I'm reading rigid and formal geometry book of Bosch and in the proof of the proposition 4.2.3 it claims that if $f:A\to B$ is a finite local morphism between two local Noetherian rings such that the $m_A$-adic topology on $B$ and it's $m_B$-adic topology are the same then the image of $A$ in $B$ is closed.
the book says this is a consequence of krull intersection theorem but I can't understand why.any idea?
Pick some $b\in B\setminus f(A)$. In order to show that $f(A)$ is closed in $B$ it suffices to show that there exists $n\in \mathbb{N}$ such that $b+\mathfrak{m}^n_B\subseteq B\setminus f(A)$. Since $\mathfrak{m}_B$-adic topology is the same as $\mathfrak{m}_A\cdot B$-adic topology on $B$, we derive that it suffices to prove that there exists $n\in \mathbb{N}$ such that $b+\mathfrak{m}^n_A\cdot B\subseteq B\setminus f(A)$. So it suffices to show that there exists $n\in \mathbb{N}$ such that $$\left(b+\mathfrak{m}^n_A\cdot B\right) \cap f(A) = \emptyset$$ and this means that $$b\neq \mathfrak{m}_A^n\cdot B\,\,\textbf{mod}\,f(A)$$ Consider now $N = B/f(A)$ and $x = b+ f(A)$. Since $b\not \in f(A)$, we derive that $x\neq 0$. $N$ is a finitely generated $A$-module. By Krull's interesection theorem we deduce that $$\bigcap_{n\in \mathbb{N}}\mathfrak{m}_A^n\cdot N = 0$$ Hence there exists $n\in \mathbb{N}$ such that $x\not \in \mathfrak{m}^n_A\cdot N$. This is exactly what we wanted above.