Classification of group schemes of order 2: Why is $sx=x\otimes 1+1\otimes x-cx\otimes x$?

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In their paper Group Schemes of Prime Order, Tate and Oort state

Suppose $$G=\operatorname{Spec}(A)\text,\quad S=\operatorname{Spec}(R) \text,$$ and suppose the augmentation ideal $I=\operatorname{Ker}(A\to R)$ is free of rank one over $R$ (so $G$ is of order $p=2$), say $I=Rx$; then there exist elements $a$ and $c$ in $R$ such that $x^2=ax$ and such that the group structure on $G$ is defined by $sx=x\otimes 1+1\otimes x-cx\otimes x$. One easily checks that $ac=2$; [...]

Then they proceed to take this claim for granted without losing a word about how to obtain it. How does one prove this? I tried applying the Hopf algebra axioms to a general linear map (which is easy since the involved modules are free), but the resulting equations are still too general. In particular, with the comultiplication $$s\colon A\to A\otimes_RA,\;\begin{cases}1\mapsto 1\otimes 1\\x\mapsto \alpha(1\otimes 1)+\beta(1\otimes x)+\gamma(x\otimes 1)+\delta(x\otimes x)\end{cases}$$ and coinverse $$r\colon A\to A,\;\begin{cases}1\mapsto 1\\x\mapsto \eta+\xi x\end{cases}$$ I obtained the equations $$ \beta=\gamma \\ \alpha+\beta\eta=0 \\ \gamma+\delta\eta+\beta\xi-\delta\tau a=0$$ but these do not give enough information to conclude $\beta=\gamma=1$ and $\alpha=0$ as desired, unless I miss something. Does anybody have a clue how this works? Thank you.

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The augmentation ideal $I$ is indeed of rank one over $R$. This is not trivial in itself, but we skip the verification. Since $A$ is a Hopf algebra, we must have $$ (\epsilon \otimes \operatorname{id}_A)\circ s = \operatorname{id}_A, $$ where $\epsilon: A \to R$ is the augmentation map. Note that the corresponding group-theoretic statement is $1\times g=g$. Apply both sides to $x$ and remember that $\epsilon(x)=0$. You can now conclude $\alpha=0$ and $\beta = \gamma=1$