Exercise 2.11 of Waterhouse about Hopf algebra I asked part (c) of this exercise already in this link Categorization of Group Scheme of rank 2. But I still have some difficulty solving next parts of this exercise. So I hope anyone can help me to clear it out. Thank you

Let $A$ be a Hopf algebra over $k$ (a base ring) which is free of rank 2. I proved that $I = Ker(\epsilon) = kx$ for some $x$, and $\Delta(x) = x \otimes 1 + 1 \otimes x + b x\otimes x$ for some $b$ in k. Also, $x^2 + ax = 0$ for some $a$, and $A = k[X]/(X^2 + aX)$.
(d) Use $\Delta(x^2) = (\Delta x)^2$ to show $(2 - ab)^2=2-ab$.
(e) Show $Sx = cx$ with $c^2 = 1$. Then use (e) and $0 = (S, id)\Delta x$ to show $c = 1$ and $ab = 2$.

For part (d), after replacing the expression of $\Delta x$ into $\Delta(x^2) = (\Delta x)^2$, I got $ax(2-ab)^2 = ax(2-ab)$. Because $1, x$ is a basis of $A$, so we must have $a(2-ab)^2 = a(2-ab)$. But because $k$ may be not an integral domain, so $a$ can be a zerodivisor. In this case, I can't conclude that $(2-ab)^2 = (2-ab)$.
For part (e), the fact that $Sx = cx$ for some $c^2=1$ seems to imply that $S^2 = id$, which is not stated in the book of Waterhouse. But in the book Algebraic Groups of Milne, Exercise 3.3 (a), it states that $S^2=id$. The difference between the two definitions is, the definition of Waterhouse assumes $k$ is a general ring, while the definition of Milne assumes $k$ is a field. So, is this relation always true for all Hopf algebra? And if not, is it always true in case $k$ is a field? Thank a lot for helping me.
 A: We have
$$\begin{align*}
\Delta(x^2) &= \Delta(-ax) = -a \Delta(x) \\ 
&= -a (x \otimes 1 + 1 \otimes x + b x \otimes x), \\
\Delta(x)^2 &= (x \otimes 1 + 1 \otimes x + b x \otimes x)^2 \\
&= x^2 \otimes 1 + 1 \otimes x^2 + b^2 x^2 \otimes x^2 + 2 x \otimes x + 2b x^2 \otimes x + 2b x \otimes x^2 \\
&= -ax \otimes 1 - 1 \otimes ax + a^2 b^2 x \otimes x + 2 x \otimes x - 2ab x \otimes x - 2ab x \otimes x \\
&= -a(x \otimes 1 + 1 \otimes x) + (2 - 4ab + a^2 b^2)(x \otimes x).
\end{align*}$$
Equating the coefficients of $x \otimes x$, we have $$-ab = 2 - 4ab + a^2 b^2.$$  Simplifying, we have
$$\begin{align*}
0 &= (ab)^2 - 3ab + 2 \\
&= (2 - ab)(1 - ab) \\
&= (2 - ab)((2 - ab) - 1 \\
&= (2-ab)^2 - (2-ab),
\end{align*}$$
so that $(2-ab)^2 = 2 - ab$ as desired.
For the rest of the question, a Hopf algebra for which $S^2 = \mathrm{id}$ is called involutive. Not all Hopf algebras are involutive (even over a field), but there are general conditions that would imply that a Hopf algebra is involutive.  Googling this word will probably turn up more references.
