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I’m reading about affine group schemes by Waterhouse and in the proof of showing the (Jordan) decomposition of Abelian affine group scheme (equivalently cocommutative Hopf algebra), I came across the argument, saying:

“Since the Hopf algebra $A$ is cocommutative, the antipode $S$ is a coalgebra morphism and therefore sends $ A’ $ (a sub-bialgebra of $A$) into $A’$“

I know that the antipode is an algebra automorphism of $A$, and in this case a bialgebra automorphism of A, but I still don’t understand why does the restriction of $S$ to $A’$ would send $A’$ back to itself.

P/S: we are assuming that Hopf Algebras are commutative here.

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  • $\begingroup$ It is not per se true that a sub-bialgebra of a commutative cocommutative Hopf algebra must be a sub-Hopf algebra. After all, submonoids of abelian groups aren't always subgroups. Maybe the context says something helpful here; perhaps finiteness conditions? $\endgroup$ – darij grinberg Sep 30 '17 at 15:48
  • $\begingroup$ Yes, I should mention that $A’$ is a directed union of finite dimensional subcoalgebra of $A$. It was later shown that by construction $A’$ preserves multiplication and contains the unit - which makes it a bi-subalgebra. $\endgroup$ – AffineSpace Sep 30 '17 at 15:53
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    $\begingroup$ That's not helpful. Try the "positive part" of the group ring of $\mathbb Z $; it satisfies all your conditions. $\endgroup$ – darij grinberg Sep 30 '17 at 16:00
  • $\begingroup$ There is a given coalgebra projection $p : A \rightarrow A’$ that was also shown to be an algebra projection (and thus a bialgebra projection too). The only reasoning that I can now give is that for $a’ \in A’$, we have $p(S(a’)) = S(p(a’)) = S(a’)$, which shows that $S(A’) \subseteq A’$. But I still don’t see the logical connection that the author was trying to claim in the line he wrote. $\endgroup$ – AffineSpace Sep 30 '17 at 16:26
  • $\begingroup$ but isn't that a valid proof, given this projection $p$ ? $\endgroup$ – darij grinberg Oct 1 '17 at 1:09
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I do not have access to the book you are citing, but if i understand correctly, you are speaking about a commutative, cocommutative hopf algebra.
If your assumptions are complemented by: finite dimensional hopf algebra over an algebraically closed field of characteristic zero, then the author's claim:

Since the Hopf algebra $A$ is cocommutative, the antipode $S$ is a coalgebra morphism and therefore sends $ A’ $ (a sub-bialgebra of $A$) into $A’$

seems to be correct.
Here is my argument:

Under these assumptions (i.e.: f.d., commutative, cocommutative over alg. closed field $k$ with $chark=0$), it is relatively easy to show that the hopf algebra $H$ is actually isomorphic to the group algebra of the group $G(H)$ formed by its grouplike elements: $$ H\cong kG(H) $$ (This isomorphism can also be viewed as a direct consequence of the Cartier-Konstant-Milnor-Moore theorem applied in the finite dimensional setting.)
Thus, by the same argument, the sub-bialgebras $A'$ of $H$ -which are again f.d., commutative, cocommutative, over an algebr. closed field of $chark=0$- are group algebras $A'=kN$, for some subgroup $N$ of $G(H)$. But then, since for any grouplike $S(g)=g^{-1}$ we get that:
$$ S(\sum_{g\in N}a_g g)=\sum_{g\in N}a_g S(g)= \sum_{g\in N}a_g g^{-1}\in kN $$ thus $S(kN)\subseteq kN \ \ $ i.e.: $ \ S(A')\subseteq A'$.

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  • $\begingroup$ Why are they group algebras $kN$? $\endgroup$ – darij grinberg Oct 1 '17 at 0:05
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    $\begingroup$ By the same argument as in the initial case of $H$: The sub-bialgebras are again f.d., commutative, cocommutative hopf algebras and thus they are group algebras of some (sub)group. $\endgroup$ – KonKan Oct 1 '17 at 0:08
  • $\begingroup$ Why are they Hopf? $\endgroup$ – darij grinberg Oct 1 '17 at 0:08
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    $\begingroup$ because $S(g)=g^{-1}$ for all grouplikes. $\endgroup$ – KonKan Oct 1 '17 at 0:11
  • $\begingroup$ @darij grinberg, i've edited hoping to become more clear. $\endgroup$ – KonKan Oct 1 '17 at 0:57

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