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.



Z_p flatness and irreducible components.
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.


