Timeline for Why define curves over perfect fields?

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Jun 13, 2016 at 11:24	comment	added	Jason Starr		The example of nfdc23 also shows that, over an imperfect field $k$, a finite type $k$-scheme that is regular need not be smooth (even if it is geometrically reduced).
Jun 13, 2016 at 10:49	comment	added	nfdc23		Over a perfect field $k$, any reduced scheme $X$ of finite type is $k$-smooth on a dense open subset since for each (reduced) irreducible component $X_i$ of $X$ the function field $k(X_i)$ admits a separating transcendence basis (as for any finitely generated field over a perfect field). This fails over every imperfect field $k$ (e.g., $y^2=x^p-a$ for $a\in k-k^p$ with ${\rm{char}}(k)=p>0$). More conceptually, "reduced" is the same as "geometrically reduced over $k$" for such $X$, since $\overline{k}/k$ is a directed union of finite etale extensions of $k$ (by definition of perfectness).
Jun 13, 2016 at 9:09	history	asked	Dan	CC BY-SA 3.0