Theorem \[3.3\]. Let $P(T_i)$ denote the projective cover of $T_i$. Then $P(T_i)\cong
H(0,\mu)e_i$, where $0\leqslant i\leqslant t-1$.

Since $\xi^i$ is an eigenvalue of the action of $g$ on $T_i\cong P(T_i)/{\rm rad}(P(T_i))$, $\xi^i$ is the unique eigenvalue of the action of $g$ on $P(T_i)$. It follows that $P(T_i$) must be the unique summand of $H(0,\mu)e_i$ up to isomorphism of $H(0, \mu)$-modules. Since dim$T_i=1$, the left regular module $H(0, \mu)$ has the decomposition $H(0, \mu)\cong\bigoplus_{0\leqslant i\leqslant t-1}P(T_i)$, which forces that $P(T_i)\cong H(0,\mu)e_i$.

Now we are going to consider the case of $\lambda=1$. Let us first show the following lemma for the case of $\mu=0$.

\[3.9-1\] In the Hopf algebra $H(1, 0)$, we have $$b^mab^{p-1}=\frac{m!}{2^m}a^{m+1}b^{p-1},\ \ m\geqslant 0.$$

We prove the equation $b^mab^{p-1}=\frac{m!}{2^m}a^{m+1}b^{p-1}$ by induction on $m$. If $m=0$, it is obvious. Now let $m\geqslant 0$ and assume $b^mab^{p-1}=\frac{m!}{2^m}a^{m+1}b^{p-1}$. Since $b^p=0$, by Lemma \[2.2\](1) we have $$\begin{array}{rcl}
b^{m+1}ab^{p-1}&=&\frac{m!}{2^m}ba^{m+1}b^{p-1}\\
&=&\frac{m!}{2^m}(a^{m+1}b+\frac{m+1}{2}a^{m+2})b^{p-1}\\
&=&\frac{m!}{2^m}(a^{m+1}b^p+\frac{m+1}{2}a^{m+2}b^{p-1})\\
&=&\frac{(m+1)!}{2^{m+1}}a^{m+2}b^{p-1}.\\
\end{array}$$ This completes the proof.

\[3.9\] Let $\{S_0, S_1, \cdots, S_{t-1}\}$ be the complete set of non-isomorphic simple $H(1,\mu)$-modules described as Theorem \[3.4\]. Let $P(S_i)$ denote the projective cover of $S_i$. Then

$(1)$ $P(S_0)\cong H(1,\mu)e_0$ and ${\rm dim}P(S_0)=p^{s+2}$.

$(2)$ Let $1\leqslant i\leqslant t-1$. Then ${\rm dim}P(S_i)=p^{s+1}$. Moreover, if $\mu=0$, then $P(S_i)\cong H(1,0)b^{p-1}e_i$ 