Let $k$ be a field of characteristic $p > 0$ (algebraically closed, if you want; that doesn't make a difference). Consider a finite type $k$-group scheme $E$ that is a (central) extension of $\alpha\_p$ by $\mathbb{G}\_m$. Is $E$ necessarily commutative?


Edit: $E$ is an extension of $A$ by $B$ if it fits into a short exact sequence (which is part of the data of an extension)
$$
1 \rightarrow B \rightarrow E \rightarrow A \rightarrow 1,
$$
and the extension is central if $B$ is in the center of $E$. So in the case at hand I am looking at central extensions
$$
1 \rightarrow \mathbb{G}\_m \rightarrow E \rightarrow \alpha\_p \rightarrow 1.
$$



Are there noncommutative extensions of $\alpha_p$ by $\mathbb{G}_m$?
So, bilinear maps $\alpha\_p \times \alpha\_p \rightarrow \mathbf{G}\_m$ are classified by maps from $\alpha\_p$ to itself (since $\alpha\_p$ is Cartier self-dual). The collection of such maps is a $1$-dimensional vector space $V$ over $k$.
The group $\mathbf{Z}/ 2 \mathbf{Z}$ acts on the vector space $V$ by "swapping'' the two factors of $\alpha\_p$. This action is trivial (this follows from the fact that you can write down the Cartier self-duality on $\alpha\_p$ in a symmetric way). It follows that if $p \neq 2$, there are no nonzero skew-symmetric bilinear maps from $\alpha\_p$ to $\mathbf{G}\_m$, so any central extension of $\alpha\_p$ by $\mathbf{G}\_m$ must be commutative.


If $p = 2$, then the symmetric bilinear maps $\alpha\_2 \times \alpha\_2 \rightarrow \mathbf{G}\_m$ are also skew-symmetric, so this argument does not apply.
It is still true that there are no noncommutative central extensions of $\alpha\_2$ by $\mathbf{G}\_m$, but you have to work a little bit harder to show this. You can find an argument in Example 3.2.7 of my paper with Mike Hopkins "Ambidexterity in K(n)-Local Stable Homotopy Theory" (we need this fact to discuss "alternating powers" of 1-dimensional p-divisible groups, which arise naturally from some calculations in algebraic topology).


