Field of definition of a normal subgroup of Reductive group I have asked this question on stackexchange and have not received any answers or comments after 2 days of it being there.
I read somewhere that the following statement is correct. A proof or any hint as to how to prove it would be helpful.
Let $G$ be a connected reductive group defined over $k$(maynot be of characteristic 0). Let $H$ be a connected normal subgroup of $G\times \text{spec}\overline{k}$(apriori $H$ is only defined over $\overline{k}$). Then $H$ is defined over a finite 'separable' extension of $k$.
Also there must be counterexamples where this is not true.
Thanks.
 A: First, some easy reduction steps to pass to the case where $G$ is semisimple. Let $T$ be a maximal $k$-torus in $G$, and pass to a finite separable extension of $k$ so that $T$ is split.  We know that $H$ is the almost-direct product $\mathscr{D}(H) \cdot Z$ for a subtorus $Z \subset T_{\overline{k}}$ that is the maximal central $\overline{k}$-torus in $H$. But $T$ is $k$-split (by our initial finite separable extension on $k$), so $Z = S_{\overline{k}}$ for a unique $k$-subtorus $S \subset T$. Hence, our original task for $H$ is reduced to the same for $\mathscr{D}(H)$ that is normal in $\mathscr{D}(G_{\overline{k}}) = \mathscr{D}(G)_{\overline{k}}$. Hence, we can replace $G$ with $\mathscr{D}(G)$ and $H$ with $\mathscr{D}(H)$ to reduce to the case that $G$ is semisimple, and even $k$-split. 
In the split case, one can prove something much more precise: for each irreducible component $\Phi_i$ of the root system $\Phi = \Phi(G,T)$ (assume $G \ne 1$, so $\Phi$ is non-empty and hence admits irreducible components), the $k$-group $G_i \subset G$ generated by the root groups $U_a$ for $a \in \Phi_i$ is connected semisimple with root system (relative to its split maximal $k$-torus $T_i := T \cap G_i$) naturally identified with $\Phi_i$, and the $G_i$'s satisfy the following properties: (i) they are precisely the minimal non-trivial smooth connected normal $k$-subgroups of $G$, (ii) the $G_i$'s pairwise commute and for each subset $J$ of the set $I$ of $i$'s the map $\prod_{i\in J} G_i \rightarrow G$ is a central isogeny onto a smooth connected normal $k$-subgroup $G_J \subset G$, (iii) each smooth connected normal $k$-subgroup of $G$ coincides with $G_J$ for a unique $J$, (iv) each $G_i$ has no nontrivial smooth connected normal $k$-subgroup. 
Note that in view of the formulation, to prove (i)-(iv) it suffices to work over $\overline{k}$! And once that is done, we see via (iii) that the original question is answered in a more precise form. (One could prove a less precise version in which we only show the descent of $H$ to a $k$-subgroup by generating $H$ by some $T_{\overline{k}}$-root groups without establishing the link to subsets of the set of irreducible components of the root system. But this would not be as satisfying.) The most substantial ingredient in the proof is (iv).  For a complete proof of (i)-(iv), see 10.2 in https://www.ams.org/open-math-notes/omn-view-listing?listingId=110663  Of course, aside from issues of algebro-geometric technique, everything done there is quite classical (and 14.10 in Borel's textbook is a "classic" reference for it). It is mentioned by @anon that this is all done in Milne's new book, but I don't have a copy of that book (yet) and so don't know a precise reference within it.
A: After passing to a finite separable extension of the base field, we may suppose that the reductive group is split. After passing to a finite covering, we may suppose that it is the product of a simply connected semisimple group $G$ and a torus. The torus presents no problem. After passing to a finite separable extension, we may suppose that G is a product of absolutely almost-simple normal subgroups, so it all comes down to looking at the subgroups of the centre of $G$, but this presents no problem.
Added: All the statements used can be found, for example, in this
book
