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Mar 20, 2018 at 17:44 comment added dhy @nfdc23: You are of course right. I'm not sure what I was thinking when I wrote that...
Mar 20, 2018 at 16:47 comment added nfdc23 @dhy: in characteristic 0 all group schemes of finite type are smooth, so "non-reduced kernel" cannot occur. As you note, the scheme $Y_1$ is nothing other than the fiber over the identity for a morphism $X\to {\rm{Pic}}_{Y/k}$ (the slicker proof of the result in Mumford's book if he'd been willing to invoke Picard schemes at the cost of making his book less self-contained). So one could choose a curve $C$ in ${\rm{Pic}}_{Y/k}$ through the origin and take $X$ to be a double cover of $C$ branched over the origin.
Mar 20, 2018 at 16:21 comment added dhy Take $X$ a smooth characteristic $0$ variety for simplicity (you can probably get away with much less.) $L$ defines a map from $Y$ to the Picard scheme of $X$, and your $Y_1$ is the fiber of the identity. So if you for instance, take $Y$ to be $\operatorname{Pic}_0(X)$ and $L$ to be the square of the tautological line bundle, then this map is the multiplication by $2$ map and has a non-reduced kernel.
Mar 20, 2018 at 14:57 history asked usr0192 CC BY-SA 3.0