Suppose I have an extension of fields $L/K$, a group scheme $G\_K$ over $\operatorname {Spec} K$. Let $G\_L$ denote the pullback of $G\_K$ to $\operatorname{Spec} L$. Then, under what conditions on the extension $L/K$ can one say that we have an equality of $L$-algebras of the form $\operatorname{Lie} (G\_L) = \operatorname{Lie} (G\_K)\otimes \_K L$?



Lie algebras and pulled back group schemes
As the link in Erica's [comment](https://mathoverflow.net/questions/440500/lie-algebras-and-pulled-back-group-schemes#comment1136246_440500) shows, you can find this in SGA3 Exp. 2, but it is not so easy to extract from the very general language. Here is a rough guide: From Definition 3.9.0, the Lie algebra of $G$ over $K$ is the pullback of the total tangent space $T\_{G/K} \to G$ along the unit section $e: \operatorname{Spec} K \to G$. Base-changing this whole diagram along $\operatorname{Spec} L \to \operatorname{Spec} K$ yields the Lie algebra of $G\_L$ over $L$, once we can identify $(T\_{G/K})\_L$ with $T\_{G\_L/L}$, but this follows from Proposition 3.4. It comes down to the fact that the Hom functor (in this case, applied to the spectrum of dual numbers) commutes with arbitrary base change.


