Say $G$ is a solvable linear algebraic group over some field $k$ of characteristic 0. This means that its derived series eventually terminates with a 1. My question is:
Is "being solvable" a geometric property? By this, I mean: is it true that $G$ is solvable iff $G_{\bar{k}}$ is solvable?
I know that "being virtually solvable" is not a geometric property, but it's not clear to me that this holds also for "being solvable"
Details and references in the answers are very welcome.
Yes, this is true (I'm assuming by ``linear algebraic group" you mean smooth affine group), over any field $k$. In fact solvability of $G$ can be checked by checking solvability in the usual group-theoretic sense of $G(K)$ for any algebraically closed field $K$ containing $k$. Combining this with compatibility of the formation of the (smooth) derived $k$-subgroup scheme of $G$ gives what you want (smoothness is what ensures that, over an algebraically closed field $K$, the scheme-theoretic derived series ends in $1$ if and only if the derived series of the abstract group $G(K)$ does).