If $G$ is a connected reductive group over a perfect field $k$, why is the radical of $G$ equal to the connected component of the center? Borel's book (Linear Algebraic Groups) has a proof if $k$ is algebraically closed -- one can characterize the center of $G$ as the intersection of the Borels. But I don't know if that generalizes in some way over an arbitary field.

If $G$ is a connected reductive group over a perfect field $k$ (The definition given in Milne's "Algebraic Groups": $G$ is a connected group variety containing no non-trivial connected unipotent normal subgroup variety, even over the algebraic closure of $k$), why is the radical of $G$ (that is, the maximal connected solvable normal subgroup) equal to the connected component of the center? This is stated on page 7 of Shahidi's "Eisenstein series and automorphic L-functions." On page 357 of Milne's Algebraic Groups it says that the center of $G$ is of multiplicative type, and its largest subtorus is equal to the radical of $G$.

Borel's book (Linear Algebraic Groups) has a proof if $k$ is algebraically closed -- one can characterize the center of $G$ as the intersection of the Borels. But I don't know if that generalizes in some way over an arbitary field.

If $G$ is a reductive group over a field $k$, why is the radical of $G$ equal to the connected component of the center? Borel's book (Linear Algebraic Groups) has a proof if $k$ is algebraically closed -- one can characterize the center of $G$ as the intersection of the Borels. But I don't know if that generalizes in some way over an arbitary field.

If $G$ is a connected reductive group over a perfect field $k$, why is the radical of $G$ equal to the connected component of the center? Borel's book (Linear Algebraic Groups) has a proof if $k$ is algebraically closed -- one can characterize the center of $G$ as the intersection of the Borels. But I don't know if that generalizes in some way over an arbitary field.