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.
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:
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'$.