Unnecessary finite dimensionality requirement in Theorem 5.1.8 of Sweedler’s “Hopf Algebras” I’m currently reading Sweedler’s Hopf Algebras, but am confused by the finite dimensionality in the following theorem:

Theorem 5.1.8 A finite dimensional Hopf algebra $H$ is semi-simple as an algebra if and only if$\epsilon(\int) \neq 0$.

Sweedler seems to use the finite dimensionality of $H$ only for the implication
$$
  \text{$H$ semisimple}
  \implies
  \epsilon( \textstyle\int ) \neq 0
$$
by using that $\dim {\int} = 1$, which seems unnecessary to me.

Question: Why does Sweedler requires $H$ to be finite dimensional?

Sweedler’s argumentation is as follows:

If $H$ is semi-simple, then there is a left ideal $I$ such that
  $$
  H = \operatorname{Ker} \epsilon \oplus I.
$$
  For $x \in \operatorname{Ker} \epsilon$ and $y \in I$ we have $xy \in \operatorname{Ker} \epsilon \cap I$.
  Hence $xy = 0$.
  Then for any $h \in H$,
  $$
  h = (h - \epsilon(h)1) + \epsilon(h)1,
$$
  so $hy = \epsilon(h)y$, since $(h - \epsilon(h) 1) \in \operatorname{Ker} \epsilon$.
  Thus $I \subseteq {\int}$, a $1$-dimensional space, hence $I = {\int}$.
  Since $H = \operatorname{Ker} \epsilon \oplus I$ we conclude $\epsilon(\int) \neq 0$.

It seems to me that the inclusion $I \subseteq {\int}$ sufficies because then $\epsilon(\textstyle \int) \supseteq \epsilon(I) \neq 0$.
Remark: If I’m not mistaken then it should also follow from a later exercise (any Hopf algebra that contains a nonzero finite dimensional one-sided ideal is already finite dimensional itself) that the above theorem also holds for any infinite dimensional Hopf algebra $H$: By considering the above one-dimensional ideal $I$ we see that $H$ cannot be semisimple. On the other hand ${\int} = 0$ (and hence $\epsilon(\int) = 0$) because for every nonzero $x \in \int$ the one-dimensional span $kx$ would be a left ideal.
But this doesn’t explain why Sweedler restricts the above theorem to finite dimensional Hopf algebras in the first place.
 A: You don't need that hypothesis in the proof but you don't earn nothing by removing it, as a Hopf algebra satisfying the above conditions is automatically finite-dimensional. You may in fact prove the following result.
Theorem (Maschke Theorem for Hopf algebras) For a Hopf
algebra over a field $\Bbbk $ the following assertions are equivalent.


*

*$H$ is semisimple as a ring.

*There exists $t\in \int_{H}^{l}$ such that $%
\varepsilon \left( t\right) =1$.

*$H$ is separable as an algebra.


Proof:
To prove that $\left( 1\right) $ implies $\left( 2\right) $ consider the left $H$-linear morphism $\varepsilon :H\rightarrow \Bbbk $. Since $H$ is semisimple and $\varepsilon $ is surjective, it admits a left $H$-linear section $\sigma :\Bbbk \rightarrow H$. Set $t:=\sigma \left( 1_{\Bbbk}\right) $ and observe that for every $h\in H$ we have $ht=h\sigma \left(1_{\Bbbk }\right) =\sigma \left( h\cdot 1_{\Bbbk }\right) =\sigma \left(\varepsilon \left( h\right) 1_{\Bbbk }\right) =\varepsilon \left( h\right) t$
and that $\varepsilon \left( t\right) =\varepsilon \left( \sigma \left(1_{\Bbbk }\right) \right) =1_{\Bbbk }$.
To prove that $\left( 2\right) $ implies $\left( 3\right) $ consider the Casimir element $e=\sum t_{\left( 1\right) }\otimes S\left( t_{\left( 2\right) }\right) $. Of course, $\sum t_{\left( 1\right) }S\left( t_{\left( 2\right) }\right) =\varepsilon \left( t\right) 1_{B}=1_{B} $, whence $e$ is a separability idempotent.
Finally, to prove that $\left( 3\right) $ implies $\left( 1\right) $ pick any surjective morphism of left $H$-modules $\pi:M\rightarrow N$. Since it is in particular of $\Bbbk $-vector spaces it admits a $\Bbbk $-linear section $\sigma :N\rightarrow M$. Of course, $\sigma $ is not $H$-linear in general, but we may consider $\tau:N\rightarrow M:n\longmapsto \sum e^{\prime }\sigma \left( e^{\prime \prime
}n\right) $ where $e=\sum e'\otimes e''$ is the separability idempotent. This is $H$-linear because $\sum e^{\prime }\sigma \left(e^{\prime \prime }hn\right) =\sum he^{\prime }\sigma \left( e^{\prime \prime}n\right) $ for every $h\in H$ and it is still a section since $\pi \left(\tau \left( n\right) \right) =\sum \pi \left( e^{\prime }\sigma \left(e^{\prime \prime }n\right) \right) =\sum e^{\prime }\pi \left( \sigma \left(e^{\prime \prime }n\right) \right) =\sum e^{\prime }e^{\prime \prime }n=n$
for every $n\in N$. $\square$
Now the point is that a separable $\Bbbk$-algebra is always finite-dimensional.
[06/05/2019 - Note added] To see why $e=\sum t_{(1)}\otimes S(t_{(2)})$ is Casimir (i.e. $he=eh$ for all $h\in H$) observe that
$$
\sum ht_{(1)}\otimes S(t_{(2)}) = \sum h_{(1)}t_{(1)}\otimes S(h_{(2)}t_{(2)})h_{(3)} =  \sum t_{(1)}\otimes S(t_{(2)})h
$$
in view of the antipode condition and the fact that $t$ is a left integral.
