# The intersection of the annihilators of all finite dimensional weight modules is zero

Let's start with some definitions. Let $$U_q:= U_q(\mathfrak{gl}_n)$$ be the quantum enveloping algebra of $$\mathfrak{gl}_n$$, generated by the standard generators $$\{e_i,f_i, x_j^{\pm}\,|\; i=1,\ldots, n-1, \; j=1,\ldots, n\}$$ together with the standard relations, as in Lusztig’s book for example.

By a weight module $$V$$, I mean an $$U_q-$$module that is direct sum of all weight subspaces $$V_\lambda=\{ v\in V|\; x_i v =q^{(\lambda,\mu_i)}v\}$$, where $$\mu_i$$ is a root and $$\lambda$$ is an element of the root lattice.

And here is the context. It is well known that if $$U$$ is a cocommutative bialgebra (i.e. $$\tau \Delta =\Delta$$ with $$\tau(a\otimes b ) = b\otimes a$$) and $$\Delta$$ being the comultiplication) then for any finite dimensional $$U-$$modules $$M$$ and $$N$$, $$\tau: M\otimes N \to N\otimes M$$ is an $$U-$$module isomorphism. On the other hand if we take $$U$$ to be the quantum enveloping algebra of a semisimple Lie algebra (in my case $$\mathfrak{gl}_n$$) which is not cocommutative, then using the $$R$$-matrix we still have an isomorphism $$M\otimes N \to N\otimes M$$ for $$M$$ and $$N$$ finite dimensional.

Now I want to prove that the isomorphism still holds even if only one of them $$M$$ or $$N$$ is a finite dimensional $$U_q-$$module. To reduce the problem to the case where both are finite dimensional, I need to prove that the intersection of the annihilators of all finite dimensional weight modules is zero. Any hint or sketch of a proof. Thanks

If $$q$$ is not a root of unity, then any element which annihilate all the finite-dimensional $$U_q$$-modules is the zero element.
Indeed, any irreducible finite-dimensional $$U_q$$-module can be obtained as a quotient of a Verma module. So it's easy to see that if $$u \in U$$ annihilate all finite-dimensional modules, it should annihilate all elements in $$U$$. But this is easily seen to be impossible since $$U_q$$ has no zero-divisors by the PBW theorem.