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This is a very elementary question as I am just learning about abelian varieties by reading Mumford's book,

Let $X$ be a complete variety (irreducible and reducible over an algebraically closed field) and $Y$ a scheme, and $L$ a line bundle on $X \times Y$. In his book on Abelian Varieties, on p.85-87 Mumford shows there exists a "maximal closed subscheme $Y_1$ of $Y$ over which $L$ is trivial" (I'm omitting the precise definition of what this entails but it is in the first proposition in section 10 Theorem of Cube II).

This result is a strengthening (to the scheme-theoretic setting) of a result he proved on p.51 (Seesaw Theorem - provisional form) where he proved the result I mentioned above when $Y$ is a variety (as opposed to a scheme).

I'm trying to understand the nuances between the scheme-theoretic vs variety-theoretic results.

So what is a good example that would highlight the difference? For example, if $Y$ is non-reduced, by the result for varieties applied to $Y_{red} \subseteq Y$, we get a variety $Y_1'$. By taking the trivial line bundle on $X \times Y$ we see that $Y_1=Y$ so taking $Y$ to be non-reduced is a trivial example of the benefit of generalizing the result to schemes. Are there examples where $Y$ is a variety but $Y_1$ is non-reduced?

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    $\begingroup$ 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. $\endgroup$
    – dhy
    Commented Mar 20, 2018 at 16:21
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    $\begingroup$ @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. $\endgroup$
    – nfdc23
    Commented Mar 20, 2018 at 16:47
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    $\begingroup$ @nfdc23: You are of course right. I'm not sure what I was thinking when I wrote that... $\endgroup$
    – dhy
    Commented Mar 20, 2018 at 17:44

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