Let $Y\to X$ be a finite surjective morphism of smooth projective geometrically connected varieties over $\mathbb{Q}$. Let $k$ be a number field and consider the induced morphism 


$$f:Res\_{k/\mathbb{Q}} Y\_k\to Res\_{k/\mathbb{Q}} X\_k.$$


Let $\Delta\to Res\_{k/\mathbb{Q}}X$ be the diagonal embedding of $X$ into the Weil restriction $Res\_{k/\mathbb{Q}} X\_k$ of $X\_k$ to $\mathbb{Q}$.



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> > What are the irreducible components of the scheme-theoretic inverse image $f^{-1}(\Delta)$? Is $f^{-1}(\Delta)$ equal to $Y$, and is the morphism $f^{-1}(\Delta)\to X$ the morphism $Y\to X$ we started with?
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Here $f^{-1}(\Delta) $ is defined to be the fibre product of $\Delta$ with $Res\_{k/\mathbb{Q}} Y\_k$ over $Res\_{k/\mathbb{Q}} X\_k$.



The diagonal of the Weil restriction
Welcome to mathoverflow, Pat!


I believe that $f^{-1}(\Delta) \neq Y$. Let $Y$ be a torsor for an elliptic curve $E$ over $\mathbb Q$ and choose $k$ so that $Y\_k \cong E\_k$ or in other words $Y(k) \neq \emptyset$. Choose the torsor so that the diagram $C\_k\cong E\_k\to \mathbb P^1\_k$ commutes. I believe this can be done by letting the torsor correspond to a cocycle in $H^1(G\_\mathbb Q, E[2]) \subset H^1(G\_\mathbb Q, E(\overline{\mathbb Q}))$.


In general, for a $\mathbb Q$ scheme $S$, from the functor of points persepective, $$f^{-1}(\Delta)(S) = \{\lambda: S\_k \to Y\_k : S\_k\to Y\_k\to X\_k \text{ is defined over } \mathbb Q\}.$$


In our particular case, let's take $S = \mathbb Q$. Then, there are no maps from $S \to Y$ but there are maps from $S \to f^{-1}(\Delta)$: Simply take the origin for instance.


From the functor of points, it is at least clear that $Y$ sits inside $f^{-1}(\Delta)$. I am not sure what else is there.


