Avoiding Hilbert's Nullstellensatz to show $\mathcal{I}(Z(I)) \subseteq \sqrt{I}$ for some ideal 
Let $K$ be a field and let $I := (Y^2-X^3)_{K[X,Y]}$. Show directly that $\mathcal{I}(Z(I)) = \sqrt{I}.$
(We defined $Z$ as the set of zeroes of a given polynomial and $\mathcal{I}(A) := \{P \in K[X,Y] : P(x) = 0$ for all $x \in A \subset \mathbb{A}^2\}$.)

To clearify: I am not supposed to use Hilbert's Nullstellensatz, I have to use a direct computation.
The easy part is $\mathcal{I}(Z(I)) \supseteq \sqrt{I}$ is clear to me. However, I do not see how to do the other direction. It should go down like this:
We pick $P \in \mathcal{I}(Z(I))$ and find a $n \in \mathbb{N}$ such that $P^n \in I$. However, I do not see how to proceed here. Could you give me a hint?
 A: If $|K|<\infty$, this may not be true depending on what you mean by $\mathbb A^2$. Some author insists $\mathbb A^2=\bar K^2$, then there is no problem. But if $\mathbb A^2=K^2$, then for example, over $K=\mathbb F_2$, we have $Z(I) = \{(0,0),(1,1)\}$, therefore $X^2-X$ vanishes on $Z(I)$, but not in $\sqrt{I}$.
If we're allowed to assume $|K|=\infty$, my solution to a similar question can be used to show that $\mathcal I(Z(I)) = I$ in your case as well, and $I\subset \sqrt{I}$ is trivial.
A: In fact $\mathcal{I}(Z(I))=I$
Let $f(x,y)\in Z(I)$, then we have
$$f(t^2,t^3)=0$$ identically where $t$ is a variable. I am assuming here that $k$ is infinite.
Now modulo $y^2-x^3$ we have
$$f(x,y)=p(x)+q(x)y$$
Basically we replace higher powers of $y$ with a polynomial in $x$, using $y^2=x^3$. Let the degree of $p(x)$ be $n$ and $q(x)$ be $m$.
Now we have $$p(t^2)+q(t^2)t^3=0$$
thus the highest powers must cancel. This is impossible since
$2n=2m+3$ has no solution, so the highest terms of $p(t^2)$ and $q(t^2)t^3$ cannot have the same degree.
Thus $p(x)+q(x)y=0$ identically and $f\in I$.
