Let $A$ be an abelian subgroup of the unimodular group of degree $n$ (i.e. $GL(n,\mathbb Z)$). $A$ can be regarded as a group of automorphisms of a free abelian group of rank $n$ ($\mathbb Z^n$), and so also as a group of linear transformation of $V:=\mathbb Z^n\otimes \mathbb Q$. Assume that $A$ is irreducible as group of linear transformations (i.e. $V$ is simple as $\mathbb QA$-module). Let $C$ be the centralizer of $A$ in the ring of all $n\times n$ matrices whose coefficients are rational numbers. Then $C$ is a finite dimensional algebra over the field of rational numbers.
Now
1) How we can apply the Schur's lemma to infer that $C$ is a division algebra?
2) Assume 1) is proved. Why the center of $C$ is a finite extension of the field of rational number?
Obviously, you should be suspecting there is a relationship between $C$ and $End_{\Bbb Q A}(V)$.
Each $f\in End_{\Bbb Q A}(V)$ is a $\Bbb Q$ linear map such that $f(a(x))=a(f(x))$ for all $x\in V$. Thus $fa=af$ for all $a\in A$. Conversely any such $\Bbb Q$-linear $f$ commuting with the elements of $a$ is a $\Bbb Q A$-linear map. So: $C=End_{\Bbb Q A}(V)$.
Since $V$ is assumed to be $\Bbb Q A$-simple, $C=End_{\Bbb Q A}(V)$ is a division algebra. It's easy to check that the center of a division algebra is a field.
Finally, notice that the center is a $\Bbb Q$ subalgebra of the finite dimensional $\Bbb Q$-algebra $M_n(\Bbb Q)$: it can't have any more dimensions than $M_n(\Bbb Q)$. Since it contains $\Bbb Q$, it's a field extension of $\Bbb Q$.