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Aug 14, 2015 at 22:24 comment added Will Sawin Yes, $\mathcal X$ is uncountable. From the ring $\mathbb Z_p[x,y,z,w]/(f(x,y,z,w))$ for a homogeneous polynomial $f$ we can recover the projective variety $f(x,y,z,w)$ by blowing up, but there are uncountably many non-isomorphic hypersurfaces. In fact this trick works with any projective variety.
Aug 13, 2015 at 18:21 comment added Joël Thanks for these interesting reflexions. My guess is that $\mathcal X$ is not countable but I don't see how to prove it. That should be an easy question for specialists of classifications of singularities. (BTW, I don't really believe in the conjecture that all universal deformation rings in our context should be complete intersection. That seems wishful thinking to me.)
Aug 13, 2015 at 14:49 history edited Passerby CC BY-SA 3.0
edited conditions on rings in $\mathcal{X}$
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Aug 13, 2015 at 5:20 review First posts
Aug 13, 2015 at 5:30
Aug 13, 2015 at 5:18 history answered Passerby CC BY-SA 3.0