Let $H$ be a Hopf Algebra (over a field $K$), with comultiplication $\Delta$, counit $\varepsilon$, and antipode $S$.
A $K$-subspace V is said to be:
- A right ideal if $VH \subseteq V$
- A right coideal if $\Delta (V) \subseteq V \otimes H$ (some call this a left coideal...)
Suppose that there is a nonzero right ideal $J$ in $H$, such that $\dim_K (J)$ is finite. Then $\dim_K (H)$ is finite.
The proof breaks in two parts:
- If $V$ is a nonzero $K$-subspace of a Hopf algebra $H$ that is both a right ideal and a right coideal, then $V=H$.
- If $J\subseteq H$ is a finite-dimensional, nonzero right ideal of $H$, then $N:= H^* J = \{ f\cdot x = \sum x_{(1)}f(x_{(2)}) : f\in H^* , x\in J\}$ is finite dimensional, nonzero, and it is both a right ideal and a right coideal (thus $H=N$ is finite dimenisonal)
Now, the first part is easy (you prove that there is $x\in V$ such that $\varepsilon (x) = 1$, then $1_H = \varepsilon(x)1_H = \sum x_{(1)}S(x_{(2)}) \in V$).
My problem is the second part. I can prove that $N$ is a finite dimensional subspace: since $J$ is finite dimensional, we can assume it is contained in a finite dimensional subcoalgebra $D$ of $H$; then $N\subseteq D $.
Does anyone have suggestions on how to prove that $N$ is a right ideal and a right coideal?
I'll use Sweedler-notation without the sum symbol.
For $h\in{H},f\in{H^*},x\in{J}$ you have:
$m_{(1)}f(m_{(2)})h=m_{(1)}h_{(1)}f(m_{(2)}h_{(2)}S(h_{(3)}))=m_{(1)}h_{(1)}(S(h_{(3)})\rightharpoonup f)(m_{(2)}h_{(2)})=\\(mh_{(1)})_{(1)} (S(h_{(2)}))\rightharpoonup f)(mh_{(1)})_{(2)}\in{N},$
where I used the notation $(a\rightharpoonup g)(b)=g(ba)$.
This shows that $N$ is indeed a right ideal. The proof that $N$ is also a right-coideal should be similiar, maybe this calculation helps you a bit. In fact, $N$ is the right-coideal generated by $J$.