I need a construction in linear algebraic groups which uses taking quotient by a central finite group subscheme.
My question is, whether it goes through in ``bad'' characteristics, when this group subscheme is not smooth.
First I write this construction in a special case, and then in the general case.


Let $G$ be a connected semisimple $k$-group over a field $k$ of characteristic $p>0$.
We may assume that $k$ is algebraically closed.
Assume that the corresponding adjoint group $G^{ad}$ is $PGL\_n$.


In general, my group $G$ is not special (recall that a $k$-group $G$ is called special if $H^1(K,G)=1$ for any field extension $K/k$).
I want to construct a special $k$-group $H$ related to $G$.
For this end I consider the universal covering $G^{sc}$ of $G$, then $G^{sc}=SL\_n$. Let $Z$ denote the center of $G^{sc}$, then $Z=\mu\_n$.


We have a canonical epimorphism $\varphi\colon SL\_n \to G$. We denote by $C$ the kernel of $\varphi$.
Then $C$ is a group subscheme of $Z$, defined over $k$.


Since $Z=\mu\_n$, there is a canonical embedding $Z\hookrightarrow \mathbb{G}\_m$ into the multiplicative group $\mathbb{G}\_m$.
Thus we obtain an embedding $C\hookrightarrow \mathbb{G}\_m$. Consider the diagonal embedding
$$C\hookrightarrow SL\_n\times \mathbb{G}\_m.$$
I would like to define $H:=(SL\_n\times \mathbb{G}\_m)/C$. Is such a quotient defined, when char($k$) divides $n$ and $C$ is not smooth?


Note that $SL\_n$ embeds into $H$, and we have a short exact sequence
$$1\to SL\_n \to H \to \mathbb{G}\_m \to 1$$
In this exact sequence both $SL\_n$ and $\mathbb{G}\_m$ are special, and from the Galois cohomology exact sequence we see that $H$ is special as well.


In the general case I assume that $G^{ad}$ is a product of groups $PGL\_{n\_i}$, $i=1,\dots s$. Then $G^{sc}$ is the product of $SL\_{n\_i}$.
Let $C$ denote the kernel of the canonical epimorphism $\varphi\colon G^{sc}\to G$, then $C$ is contained in the center $Z$ of $G^{sc}$.
We have $Z=\prod\_{i=1}^s \mu\_{n\_i}$. Again we embed diagonally
$$C\hookrightarrow (\prod\_{i=1}^s SL\_{n\_i}) \times (\mathbb{G}\_m)^s $$
and denote by $H$ the quotient. Again $H$ is special (if it is defined), and again my question is, whether this construction makes sense
when char($k$) divides $n\_i$ for some $i$.


Any help is welcome!



Quotient of a  reductive group by a non-smooth central finite subgroup
This is an instance of what I believe is called the $z$-construction, and it is a very useful trick in the arithmetic theory of algebraic groups. (Small correction: your diagonal embedding should really be "anti-diagonal". You are really computing a "central pushout".) However, you may need to restrict your ground field to be local or global (with some caveats in case of real places) to get the cohomological vanishing. 


In general, let $G$ be any split connected semisimple group over a field $k$, and $G' \rightarrow G$ the $k$-split simply connected central cover. Let $Z$ be the center of $G'$.This is a $k$-group of multiplicative type, possibly not etale. It is $k$-split since $G'$ is $k$-split, so we can choose a $k$-subgroup inclusion of $Z$ into a $k$-split $k$-torus $T$ (seen using character group). Now just form the central pushout of $G'$ along this inclusion. More precisely, the antidiagonal map
$$Z \rightarrow G' \times T$$
is a central $k$-subgroup scheme, and in SGA3, VI$ \_{\rm{A}}$ quotients are constructed and studied for arbitrary finite type group schemes over a field modulo closed subgroup schemes of finite type (and even more generally). We can therefore form the quotient by this central subgroup scheme, and it has all of the reasonable properties one would want for a quotient. This pushout $H$ is a $k$-group containing $G'$ as a normal $k$-subgroup, and the quotient $H/G'$ is $T/Z$. Being a quotient of $k$-split torus, it is again a $k$-split torus. 


Provided the ground field is a local or global function field (with some further restrictions in case of a real place), all simply connected semisimple groups have vanishing degree-1 Galois cohomology. So that does the job for such fields. One can also do a variant for split connected reductive groups, and even a variant without a split hypothesis (but then the conclusion is a little different). 


