When can I say that $A \otimes_{\mathbb{Z}_p} \mathbb{Q}_p \cong B \otimes_{\mathbb{Z}_p} \mathbb{Q}_p$ implies that $ A \cong B$?

Question

I have to show that $ A \cong B$ and I know that $$A \otimes_{\mathbb{Z}_p} \mathbb{Q}_p \cong B \otimes_{\mathbb{Z}_p} \mathbb{Q}_p$$ with $A$ and $B$ commutative $\mathbb{Z}_p$-algebras that are free as $\mathbb{Z}_p$-module and of finite rank.
Can I say this? Do i need other conditions on $A$ and $B$?

Answer 1

I presume your isomorphisms are algebra isomorphisms?

There are quite easy counterexamples with $A$ and $B$ having rank $2$.

Consider $A_n=\Bbb Z_p+p^{(2n+1)/2}\Bbb Z_p$ for integers $n\ge0$. These are subrings of $\Bbb Q_p(\sqrt p)$. Then each $A_n\otimes_{\Bbb Z_p}\Bbb Q_p$ is isomorphic to $\Bbb Q_p(\sqrt p)$, but the $A_n$ are non-isomorphic as $\Bbb Z_p$-algebras.