If $k$ is an algebraically closed field of characteristic zero and $H$ is a cocommutative Hopf algebra, then
$$
H \cong U(P(H)) \ltimes kG(H).
$$
What happens if the field is not algebraically closed? Is the theorem still true or is there any counterexample? What about characteristic different from zero?



Cartier-Kostant-Milnor-Moore theorem
When $k$ fails to be algebraically closed the theorem is false but the discrepancy can be understood in terms of Galois descent and so in principle understood in terms of Galois cohomology. 


Suppose $H$ is a cocommutative Hopf algebra over $k$, not algebraically closed but characteristic $0$. Then the classification theorem applies to the base change $H \otimes\_k \bar{k}$ of $H$ to the algebraic closure, which must therefore take the form of the semidirect product of a universal enveloping algebra and a group algebra. So $H$ is a $k$-form of such a thing, but need not be such a thing itself.


For example, suppose $H$ is finite-dimensional, so $H \otimes\_k \bar{k}$ is a group algebra $\bar{k}[\Lambda]$ of a finite group $\Lambda$. The most obvious $k$-form of $\bar{k}[\Lambda]$ is, of course, $k[\Lambda]$, but other $k$-forms are possible. In this setting $H$ necessarily splits over a finite Galois extension $L$ in the sense that we must already have


$$H \otimes\_k L \cong L[\Lambda].$$


Now, the Galois group $G = \text{Gal}(L/k)$ acts on the LHS by Hopf algebra automorphisms (over $k$), so also acts on the RHS by Hopf algebra automorphisms (over $k$), and hence acts on the group of grouplike elements $\Lambda$ on the RHS. If we already had $H \cong k[\Lambda]$ then this action would be trivial, but in fact every possible action of $G$ on $\Lambda$ occurs, and conjugacy classes of such actions classify the possible choices of $H$. Given such an action $\rho : H \to \text{Aut}(\Lambda)$ we can recover $H$ as


$$H \cong L[\Lambda]^G$$


where the action of $G$ on $L[\Lambda]$ extends the action of $G$ on $\Lambda$.


Very explicitly, let $k = \mathbb{R}, L = \mathbb{C}$, and $\Lambda = C\_3$, where the Galois group $G = C\_2$ acts via inversion. Then we have


$$H \otimes\_{\mathbb{R}} \mathbb{C} \cong \mathbb{C}[C\_3] \cong \mathbb{C}[x]/(x^3 - 1)$$


where $x$ is grouplike but the action of complex conjugation sends $x$ to $x^{-1} = x^2$. The fixed point subalgebra $H$ has a basis given by the identity $1$ and the "real and imaginary parts"


$$c = \frac{x + x^{-1}}{2}$$
$$s = \frac{x - x^{-1}}{2i}$$


(it helps to think of $x$ as secretly being $\omega = e^{\frac{2\pi i}{3}}$ here) of $x$, with comultiplication given by the sine and cosine angle addition formulas


$$\Delta(c) = c \otimes c - s \otimes s$$
$$\Delta(s) = c \otimes s + s \otimes c.$$


So neither $c$ nor $s$ is grouplike, and in fact $H$ has no nontrivial grouplike elements except the identity. In $H \otimes\_{\mathbb{R}} \mathbb{C}$ the nontrivial grouplike elements are $c + is$ and $c - is$. 


