Let $H$ be a commutative and cocommutative Hopf algebra over an algebraically closed field $k$. I've read that if $H$ is grouplike in the sense that it has no nonzero primitive elements, then $H$ is isomorphic to the group ring on its grouplike elements,but no reference or proof was given.
What is a proof of this fact?
What is a good reference?
Is there differing terminology I should be aware of?
Are the commutativity and cocommutativity hypotheses necessary?
My source is the discussion after remark 1.0.7 in Hopkins and Lurie.
Best Answer
Cocommutative is necessary, because else we can take something like $H=\mathcal{O}(\mathbf{GL}_n)$.
The following proof is adapted from Milne's book Algebraic Groups. There are some finiteness assumptions in this book, but they do not matter for this proof.
First note that in any coalgebra over a field $k$, the grouplike elements are linearly independent. The proof is similar to the classical proof for Dedekind's result on the linear independence of distinct characters.
For any commutative Hopf algebra $H$ with group of grouplike elements $G$, the inclusion $G \hookrightarrow H$ thus extends to an injective morphism of Hopf algebras $k[G] \hookrightarrow H$. Thus it is sufficient to show that this morphism is surjective, i.e. that the grouplike elements span $H$.
Now for the actual proof, we need a small lemma.
Lemma If $A$ is a finite-dimensional algebra over an algebraically closed field $k$ that is not étale, then there exists a surjective homomorphism $A \to k[\varepsilon]=k[x]/(x^2)$
Proof We can assume that $A$ is local, because in general $A$ is a finite product of local rings and it is étale iff all the factors are étale, so we can project on a non-étale factor. In this case, the maximal ideal $\mathfrak{m}_A$ is nilpotent and nonzero. We have $\mathfrak{m}_A\neq \mathfrak{m}_A^2$ by Nakayama. We can project onto $A/\mathfrak{m}_A^2$ and can thus assume that $\mathfrak{m}_A^2=0$. Now take a vector space basis $v_1, \ldots, v_n$ for $\mathfrak{m}_A$. Then $1,v_1,\ldots,v_n$ is a basis for $A$. The unique $k$-linear map $A \to k[\varepsilon]$ given by $1\mapsto 1, v_i \mapsto \varepsilon$ is actually a morphism of algebras because $\mathfrak{m}_A^2=0$.
Now a definition.
Definition A finite-dimensional $k$-coalgebra $C$ is called coétale if $C^\vee$ is étale. A general $k$-colagebra is called coétale if it is the directed union of coétale finite-dimensional sub-coalgebras.
Now we can finally give the argument:
Theorem For a commutative cocommutative Hopf algebra $H$ over an algebraically closed field $k$, the following are equivalent:
Proof
$1. \implies 2.$ is a simple computation.
$2. \implies 3.$ We proceed by contraposition. If $H$ is not coétale, there's some finite-dimensional sub-coalgebra $C \subset H$ such that $C$ is not coétale. By the lemma, this means that there's a surjective homomorphism $C^\vee \to k[\varepsilon]$ which upon dualizing gives an injective homomorphism of coalgebras $k[\varepsilon]^\vee \to C$. One now checks that the comultiplication on $k[\varepsilon]^\vee$ satisfies $\Delta(\varepsilon)=1\otimes \varepsilon+\varepsilon\otimes 1$, thus $\varepsilon$ is primitive. Furthermore $1$ is grouplike. Denote the image of $\varepsilon$ under $k[\varepsilon]^\vee \to C\subset H$ by $a$ and the image of $1$ by $b$. We get that $b$ is grouplike (hence invertible) and $a$ satisfies $\Delta(a)=a\otimes b +b\otimes a$. Hence $b^{-1}a$ is primitive and nonzero.
$3\implies 1$ Let $C\subset H$ be finite-dimensional sub-coalgebra. By assumption $C^\vee \cong k^n$ as algebras. This isomorphism can be realized via the perfect pairing $C\times C^\vee \to k$ by an $n$-tuple of elements from $C$, i.e. the isomorphism is given by $x \mapsto (x(c_1), \ldots x(c_n))$ for some $c_1, \ldots, c_n \in C$. These $c_i$ span $C$. The fact that the isomorphism a map of algebras implies that the $c_i$ are grouplike. Thus $C$ is spanned by its grouplike elements. Because $H$ is a union of its finite-dimensional sub-coalgebras, $H$ is also spanned by its grouplike elements.