Characterzation of $I$-adic topologies Let $I,J$ be two ideals of a ring $A$.
We can define the $I$-adic topology on $A$ by declaring $$\{x+ I^n \}_{x \in A, n \ge 0 }$$
be a basis of open neighboord for $A$. (similarly for $J$.
It seems to me that: the $I$-adic topology aand $J$-adic topology on $A$ coincide $\Leftrightarrow Spec A/I $  is equal as sets to $Spec A/J$.
I can show $\Rightarrow$. : This follows aas exists $n$ $I^n \subset J, J^n \subset I$. Can one deduce the converse?
 A: The converse isn't true in general: take $V$ a non-noetherian valuation ring of rank $1$ (if you don't know what rank $1$ means, just take concretely for instance $V=\mathcal O_{\mathbb C_p}$ the ring of integers in the p-adic complex numbers). Any such ring has only two prime ideals, $0$ and the maximal ideal $\mathfrak m$. In the explicit example $V=\mathcal O_{\mathbb C_p}$ we can give a quick proof: another prime ideal must satisfy $0\subset\mathfrak p\subset\mathfrak m$ (since $\mathfrak m$ is the unique maximal ideal) with strict inclusions. Take some $x\in\mathfrak p\smallsetminus\{0\}$ and some $y\in\mathfrak m\smallsetminus\mathfrak p$. In the former we have $|x|_{\mathbb C_p}\neq0$ and in the latter case we have $|y|_{\mathbb C_p}<1$, and then you can take some $n$ so that $|y^n|_{\mathbb C_p}=|y|_{\mathbb C_p}^n<|x|_{\mathbb C_p}$, but then $y^n\in(x)\subseteq\mathfrak p$ (because $y^n=x(\frac{y^n}{x})$ and $|\frac{y^n}x|_{\mathbb C_p}\le 1$), and this is impossible unless $y\in\mathfrak p$ itself.
I also claim that because $V$ is non-noetherian one has $\mathfrak m^2=\mathfrak m$. Again in the concrete case $V=\mathcal O_{\mathbb C_p}$ you can see this easily: if $x\in\mathfrak m$, which again means $|x|_{\mathbb C_p}<1$, then you can take some element $y\in\mathbb C_p$ with $y^2=x$ since $\mathbb C_p$ is algebraically closed, but then $|x|_{\mathbb C_p}=|y|_{\mathbb C_p}^2$ and in particular $y\in\mathfrak m$ with $x\in(y^2)\subseteq\mathfrak m^2$, which shows $\mathfrak m\subseteq\mathfrak m^2$.
Now if you take any nonzero element $\pi\in\mathfrak m$ and let $I=(\pi)$, then you can see that $\operatorname{Spec}V/I=\operatorname{Spec}V/\mathfrak m=\{\mathfrak m\}$, but the $I$-adic and $\mathfrak m$-adic topologies do not coincide since $\mathfrak m^2=\mathfrak m$ implies $\mathfrak m^n=\mathfrak m\not\subseteq I$ for all $n$.
The statement does hold if $I$ and $J$ are finitely generated, as suggested by Alex in the comments. To see this, notice that the hypothesis gives us
$$\sqrt I=\bigcap_{\mathfrak p\in\operatorname{Spec}(A/I)}\mathfrak p=\bigcap_{\mathfrak p\in\operatorname{Spec}(A/J)}\mathfrak p=\sqrt{J}.$$
Thus $I\subseteq\sqrt I=\sqrt J$, and using the fact that $I$ is finitely generated this implies that $I^n\subseteq J$ for some $n$ (specifically, if $x_1,\dots,x_k$ are generators of $I$, choose some $N$ for which $x_i^N\in J$ for each $i$, then take $n=kN$). Similarly you can find $J^m\subseteq I$ for some $m$.
