MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.

share|cite|improve this question
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. – BCnrd May 4 '10 at 23:04
"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. – Matthieu Romagny May 8 '10 at 8:11
up vote 9 down vote accepted

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.

share|cite|improve this answer

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).

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.