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First, in general it is not true that $\Omega^1_{B/A}$ is a free module. In fact, assuming $A \to B$ is flat and finitely generated, $\Omega^1_{B/A}$ is locally free if and only if $A \to B$ is a smooth ring map.

If in your situation (or any other adic formally smooth situation) you take the completion of $\Omega^1_{B/A}$ then you are guaranteed to get a projective module of the correct rank (and in your particular case you would indeed get the free module of rank 1 you expected). This is proved in EGA 0.IV, theorem 20.4.9. I think that the module of finite derivations should coincide with this completion. I recommend you check the book "Kahler Differentials" by Ernst Kunz which discusses the module of universal finite derivations.

Edit: remark 1.8 in http://arxiv.org/PS_cache/alg-geom/pdf/9510/9510007v4.pdf claims that indeed this completion coincides with the module of universal finite derivations, thus providing a positive answer to your question.

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First, in general it is not true that $\Omega^1_{B/A}$ is a free module. In fact, assuming $A \to B$ is flat and finitely generated, $\Omega^1_{B/A}$ is locally free if and only if $A \to B$ is a smooth ring map.
If in your situation (or any other adic formally smooth situation) you take the completion of $\Omega^1_{B/A}$ then you are guaranteed to get a projective module of the correct rank (and in your particular case you would indeed get the free module of rank 1 you expected). This is proved in EGA 0.IV, theorem 20.4.9. I think that the module of finite derivations should coincide with this completion. I recommend you check the book "Kahler Differentials" by Ernst Kunz which discusses the module of universal finite derivations.