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There is a little question haunted me for few days. I will be grateful to anyone who can give me any clue how to solve it.

Let $V$ be a nontrivial module of $\mathrm{U}_q(\mathfrak{g})$ (the quantum group of adjoint type for a simple finite dimensional Lie algebra $\mathfrak{g}$), $0\neq x\in\mathrm{U}_q(\mathfrak{g})$. Is $x.\mathrm{T}(V)\neq 0$ where $\mathrm{T}(V)$ is a tensor product module?

I only guess it is right. In fact maybe not.

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  • $\begingroup$ What do you mean by "a tensor product module"? $\endgroup$
    – Mariano Suárez-Álvarez
    Commented May 26, 2011 at 4:05
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    $\begingroup$ I believe your question is the following. You have a Hopf algebra $H$, which in your case is a very specific Hopf algebra over the field of rational functions in a variable $q$ over characteristic $0$. You have an $H$-module $V$; then you can form the tensor algebra $T(V) = \bigoplus V^{\otimes \bullet}$ in the category of $H$-modules. You have a non-zero element $x \in H$, and ask if $x$ necessarily acts nontrivially on $T(V)$. In general, the answer is obviously "no", but your question is if it happens to be "yes" for the specific case of quantum groups. Is this right? $\endgroup$
    – Theo Johnson-Freyd
    Commented May 26, 2011 at 4:18
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    $\begingroup$ Incidentally, all of your questions so far have had titles of the form "A simple question about ---". This is not a great titling convention for MO. MathOverflow titles can be quite long, and there's no reason not to include a short version of your question in the title. Your questions would also be improved with more background and motivation: what do you know, why do you want to know what you don't know. Please be sure to read mathoverflow.net/howtoask , and maybe also spend some more time reading MO for a sense of the style of the best questions. $\endgroup$
    – Theo Johnson-Freyd
    Commented May 26, 2011 at 4:22
  • $\begingroup$ There are easy counterexamples if $g$ is not simple or $U_q (g)$ is not of adjont type. I will edit to add these conditions as well. $\endgroup$
    – Bugs Bunny
    Commented May 26, 2011 at 6:21
  • $\begingroup$ Theo Johnson-Freyd: You are right.Thank you.And the same to Bugs Bunny... $\endgroup$
    – X---
    Commented May 26, 2011 at 9:33

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