TITLE: Are there always nontrivial primitive elements in a Hopf algebra?
QUESTION [6 upvotes]: Let $k$ be an algebraically closed field of arbitrary characteristic.  Let $H$ be a Hopf algebra over $k$.  We say $x\in H$ is a primitive element if $\Delta(x)=1\otimes x+x\otimes 1$, where $\Delta$ is the comultiplication in $H$.  The set of primitive elements in $H$, denoted $P(H)$, is a Lie subalgebra, with bracket given by the commutator.
Is $P(H)$ always nontrivial?  If not, what is an example of a Hopf algebra with $P(H)=0$?  What if I require $H$ to be finite-dimensional, or finite-dimensional and local?  Do we have examples of such Hopf algebras with no nontrivial primitive elements?

REPLY [4 votes]: In any group hopf algebra $\mathbb{k}G$ (where $\mathbb{k}$ is a field of characteristic zero), we have a trivial set of primitives:  $P(\mathbb{k}G)=0$, because:
$$
\Delta(\Sigma_{i} k_ig_i)=\Sigma k_i\Delta(g_i)=\Sigma k_ig_i\otimes g_i \neq 1\otimes \Sigma_{i} k_ig_i + \Sigma_{i} k_ig_i \otimes 1 = \Sigma_{i} k_i(1\otimes g_i + g_i \otimes 1)
$$
Note that $\Sigma k_ig_i\otimes g_i$ cannot equal $\Sigma_{i} k_i(1\otimes g_i + g_i \otimes 1)$, due to the fact that $g_i \otimes 1$ and $g_i\otimes g_i$ (for $g_i\neq 1$) are linearly independent in $\mathbb{k}G\otimes\mathbb{k}G$.
And if you want a finite dimensional example, you can always use a finite group $G$. 
More generally, it can be shown that: 
"If $H$ is a $\mathbb{k}$-Hopf algebra of finite $\mathbb{k}$-dimension and $\mathbb{k}$ is a field of characteristic zero, then $P(H)=0$" 
The proof of the above fact, amounts to showing -inductively- that, if there is $0\neq x \in P(H)$ then $x^{n}$ (for all positive integers $n$) are linearly independent, which leads us to an obvious contradiction to the finite dimensionality of $H$. Details on the proof can be found for example at Hopf algebras - an introduction (Dascalescu-Nastasescu-Raianu), exercise 4.2.16, page 174.