# Weil restriction of a projective variety to a finite extension over $\mathbb{Q}$

My question maybe a bit stupid, but I can't quite grasp what the Weil restriction actually does... In particular:
Given a projective variety $$V$$ defined over $$L$$ algebraically closed, of characteristic 0, given a finite extension $$K$$ of $$\mathbb{Q}$$, is the Weil restriction of $$V$$ to $$K$$ a model of $$V$$ over K?

• The Weil restriction is defined for a finite extension $L/K$. – abx Mar 16 at 22:09
• And cannot be a model for the original abelian variety, since it has higher dimension (if the extension is non-trivial). – Alex B. Mar 17 at 1:19

I thought I'd answer this, because Weil restriction is confusing. The best treatment I've seen is in Milne's notes on algebraic geometry, available from his website. (I'd be happy to have other references*.) Given field extension $$L/K$$, one gets a functor from schemes over $$K$$ to schemes over $$L$$ given by base change, that I'll denote by $$(-)_L$$. Given a scheme $$V/L$$, the Weil restriction -- if it exists -- is the $$K$$-scheme $$R_{L/K}V$$ characterized by $$Hom_L(T_L, V)\cong Hom_K(T, R_{L/K}V)$$ So it is sort of a right adjoint to base change. In particular, the $$K$$-rational points of $$R_{L/K}V$$ are the $$L$$-rational points of $$V$$. Milne shows that $$R_{L/K}V$$ exists when $$L/K$$ is finite separable and $$V$$ a quasiprojective variety. If you look at the construction, you'll see that certain geometric properties such as dimension can change, as was commented above. The construction is probably most often used when $$V$$ is an algebraic group, in which case so is the restriction.