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?