All throughout, we may assume the rings are Noetherian if required.
Let $(A,\mathfrak{m})$ be a local ring and let $B$ be an $A$-algebra which is separated for the $\mathfrak{m}$-adic topology. Does it then follow that $B$ with the $\mathfrak{m}$-adic topology is a Zariski ring (which is to say that $\mathfrak{m}B \subseteq J(B)$, where the latter is Jacobson Radical of $B$).
My question is derived from Kollár's Rational Curves on Algebraic Varieties, where this is claimed in the statement of Proposition 7.4.1. In reality I am interested in this result, so I will include the statement and proof as it may be that I have missed an assumption when asking a more general question.
Proposition 7.4.1:
Let $(A,\mathfrak{m})$ be a local ring and $B$ an $A$-algebra which is separated in the $\mathfrak{m}$-adic topology (in particular, $\mathfrak{m}B$ is contained in every maximal ideal of $B$). Let $M,N$ be finite $B$-modules, and assume that $N$ is flat over $A$. Let $g\colon M\to N$ be a morphism and set $Q = \mathrm{coker}\ g$.
(7.4.1.1) The following are equivalent:
(7.4.1.1.1) $g$ is injective and $Q$ is flat over $A$;
(7.4.1.1.2) $g\otimes (A/\mathfrak{m})$ is injective.
(7.4.1.2) $g$ is an isomorphism iff $g\otimes (A/\mathfrak{m})$ is an isomorphism.
I understand the proof of (7.4.1.1), but the proof of (7.4.1.2) proceeds as follows:
Only one direction of the second part is nontrivial. If $g\otimes_A (A/\mathfrak{m})$ is an isomorphism, then $g$ is injective by the first part. By the Nakayama lemma $g$ is also surjective.
In order to apply Nakayama's lemma I believe you would need to know that $\mathfrak{m}B\subseteq J(B)$, hence the question posed in the title, but I am unable to prove that this is true. In fact, I believe that it is not true unless you include some sort of finiteness hypothesis on $B$ as an $A$-algebra, but I am unable to produce a counterexample.