I just used the following.

Lemma. Let $A$ be a $\mathbb{Z}_p$-flat ring, of finite type over $\mathbb{Z}_p$, and suppose that $A \otimes \mathbb{F}_p$ is a domain. Then $A$ is a domain.

Proof: Suppose $ab = 0$ in $A$. Then one of $a, b$ must lie in $pA$, so we can write (without loss of generality) $a = p a_1$. Then by flatness $a_1 b=0$.

Continuing in this manner, we find that one of $a$ and $b$ must be infinitely divisible by $p$. But the finite type hypothesis implies that this is impossible unless one of $a$ or $b$ is in fact zero.

Given the statement, it seems like there should be a more conceptual reason why this should be true. Can anyone supply one? (A proof using more general facts in EGA counts as conceptual).

Edit: Kevin Buzzard gives a compelling reason why I have never seen this "fact" used before. Thank you both for your answers.

Edit 2: I suppose that replacing "finite type" with "p in the radical" would work (with an application of the Krull intersection theorem). In particular, the result is true as stated for a local $\mathbb{Z}_p$-algebra.

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    $\begingroup$ It's true (in terms of integral schemes) if you work with a proper flat scheme over a dvr, since the only open set which contains the special fiber is the entire space (in contrast with the disconnected counterexample suggested by Kevin). Ditto for the property of reducedness. $\endgroup$ – BCnrd May 4 '10 at 23:04
  • $\begingroup$ "But the finite type hypothesis implies that this is impossible" : you need a little more assumptions in order to use the Krull intersection theorem, check Matsumura, Commutative Ring Theory, thm. 8.10. $\endgroup$ – Matthieu Romagny May 8 '10 at 8:11

Your proof seems wrong to me. I might be misunderstanding some things you wrote, but surely $\mathbf{Q}{}_p=\mathbf{Z}_p[X]/(pX-1)$ is finite type over $\mathbf{Z}_p$, and contains many elements which are infinitely divisible by $p$. Again if I've understood your definitions correctly, $\mathbf{Q}{}_p\times\mathbf{Z}_p$ is a counterexample to your assertion.


With assumptions of finite type, the relevant notion is purity (N.B. proper implies pure). Let me explain briefly.

Put $R=\mathbb{Z}_p$. Then one says that a flat finite type $R$-scheme $X$ is pure if the closure of any associated point of the generic fibre of $X$ meets the special fibre. If $X$ has reduced fibres, this essentially means that the valuative criterion of properness is satisfied at generic points of the generic fibre : thus the relation with irreducible components is quite clear.

Let us see what happens in the affine case. In the proof of your lemma, what you need is your ring $A$ to be separated for the $p$-adic topology, so that an element $a$ lying in the intersection of the ideals $(p^n)$ is $0$. Now a deep theorem of Raynaud and Gruson says that (in the particular case of a d.v.r. base ring) the following three conditions are equivalent :

  1. $A$ is separated for the $p$-adic topology
  2. $A$ is free as an $R$-module
  3. $X=Spec(A)$ is pure over $R$.

The notion of purity exists over a general base, and you can prove the following kind of statement : for a morphism of schemes $X\rightarrow S$ which is of finite presentation, flat and pure, the locus of points of $S$ where the geometric fibre is integral is open. See my paper "Effective models of group schemes", especially theorem 2.2.1 (on arxiv or on my webpage for most recent version).


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