Let $R$ be a local and smooth $\mathbb{Z}_p$-algebra and $B$ an $R$-algebra of finite type which is an integral domain with $\operatorname{dim}B\leq \operatorname{dim}R$ such that

- $B/(p)$ is non-zero and finitely generated as an $R/(p)$-module and
- $B[\frac{1}{p}]$ is non-zero and finitely generated as an $R[\frac{1}{p}]$-module

(where both module structures are induced by the structure morphism $R\to B$).

Let $f\colon R[x]\to B$ be an $R$-algebra morphism such that $f/(p)\colon R/(p)[x]\to B/(p)$ is surjective, i.e. $\operatorname{Spec}f$ is a closed immersion on the special fibre.

Is $f[\frac{1}{p}]$ surjective? Is probably even $f$ surjective?

If $B$ was finitely generated as an $R$-module, the surjectivity of $f$ would follow from Nakayama's lemma by considering the $R$-module cokernel of $f$. If $R$ was complete (e.g. in the special case $R=\mathbb{Z}_p)$ a topological version of Nakayama's lemma (see e.g. Eisenbud's ''Commutative Algebra with a view...'' Exercise 7.2. or MO) would apply. Thank you for any help.