TITLE: For algebraic $\alpha$, does $G(\mathbb Z[d\alpha])$ have a finite index in $G(\mathbb Z[\alpha])$?
QUESTION [5 upvotes]: Let $\alpha$ be an algebraic number and $G$ be a connected $\mathbb Q(\alpha)$-algebraic group, and $d\in\mathbb Z^+$.
We fix a faithful representation of $G$ in some $GL_n$ and identify $G$ with its image under this representation. For a subring $R$ of $\mathbb Q(α)$, $G(R):=G\cap GL_n(R)$.
Does $G(\mathbb Z[d\alpha])$ have a finite index in $G(\mathbb Z[\alpha])$?
What if $d\alpha$ is an integral element?

REPLY [6 votes]: The answer is yes for $\alpha$ an integer and no for $\alpha$ a general algebraic number.
To see the problem with algebraic numbers, take $\alpha=1/2, d=2$. Then almost any algebraic group, in particular $\mathbb G_m$ or $\mathbb G_a$, will not have its $\mathbb Z$ points a finite-index subgroup of its $\mathbb Z[1/2]$-points.
If $\alpha$ satisfies a monic polynomial equation of degree $k$, then $\mathbb Z[d\alpha]$ contains all elements in $\mathbb Z[\alpha]$ that are congruent to $0$ or $1$ mod $d^{k-1}$, so $G(\mathbb Z[d \alpha])$ contains the kernel of the natural homomorphism from $G( \mathbb Z[\alpha])$ to $G(\mathbb Z[\alpha]/d^{k-1})$. Because the target is finite, its kernel has finite index.

Edit: Assuming that every element of $\mathbb Z[\alpha]$ shows up as an entry of a matrix in $G(\mathbb Z[\alpha])$. Then we can actually prove a converse - if $G(\mathbb Z[d\alpha])$ is finite-index, then $\mathbb Z[d \alpha]$ contains a finite-index ideal in $\mathbb Z[\alpha]$.  
To do this, take coset representatives for each of the finitely many cosets, and observe that every element of $\mathbb Z[\alpha]$ arises as a matrix entry of a product of one of these representatives with a matrix over $\mathbb Z[d\alpha]$, which implies the entries are a finite basis for $\mathbb Z[\alpha]$ as a $\mathbb Z[d\alpha]$-module. Each of these basis elements can be written as a polynomial in $\alpha$, hence its multiple by some power of $d$ is a polynomial in $d\alpha$. So altogether there is some fixed power of $d$ all whose multiples in $\mathbb Z[\alpha]$ lie in $\mathbb Z[d\alpha]$.