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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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I think Hartshorne in "Residues and Duality" calls it "embeddable". – Sasha Feb 23 '13 at 7:33
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. – Andrew Stout Feb 23 '13 at 7:56
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. – anon Feb 23 '13 at 8:23
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. – Jacob Bell Feb 23 '13 at 14:53

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