Why is $\mathcal{O}_F \otimes \mathbb{Z}_p$ a Dedekind ring?

Question

In the paper Compactifications de l'espace de modules de Hilbert-Blumenthal (Compositio. '78), M. Rapoport relates some assertions from a letter of Deligne to Serre, dated 1972. At the very beginning, the following claim is made:

Let $X$ be an abelian variety over a perfect field $k$ of characteristic $p>0$, with real multiplication by the ring of integers $\mathcal{O}_F$ of a totally real number field $F$. (This includes the condition that $\mathrm{Lie}(X_{k^{alg}})$ is free of rank 1 over $\mathcal{O}_F\otimes k^{alg}$.) $H^1_{cris}(X)$ is a free module of rank $2g$ over the ring of Witt vectors $W(k)$ of $k$ (so far so good). Then $H^1_{cris}(X)$, being torsion-free over $\mathcal{O}_F\otimes W(k)$, is projective because $\mathcal{O}_F \otimes W(k)$ is a Dedekind ring.

My question is: How or why is $\mathcal{O}_F \otimes W(k)$ a Dedekind ring? Take $k=\mathbb{F}_p$, so $W(k)=\mathbb{Z}_p$, and $F$ any totally real number field in which $p$ splits. Then it seems to me that $\mathcal{O}_F \otimes \mathbb{Z}_p$ is not even an integral domain. Is there some theorem about real multiplication of abelian varieties that stops $F$ from splitting over $p$? Is this some older definition of a Dedekind ring? What am I misunderstanding?

Later on, to check that $H^1_{cris}(X)$ is free over $\mathcal{O}_F \otimes W(k)$, the claim is made that it's okay to check if $H^1_{cris}(X)\otimes \mathbb{Q}_p$ is free over $F\otimes W(k)$. It may be obvious, but I'm not seeing why. Any clarification there would also be appreciated.

EDIT: As David Loeffler kindly points out in his answer, it would make sense if a Dedekind ring was a finite product of Dedekind domains.

There is at least one other definition of a Dedekind ring in the literature, in a paper of M. Bhargava (P-orderings and polynomial functions on arbitrary subsets of Dedekind rings, J. reine angew. 1997) as a "Noetherian, locally principal ring in which all non-zero primes are maximal." But according to this definition, the product of multiple Dedekind domains is never a Dedekind ring, since for example $\{0\}\times \mathbb{Z}_p$ is a non-maximal prime ideal of $\mathbb{Z}_p \times \mathbb{Z}_p$.

I will leave this question open for a little while longer, in case someone can clarify the exact sense of a Dedekind ring in M. Rapoport's paper, ideally with a reference.

Answer 1

I don't have the original reference to hand, but notice that your quote has Dedekind ring, not Dedekind domain. It's possible that what was meant might be that the ring $R = O_F \otimes \mathbf{Z}_p$ is a finite direct product of Dedekind domains (indeed of DVRs), and hence any torsion-free $R$-module is projective.

For your second question: since the Dedekind domain factors $R_i$ of $R$ (which are just the completions of $O_F$ at the primes above $p$) are all discrete valuation rings, one sees that if $M$ is a projective $R$-module, it has the form $\bigoplus_i M_i$ where $M_i = M \otimes_R R_i$ and the $M_i$ are all free; so $M$ is free if and only if the $M_i$ all have the same rank, which is something you can check after inverting $p$ (or any other element you like).

By the way, examples where $p$ is split in $F$ must exist, because abelian varieties over number fields $K$ with real multiplication by $O_F$ certainly do exist, and the reduction of such an AV at all but finitely many primes of $K$ will give AV's over finite fields with $O_F$-multiplication. In particular, you can pick your prime of $K$ such that it lies above a rational prime $p$ split in $F$.