Definitions for almost-simple algebraic group

Question

In J.S. Milne's book Algebraic Groups (2017), he defines an almost-simple algebraic group as follows:

1. An algebraic group over $k$ is almost-simple if it is semisimple and noncommutative, and every proper normal algebraic subgroup is finite.

He later changed this definition on 2022 to:

2. An algebraic group $G$ over $k$ is almost-simple if it is semisimple and noncommutative, and its only smooth connected normal subgroups are $G$ and $e$.

If we assume $k$ to be of characteristic 0 both definitions should be equivalent, since very algebraic group is smooth for this characteristic. That 1 implies 2 is obvious. For the other direction: isn't it possible for $G$ to contain some infinite but not connected normal subgroup?

I have recently started studying algebraic groups, so I am probably missing something.

Answer 1

Assume $G$ has an infinite algebraic subgroup $H_0$ that is not connected.

For smooth algebraic groups (every algebraic group, since we are working on characteristic 0), we know being connected is equivalent to not having any proper algebraic group of finite index. As such, $H_0$ does contain some normal $H_1$ such that $H_0/H_1$ is finite. This is true because every single one of our algebraic groups is smooth, and in such conditions $(H_0)^\text{o}$ (which is normal) is of finite index, and if $(H_0)^\text{o}$ was trivial then $(H_0)^\text{dev}$ would contain it and be of finite index too, and so either we take $H_1=(H_0)^\text{dev}$ (if it is nontrivial) or take $H_1$ to be any proper algebraic subgroup of $H_0$ (since if $(H_0)^\text{dev}$ is trivial then $H_0$ is commutative).

Since $\dim H_0= \dim H_1 + \dim (H_0/H_1)=\dim H_1$, our $H_1$ cannot be finite. If it was connected, then it would be trivial by (2.) and we would have a contradiction (namely $H_0$ would be connected), so it must be not connected and contain some normal $H_2$ of finite index.

Iterating this process we get a descending sequence $H_0\supsetneq H_1 \supsetneq H_2...$, but every algebraic group, being of finite type over a field, is Noetherian, so this is not possible.