Primitive Elements in Hopf Algebras

Question

Let $H(\nabla ,\eta,\Delta, \epsilon,S)$ be a Hopf algebra where the notations have their usual meaning. A primitive element is $x$ satisfying $\Delta(x)=1 \otimes x + x\otimes 1$.From some lecture notes it seems that $\epsilon(x)=0$ for $x$ primitive.It may be obvious but I am having difficulty in proving this. What I could see(from the commutative diagram for a Hopf algebra) is that $\Delta(S(1)\otimes x)=\Delta(x\otimes S(1))=\eta\epsilon(x)=c$ where $c$ is a constant $c.1$ and $1$ is the product($\nabla$) identity. So perhaps I need to show $c=0$. Need help.

Answer 1

This is true for bialgebras, not necessarily for Hopf algebras, and is an immediate consequence of the definition.

Remember that in any coalgebra you have $(\operatorname{id} \otimes \epsilon) \Delta = \operatorname{id}$. Applying this on a primitive element $x$ you get

$$(\operatorname{id} \otimes \epsilon) (1 \otimes x + x \otimes 1) = x ,$$

whence $1 \cdot \epsilon(x) + x \cdot \epsilon(1) = x$.

In any bialgebra, the compatibility conditions ensure that $\epsilon(1) = 1$, therefore the previous relation simplifies down to $1 \cdot \epsilon (x) = 0$, whence immediately $\epsilon(x) = 0$.