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Apr 18, 2011 at 2:59 vote accept Zev Chonoles
Apr 17, 2011 at 20:33 comment added Qiaochu Yuan Regarding intuition, the problem is that $k$-points don't tell you everything about a $k$-variety when $k$ isn't algebraically closed, so there's no reason to expect the condition on the right to behave well for general $k$ as Donu says. Looking at $k$-points is a bad choice of "concretization" of $k$-varieties. A much better one is the functor which sends a $k$-variety to its $\bar{k}$-points equipped with the natural action of $\text{Gal}(\bar{k}/k)$.
Apr 17, 2011 at 11:38 comment added Donu Arapura If $k$ is not algebraically closed then $f(p)=0$ may be empty, so that the Jacobian condition on the right would be vacuous. The condition on the left, called smoothness, would still have content however. Notice that smoothness is stable under field extensions. So it is equivalent to the Jacobian condition over $\bar k$.
Apr 17, 2011 at 5:28 answer added solbap timeline score: 2
Apr 17, 2011 at 4:36 history asked Zev Chonoles CC BY-SA 3.0