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Let $U_q(\mathfrak{g})$ (defined over $\mathbb{C}(q)$) be the quantized universal enveloping algebra of a simple Lie algebra $\mathfrak{g}$. Let $M$ a finite-dimensional simple left $U_q(\mathfrak{g})$-module. Is it true that $\dim_{\mathbb{C}(q)}\mathrm{End}_{U_q(\mathfrak{g})}(M)=1$? How to sketch a proof? The problem is that the field $\mathbb{C}(q)$ is not algebraically closed.

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This actually requires a fair amount of the structure theory of the representations of quantum groups. If you have a simple, then after a finite field extension, you can diagonalize the operators $K_i$. This shows that there must be a highest weight vector $v$ of a given weight. Thus your simple (after a field extension) is the unique simple quotient of a Verma module. Therefore, your original simple is defined over $\mathbb{C}(q)$ if and only if the weight of $K_i$ lies in $\mathbb{C}(q)$. This happens because if you're finite dimensional, then $E^{(n)}F^{(n)}$ acts by the quantum binomial coefficient $\binom{K_i}{n}$. This splits completely as a polynomial in $K_i$, so the simple is defined over $\mathbb{C}(q)$.

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  • $\begingroup$ ♦ Thank you very much! I got the idea. $\endgroup$ – Zhihua Chang Feb 12 '18 at 1:21

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