots,x_n]}(\mathcal{V}_{\overline{K}^n}(I))$. We notice that $f\in \mathcal{I}_{\overline{K}[x_1,\cdots,x_n]}(\mathcal{V}_{\overline{K}^n}(J))$. By the usual Nullstellensatz, this means that $f\in\sqrt{J}$, so there's some positive integer $k$ so that $f^k\in J$. Since $f\in K[x_1,\cdots,x_n]\subset \overline{K}[x_1,\cdots,x_n]$, then any power of $f$ is also in $K[x_1,\cdots,x_n]$, so we get $f^k\in J\cap K[x_1,\cdots,x_n]$. 
Since $\overline{K}$ is a $K$-vector space, we can find a splitting of the canonical injection $K\to \overline{K}$, also known as a $K$-linear projection $\overline{K}\to K$. Now we can extend this to a projection $\overline{K}[x_1,\cdots,x_n]\to K[x_1,\cdots,x_n]$ by applying it to each coefficient of a polynomial. Call the resulting map $\varphi$: it is a $K[x_1,\cdots,x_n]$-linear map of $K[x_1,\cdots,x_n]$-modules.
Now we use the definition of $J$ as $\overline{K}[x_1,\cdots,x_n]$-linear combinations of elements from $I$ to write $f^k=\sum_i^n \alpha_i h_i$ where $\alpha_i\in \overline{K}[x_1,\cdots,x_n]$ and $h_i\in I$. Apply $\varphi$ to both sides. Since $f^k\in K[x_1,\cdots,x_n]$, the LHS is unchanged, and similarly as $h_i\in I\subset K[x_1,\cdots,x_n]$, the $h_i$ are preserved as well. So our relation becomes $f^k = \sum_i^n \varphi(\alpha_i)h_i$, which demonstrates that $f^k\in I$ since the $\varphi(\alpha_i)$ are elements of $K[x_1,\cdots,x_n]$.
This proof is how I would solve this if I were only working out of this text and hadn't developed much in the way of other tools.

I think there's a more intuitive way, but we'll need to gather some extra ingredients first and it might require a slightly longer explanation. It's somewhat similar to your recent attempt.
First, we identify the maximal ideals of $K[x_1,\cdots,x_n]$: ever maximal ideal of this ring is of the form $\{f\in K[x_1,\cdots,x_n]\mid f(\alpha)=0\}$ for some $\alpha\in \overline{K}^n$. (See this previous answer of mine for a full proof.)
Next, we need some additional ring-theoretic technology. A (commutative) Jacobson ring is a ring in which every prime ideal is an intersection of maximal ideals. This means, in particular, that if we want to calculate the radical of an ideal in a Jacobson ring, it suffices to take the intersection of all maximal ideals containing this ideal. (Note this is very badly not true for non-Jacobson rings like $K[x,y]_{(x,y)}$.) It turns out that all finitely generated algebras over a field are Jacobson, so we can use this idea to give an explanation.
We start by writing $\mathcal{I}_{K[x_1,\cdots,x_n]}(\mathcal{V}_{\overline{K}^n}(I))=K[x_1,\cdots,x_n]\bigcap\mathcal{I}_{\overline{K}[x_1,\cdots,x_n]}(\mathcal{V}_{\overline{K}^n}(J))$ where again $J$ is the ideal generated by $I$. Now, the right hand side on the right of the intersection is the radical of $J$ by the Nullstellensatz, so we're considering $K[x_1,\cdots,x_n]\bigcap\sqrt{J}$ now. On the other hand, by the fact about Jacobson rings above, we can rewrite $\sqrt{J}$ as the intersection of all maximal ideals containing $J$, so now we're considering $K[x_1,\cdots,x_n]\cap \left(\bigcap_{J\subset m} m\right)$.
Now we just swap the order of taking the intersections: $K[x_1,\cdots,x_n]\cap \left(\bigcap_{J\subset m} m\right) = \bigcap_{J\subset m} \left(K[x_1,\cdots,x_m]\cap m\right)$. Now we note three things: first, that $K[x_1,\cdots,x_n]\cap m$ is a maximal ideal of $K[x_1,\cdots,x_n]$; second, that it contains $J\cap K[x_1,\cdots,x_n]=I$; and third, that every maximal ideal of $K[x_1,\cdots,x_n]$ containing $I$ can be obtained this way. This means this latter intersection is all the maximal ideals of $K[x_1,\cdots,x_n]$ which contain $I$, so it is exactly $\sqrt{I}$ by our earlier work. Thus the claim is proven.
