Reduction modulo p of a linear group over the rational numbers

Question

A paper (http://arxiv.org/pdf/1407.3158v2.pdf) contains the following theorem:

Suppose $\mathbb{G}$ is a connected, simply connected, semisimple algebraic group defined over $\mathbb{Q}$, and let $\Gamma \leq \mathbb{G}(\mathbb{Q})$ be a finitely generated Zariski-dense subgroup. Denote by $\mathbb{G}_p$ the smooth reduction of $\mathbb{G}$ over $\mathbb{F}_p$. Then for all sufficiently large prime numbers $p$, the reduction $\Gamma_p$ of $\Gamma$ is equal to $\mathbb{G}_p (\mathbb{F}_p)$.

What confuses me is the so-called smooth reduction happening. I would understand this if $\mathbb{G}$ were defined over $\mathbb{Z}$ because then we can just take all the matrix entries modulo $p$, but what can it mean for a rational number to be reduced in this way, if that is indeed what is happening?

EDIT

It would help to have also an example of how this reduction would apply to an element of $\operatorname{GL}_n(\mathbb{Q})$.

EDIT 2

I received the following answer to this question in a chatroom, but it is far too complicated for me to understand. In particular, I know little to no algebraic geometry, and I am only beginning to learn category theory.

$\mathbb{G}$ is defined over $\operatorname{Spec} \mathbb{Q}$, so it's of finite type (over $\operatorname{Spec} \mathbb{Q}$), and so you can take the ideal sheaf locally and restrict it to $\mathbb{Z}$.

-- hodgeclass

Answer 1

The top of page 2 in the linked paper on the arXiv seems to suggest this is the definition:

Let $\mathbb{G}$ be a connected simply connected, semisimple algebraic group defined over $\mathbb{Q}$ with a fixed embedding $\mathbb{G} \hookrightarrow GL_n$. Then $$\mathbb{G}(\mathbb{Z}) = \mathbb{G}(\mathbb{Q}) \cap GL_n(\mathbb{Z}).$$

This allows one to consider the reductions. I do not think that $\mathbb{G}(\mathbb{Z})$ is independent of the embedding chosen, but again the arXiv paper seems to suggest it is for primes outside of some finite set.