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May 26, 2011 at 21:58 comment added Theo Johnson-Freyd Annihilator. Annulator?!?
May 26, 2011 at 9:33 comment added X--- Theo Johnson-Freyd: You are right.Thank you.And the same to Bugs Bunny...
May 26, 2011 at 6:23 history edited Bugs Bunny CC BY-SA 3.0
ENglish+conditions on g
May 26, 2011 at 6:21 comment added Bugs Bunny 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.
May 26, 2011 at 6:02 history edited Bugs Bunny CC BY-SA 3.0
deleted 1 characters in body; edited tags; edited title
May 26, 2011 at 4:22 comment added Theo Johnson-Freyd 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.
May 26, 2011 at 4:18 comment added Theo Johnson-Freyd 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?
May 26, 2011 at 4:05 comment added Mariano Suárez-Álvarez What do you mean by "a tensor product module"?
May 26, 2011 at 2:45 history asked X--- CC BY-SA 3.0