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.

  • 1
    $\begingroup$ Take $R=\mathbb{Z}_p$, $B=$ a nontrivial finite extension of $\mathbb{Q}_p$, $f$ mapping $x$ to zero. The image of $f[1/p]$ is $\mathbb{Q}_p$ but $f\bmod p$ is surjective since $B/(p)$ is zero. $\endgroup$ Commented Apr 11, 2015 at 13:47
  • $\begingroup$ Isn't $B = R \times R[1/p]$ now a counterexample, just sending $R[x] \to R \times 0$ by some reasonable map? $\endgroup$
    – Will Sawin
    Commented Apr 11, 2015 at 23:00
  • $\begingroup$ Dear Laurent Moret-Bailly and Will Sawin, thank you both very much for your comments. Unfortunately, I was very sloppy in the formulation of the question and forgot the condition that $B/(p)$ is non-zero and $B$ is an integral domain. I made an edit to the question. Sorry again for the confusion. $\endgroup$
    – Florian
    Commented Apr 12, 2015 at 19:34

1 Answer 1


Still no. Consider the map $\mathbb Z_p[s] \to \mathbb Z_p[t]$, $s=t+pt^2$. This is a degree $1$ map on the special fiber but degree $2$ on the generic fiber. Let $R$ be the localization of $\mathbb Z_p[s]$ at $s=1+p$ and let $B$ be the localization of $\mathbb Z_p[t]$ at $t=1$.

Then this satisfies all your conditions: $B$ is finitely generated, is integral, is $1$-dimensional, $B/p$ is generated by one element and $B[1/p]$ by two, and so on. But the map $R \to B$ is surjective mod $p$ but not upon inverting $p$. By sending $x$ to zero, we get a counterexample.

Geometrically we have a family of curves parameterized by $\mathbb Z_p$ that is smooth on the generic fiber and degenerates to a reducible curve on the special fiber. By localizing, we have removed one of the components on the special fiber, so the special fiber no longer controls the general fiber. (The family of curves I am thinking of is $\mathbb Z_p[s,t']/\left( t'^2 +t'-ps\right)$, which is a parabola when $p \neq 0$ and two lines when $p=0$. Adjoining $t=t'/p$ gives back the equation I wrote earlier.)


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