The ideals $I=(X,Y)$ and $J=(X^2+Y^2)$ in $\mathbb R[X,Y]$ are such that $V(I)=V(J)$ and their radicals aren't the same contradicting the Nullstellensatz (in case it was true for arbitrary fields). However, this shouldn't be a surprise, if we look at their varieties in $\mathbb C$ we find that they aren't the same as the second has a pair of lines that were hidden.
My question is if it is true the other way around: Suppose we have two ideals $I$ and $J$ of $\mathbb K[X_1,\ldots,X_n]$ where $\mathbb K$ is an arbitrary field. Such that $V(I)=V(J)$ and such that ${\bf V}(I)={\bf V}(J)=V(I)$; where ${\bf V}$ means the variety in the affine space of dimension $n$ over $ \mathbb{\bar K}$, the algebraic closure of $\mathbb K$. That is, there are no additional points hidden in the algebraic closure. Is it true then that $\sqrt I=\sqrt J$?
My motivation for the question is a problem of primary descomposition, where we had to find one for $J=(X + Y − X^2 + XY − Y^2,X(X + Y − 1))$ in $\mathbb K[X,Y]$. We had already proved that $V(J)=\{(0,0),(1,0),(0,1)\}$ over $\mathbb{ A}_{\mathbb K}^2$ for any field $\mathbb K$. It made some computations easier in the case of algebraically closed fields because we had automatically that $\sqrt J=I$ where $I$ was the ideal of the three points $I=(X,Y)\cap(X-1,Y)\cap (X,Y-1)$.