In my preprint I propose to call $X/S$ quasi-smooth if $X$ can be embedded into a smooth $X'/S$. Does this sound fine?
Upd. So, smoothly embeddable is better? Is it ok to call a morphism smoothly embeddable?
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asked Feb 23, 2013 at 6:32
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1$\begingroup$ I think Hartshorne in "Residues and Duality" calls it "embeddable". $\endgroup$– SashaCommented Feb 23, 2013 at 7:33
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$\begingroup$ I've seen the following definition of quasi-smooth floating around: locally free kähler differentials $\Omega_{X/S}$ and $X/S$ flat (where X/S has no finiteness condition -- e.g., $X/S$ may not be of finite type). Perhaps a better name for this is pro-smooth as I could imagine the central question here is when is $X$ pro-representable by smooth schemes of finite type over $S$. Nevertheless, embeddable is better. $\endgroup$– Andrew StoutCommented Feb 23, 2013 at 7:56
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1$\begingroup$ Not good --- google came up with 25,300,000 references, the first of which defined a projective variety to be quasismooth if its affine cone is smooth away from 0. "Smoothly embeddable", abbreviated to "emmeddable" if you wish, sounds better. $\endgroup$– anonCommented Feb 23, 2013 at 8:23
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$\begingroup$ I think quasismooth has a meaning in derived geometry as well, like perfect cotangent complex concentrated in degrees -1 and 0. Also, Kai Behrend in his Annals paper "DT-type invariants via microlocal geometry" calls your varieties embeddable as well. $\endgroup$– Jacob BellCommented Feb 23, 2013 at 14:53
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