# Quantum groups at roots of 1: problem with ground ring

I must be making a basic error in my reading of Lusztig's Quantum Groups at Roots of 1, and I hope someone can show me what it is. Here is the setup: $v$ is an indeterminate, $\mathbb{Q}(v)$ is the quotient field of $\mathbb{Z}[v, v^{-1}]$, and $\mathbb{U}$ is a certain $\mathbb{Q}(v)$-algebra (it doesn't matter what for the purposes of my question). Lusztig says "regard $\mathbb{Q}$ as a $\mathbb{Q}(v)$-algebra with $v$ acting as $1$" (p.91 sec 1.5), then forms the $\mathbb{Q}$-algebra $U_{\mathbb{Q}} = \mathbb{U} \otimes _{\mathbb{Q}(v)} \mathbb{Q}$.

I think my problem is that I don't understand what "regard $\mathbb{Q}$ as a $\mathbb{Q}(v)$-algebra with $v$ acting as $1$" means. My usual interpretation of "$B$ is an $A$ algebra" is that there is a ring map $A\to B$ and $B$ becomes an $A$-module via this map. This isn't what Lusztig is doing: $v \mapsto 1$ does not induce a ring map $\mathbb{Q}(v) \to \mathbb{Q}$ as we can't make sense of the image of something like $1/(v-v^{-1})$.

One attempt to resolve this would be to replace $\mathbb{U} \otimes _{\mathbb{Q}(v)} \mathbb{Q}$ with $\mathbb{U} \otimes _{\mathbb{Z}[v,v^{-1}]} \mathbb{Q}$. I don't believe this works: $\mathbb{U}$ still contains elements like $x= 1/(v-v^{-1})$, so in $\mathbb{U} \otimes _{\mathbb{Z}[v,v^{-1}]} \mathbb{Q}$ we have $$0 = x\otimes 1 - x \otimes 1 = x \otimes v\cdot 1 - x \otimes v^{-1}\cdot 1 = (xv - xv^{-1}) \otimes 1 = 1\otimes 1$$ and the whole thing collapses.

My questions are: how should $U_{\mathbb{Q}}$ be defined so that it does not suffer from the problem above, and what does "$\mathbb{Q}$ is a $\mathbb{Q}(v)$-algebra" mean here?

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I think most of the defining relations of the algebra make sense when v=1 (just turn Gaussian [] things into classical ones), but what would (a3) on page 90 even mean? Should one clear denominators and say Ki = Ki^-1? that seems extreme. – Jack Schmidt Jun 8 '11 at 18:47
@Jack It is certainly not as simple as setting $v=1$ for the reason you suggest. In some sense the "$v \to 1$ limit" should be the ordinary enveloping algebra. – m_t_ Jun 9 '11 at 16:27
@Jack, @mt_ As for the limit of (a3), Wikipedia gives the right answer, perhaps. – Grigory M Jun 11 '11 at 20:30
@Grigory to expand on the wiki explanation, what we want is a connection between the ordinary enveloping algebra (with $H_i$s) and the "quantum group with $v \to 1$" (with $K_i$s). You can get this by taking the base ring to be $\mathbb{Z}[v^{\pm}]$ then tensoring with $\mathbb{F}_p$ with $v$ acting as $1$. Then my manipulations in the question show $K_i$ and its inverse have image $1$ (if also $K_i^p=1$), but that doesn't mean $(K_i-K_i^{-1})/(v-v^{-1})$ goes to zero - in fact it's easy to show that this element obeys the relations for $H_i$ in the ordinary enveloping algebra.... – m_t_ Jun 12 '11 at 10:41
...That's what is behind one of the isomorphisms in Lusztig's other paper "Finite dimensional Hopf algebras arising...". But this doesn't solve the problem in my question unfortunately. – m_t_ Jun 12 '11 at 10:44