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This is exercise $VI.6.2$ in Kassel's Quantum Groups. My observations are the following:

If $V(\lambda)$ is not simple, then there is a surjective module homomorphism $\psi:V(L)\rightarrow V$ to some $U_q$-module such that $V(L)/\ker\psi \cong V$. Since $\psi$ is onto, there exists some $v\in V$ s.t. $\psi(v_0)=v$. We then have $$Kv = K\psi(v_0) = \psi(Kv_0)=\lambda v$$ and $$Ev = E \psi(v_0) = 0.$$ Therefore $v$ is a highest weight vector with weight $\lambda$.

If we could show that $\dim V<\infty$, we would be done by theorem $VI.3.5$, but maybe this is not the right approach at all.

I would be glad if you could give me some hints (no full answers if possible) for this!

Jo Mo
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    It is not difficult to show that all submodules of Verma modules have finite codimension. Start by showing that all sumodules are spanned by eigenvectors of K. – Mariano Suárez-Álvarez Feb 02 '18 at 20:49
  • Well, $V(\lambda)$ is spanned by eigenvectors of $K$, namely $$v_p = \frac{1}{[p]!}F^pv_0,, p\in \mathbb{N}.$$Consequentely, a proper submodule $M$, being a subspace, will be spanned by a subset of these vectors, say with indexing set $I$. $0$ cannot be in $I$, because $V(\lambda)$ is generated by $v_0$. $M$ is closed under $U_q$-action, so there exists some $n>0$ s.t. $v_{n+1}\in M$ and $Ev_{n+1}=0$. Equivalently, $q^{-n}\lambda = q^n\lambda^{-1}$, so $\lambda^2 = q^{2n}$ and we're done. Is this correct? – Jo Mo Feb 03 '18 at 07:41
  • No, you did not justify in any way that submodule is spanned by eigenvectors. – Mariano Suárez-Álvarez Feb 03 '18 at 17:42
  • That is true, but we can simply use this answer, which works for infinite dimensional modules as well. I suppose? – Jo Mo Feb 03 '18 at 17:48
  • Yes. Notice that that is precisely the point of weight modules. – Mariano Suárez-Álvarez Feb 03 '18 at 19:11

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