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For 1-dimensional case, regular local implies PID and hence UFD. That is clear and geometry wise it is basically looking at the germs of smooth functions at a point.

For higher dimensional case, what is the geometrical reason that causes regular local ring being UFD? I do not expect this has to be the case as a lot of things could go wrong with additional degree of freedom added but it did not happen.

user26857
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user45765
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    A local ring is a UFD if its completion is. In the geometric case, completion of a regular local ring is just a power series ring which is a UFD. – Mohan Jul 11 '17 at 03:51

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It is an easy exercise to show that a noetherian domain is an ufd if and only if height one ideals are principal.

Hence, you can interpret geometrically the fact that regular local domains are ufd by saying that, on a regular variety, codimension 1 subvarieties are defined locally by one equation.

Giulio Bresciani
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