There is an exercise on page 142 of Humphreys' Linear Algebraic Groups:
Ex.1 Let $G$ be a connected algebraic group, $x \in G$ is semisimple. Must $C_G(x)$ be connected?
When $G$ is solvable, I think of another fact whose correctedness is proved on the book:
Let $H$ be a subgroup (not necessarily closed) of a connected solvable group $G$, $H$ consisting of semisimple elements. Then $C_G(H) = N_G(H)$ is connected.
So, suppose that $G$ is solvable, set $H = \langle x \rangle$. It appears that $C_G(x)$ is connected.
I think the general case could be reduced to the solvable case if for any $y \in C_G(x)$, I can find a Borel subgroup of $G$ containing both $x$ and $y$. (Then $y$ must be in $C_B(x)$ which is connected.)
I think the Borel subgroup could be found, may be through the method of Borel variety. But I have difficulty in this.
Another exercise on the same page is:
Ex.2 Let $G$ be a connected algebraic group. If $x \in G$ has semisimple part $x_s$, then $x$ is contained in the identity component of $C_G(x_s)$.
If the answer to Ex.1 is affirmative, then Ex.2 would be obvious, since $x \in C_G(x_s)$. But I am not sure.
So, would you please tell me the answer to Ex.1, or help me with the proof or counterexample? If the centralizer could be not connected, how can I give Ex.2 a proof?