Showing that $M$ is free module over $(O_K \otimes_{\mathbb{Z}_p}U)$

Let $K$ be a finite extension of the $p$-adic field $\mathbb{Q}_p$ with ring of integers $O_K$. Let $U$ be a finite local $\mathbb{Z}_p$-algebra and $m$ be its unique maximal ideal. Let $M$ be a finite module over $O_K \otimes_{\mathbb{Z}_p}U$. Assume further that $M$ is flat module over $U$.

Show that $M$ is free $(O_K \otimes_{\mathbb{Z}_p}U)$-module if $M \otimes_U U/m$ is free module over $O_K \otimes_{\mathbb{Z}_p}U/m$.

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First of all, for simplicity, I think we have $M \otimes_U U/m\cong M/mM$.

Thus we have to show as follows:

$M$ is free $(O_K \otimes_{\mathbb{Z}_p}U)$-module if its reduction $M/mM$ is free over $O_K \otimes_{\mathbb{Z}_p} U/m$.

Now $U=\mathbb{Z}_p^n$, for some $n$. Therefore $O_K \otimes_{\mathbb{Z}_p} U \cong O_K \otimes_{\mathbb{Z}_p} \mathbb{Z}_p^n \cong O_K^n$.

Thus what can we say about $O_K \otimes_{\mathbb{Z}_p}U/m$ ?

Is it $O_K \otimes_{\mathbb{Z}_p}U/m \cong (O_K/m)^n$ ?

If so is the case, then the original problem is reduced to the following:

$M$ is free $(O_K \otimes_{\mathbb{Z}_p}U)$-module if its reduction $M/mM$ is free over $(O_K/m)^n$;

and I think this follows from Local version of Nakayama Lemma.

Thanks

Solution 1:

Let $e_i \in M$ such that their reductions mod $mM$ are a basis of $M/mM$ over $O_K \otimes_{\mathbb{Z}_p} U/m$, assume $M$ is finitely generated.

Let $(h_j)$ be a basis of $O_K$, then the $g_{ij}=(h_j\otimes 1)e_i$ generate $M$ mod $mM$, so by Nakayama they generate $M$, so that $M=\sum_{i,j}{Ug_{ij}}=\sum_i{(O_K\otimes U)e_i}$.

So $M$ is flat finitely generated over $U$ (local) so is free over $U$. Moreover, $O_K \otimes U/m=\bigoplus_j{h_j \otimes U/m}$ and it follows that the $g_{ij}$ are a basis of $M/mM$ over $U/m$.

So $M$ is a free $U$-module with $U$ local and the $g_{ij}$ are a basis mod $mM$, so (consider its determinant in an actual basis) it’s a basis of $M$ over $U$. Therefore, $e_i$ is a basis of $M$ over $O_K \otimes U$.