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flitwick
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Example of a nonsmoothable scheme
I think my problem was that I didnt realize that $X\rightarrow S$ is finite - why does this hold? But ok, if this map is finite, then the fibers constist of finitely many points. I have one fiber which is isomorphic to $R$ and if $R$ is smoothable, there is at least one other fiber which is smooth over $k$. You said that this other fiber is finite the (spectrum of a) ring of functions on finitely many points, because it is smooth and finite dimensional, but why is it finite dimensional?
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Example of a nonsmoothable scheme
Thank you for your answer! The first three questions are now clear to me, but I am still confused about 4. You are talking about smooth algebras, but in which way does this algebra occur in the deformation diagram, is it the ring of global sections of the total space?
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