Let $x$ be an element of an adic ring $A$, and $I$ be an ideal of definition such that the image of $x$ in $A/I$ is nilpotent, then there exists an ideal of definition $J$, which contains $x$.
I can't understand the proof. The proof is: Let $I$ be an ideal of definition such that $x+I$ is nilpotent in $A/I$ and $n$ be a positive integer such that $x^n \in I$ . Let $J=I+xA$. $J$ is an open ideal of $A$ (why is $J$ open? is it because it contains $I$ ?). $J^n \subset I$. The $I$-adic and $J$-adic topologies coincide, so $J$ is an ideal of definition (why?).