Why does equality between radicals imply this?

Question

Let $R$ be a commutative ring.

Given an ideal $I\subset R$, we define the radical $\mathfrak{r}(I)$ to be the intersection of all prime ideals $J\subsetneq R$ where $I \subseteq J$.

Given an ideal $I \subset R$, we define $V(I)$ to be the set of prime ideals containing $I$.

I'm reading notes and the following is stated without proof,

Let $I,J\subset R$ be ideals. If $\mathfrak{r}(I)=\mathfrak{r}(J)$ then $V(I)=V(J)$.

how would one prove it?

Answer 1

Oh I see the answer now. Obviously by assumption we have that $V(\mathfrak{r}(I)) = V(\mathfrak{r}(I))$. We are done by noting that for any ideal $I$ we have that $V(\mathfrak{r}(I))\setminus V(I) = \emptyset$.