Assume $G=\text{Spec}A$ is a group scheme of order $2$ over a ring $R$, I think it is well-known that these group schemes are $G_{a,b}=\text{Spec}\frac{R[T]}{T^2+aT}$ where $ab=2, a,b\in R$, and the group law is $T\rightarrow U+V+bUV$.
But when I see a proof of this, why does it assume $A=\frac{R[T]}{T^2+\alpha T+\beta}$ at first? I remember the definition of the order is rank of $\mathcal{O}_G$ as a locally free module sheaf over $\mathcal{O}_{\text{Spec}R}$.
Thanks !
If $A$ is an $R$-algebra of rank $2$, then it has an $R$-basis of size 2. Since $A$ is an algebra with unit, we can take one of these elements to be the unit $1$ of $A$, and call the other element $T$. When multiplying elements, we will have $T^2 = aT + b1$ for some $a, b \in R$, simply because $T^2$ is again an element of $A$. This is why the proof is starting off by assuming that $A = R[T] / (T^2 - aT - b)$ or similar.