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.
Best Answer
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.
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.