Why define Chevalley groups over $\mathbb{Z}$ I'm reading a lot about the construction of Chevalley groups from simple Lie algebras and I understand that Carter in his book 'Simple Groups of Lie Type' makes a big point of wanting the generators $x_r(\zeta) := \exp(\zeta \text{ad}(e_r))$ to act on the Chevalley basis, such that the action results in sums of Chevalley basis elements with integer coefficients.
As I understand it, the purpose of this is so we can define a Lie algebra over $\mathbb{Z}$ with a Chevalley basis and then we can extend the field by means of the tensor product.
My question is fairly simple, but I really don't have much of a background in category theory, galois theory or field theory (I'm an undergraduate).

Why exactly do we do all this process of making the coefficients be in $\mathbb{Z}$. Would it be possible to extend the field otherwise? Additionally, is it true that $F\otimes_{\mathbb{Z}} \mathbb{Z} = F$ and that's the real reason that we do all this rigmarole. 

 A: As far as I understand, it's as follows: to extend the scalars of a Lie group or Lie algebra is pretty easy. Well OK, I remember when I was a beginner tensor products freaked me out, and to be honest base extensions of Lie/algebraic groups still do sometimes, but in spirit it's a pretty straightforward concept. You allow more coefficients. It's as if you've come to like some person, and now you see them dressed up in fancy clothes and with some jewellery; that can be nice and interesting, but underneath you know it's still the person you got to know and like when they wore a shirt and jeans.
But the backward process is hard. (There is a very easy thing called restriction of scalars, but in a way this is not the "right" reversal of that scalar extension that we are looking for.) Imagine we are given some Lie group or algebra $S$ over $\mathbb C$, all fancy with imaginary units and stuff. But is it already defined over $\mathbb R$ -- meaning very roughly, is there some object $R$ defined entirely with plain old real numbers, such that $S$ is just the scalar extension (dressed up version) of $R$? This is in general a very hard question. If you've only met some person in a fancy dress, can you imagine them in streetclothes?
But to find that such a group or algebra is actually defined over $\mathbb Z$ is like to be able to see that person without any clothes.
To steer away from a not necessarily too good metaphor to something more mathematical, the ring $\mathbb Z$ is the basis of all rings (technically, it is the initial object in the category of rings), meaning that if you have something that is defined over $\mathbb Z$, then from there you can go "up" again to any other ring. (One last flashback to my metaphor: Now you can dress them up with any clothes you like.)
For example, say you want to study Lie-type groups over finite fields. To do a base change between two fields, you need a homomorphism between those fields. More generally for changing scalars from some ring $R$ to some other ring $S$, you need some homomorphism $R \rightarrow S$; which, for fields, will always be plain inclusions $K \subset L$.
So it's easy to change scalars from $\mathbb R$ to $\mathbb C$. Or from $\mathbb C$ to $\mathbb C(t)$. Or from $\mathbb Q$ to $\mathbb R$. But not the other way around. And from none of them is there any chance to get to $\mathbb F_p$.
But now imagine you started with something over $\mathbb C$ or $\mathbb R$ and you have managed to strip it down to something over $\mathbb Z$. Or Chevalley did it for you. Well now you have a map $\mathbb Z \rightarrow \mathbb F_p$, so now you can change its base to your favourite finite field! As far as I understand, that was a big thing which was now allowed through Chevalley's $\mathbb Z$-basis, e.g. Steinberg then constructed analogues of some classically "real" stuff over finite fields, giving some cool new constructions in the theory of finite groups.
And even if you're not interested in finite groups and all those base change possibilities, still having something defined over $\mathbb Z$ somehow feels as if you can see it clearly, with all clutter of $\sqrt2$'s and $\pi$'s and $i$'s removed.
