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.