Does the notion of a simply connected algebraic group depend on the base field? Let $G$ be a connected semisimple algebraic group over a field $k$ with $\text{char}(k) = 0$.
One usually defines $G$ to be simply connected if every isogeny $G' \to G$ for a connected algebraic group $G'$ is an isomorphism. As far as I understand, the isogeny and $G'$ need not be defined over $k$.
Question: Sometimes people require in this definition that instead every $k$-isogeny $G' \to G$ for a connected $k$-group $G'$ is a $k$-isomorphism. Does this give the same notion of a simply connected algebraic group and if so, why?
 A: $\newcommand{\sc}{\mathrm{sc}}$$\newcommand{\Gal}{\mathrm{Gal}}$You don't even need $\mathrm{char}(k)=0$. 

Theorem: Let $k$ be a perfect field and let $G$ be a semisimple algebraic group over $k$. Then, there exists some semisimple algebraic group $G^\sc$ and a central isogeny $G^\sc\to G$ with the following universal property: if $H$ is a semisimple group over $k$ and $H\to G$ is a central isogeny, then the map $G^\sc\to G$ factors uniquely through $H$.

Proof: If $G$ is split this is well-known. Suppose now that $G$ is arbitrary. Then, we can find some finite Galois extension $K/k$ such that $G_K$ becomes split. Consider then $U:=(G_K)^\sc$. Note then that for every $\sigma\in\Gal(K/k)$ we have that by pulling back along $\sigma$ gives us a map 
$$U^\sigma \to (G_K)^\sigma\cong G_K$$
which is evidently a central isogeny. Then, by definition, there exists a unique central isogeny $U\to U^\sigma$ factorizing this map. Note though that since the kernel of $U^\sigma\to G_K$ is the same degree as $U\to G$ we have that $U\to U^\sigma$ is an isomorphism. It's not hard to see that this gives a $\Gal(K/k)$-descent datum on $U$ and thus $U$ descends uniquely to some $\widetilde{G}$ over $k$ equipped with a map $\widetilde{G}\to G$. 
Suppose now that $H\to G$ is a central isogeny. Note then that we get a central isogeny $H_K\to G_K$ and thus a unique factorization $U\to H_K$. Note then that by construction, and the fact that $(H_K)^\sigma\cong H$ naturally, shows that this descends uniquely to a map $\widetilde{G}\to H$. Thus, $\widetilde{G}=G^\sc$ as desired. $\blacksquare$
Note then that saying that $G$ is simply-connected is merely the same as saying that $G^\sc=G$. By the above construction it's clear that $(G^\sc)_K=(G_K)^\sc$ so that this statement is independent of which extension of $k$ you consider.
