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It seems that the normalizer of $H=\mathrm{GL}(n,\mathbf Z)$ in $G=\mathrm{GL}(n,\mathbf Q)$ is "almost" equal to itself, that is, $$ N_G(\mathrm{GL}(n,\mathbf Z))=Z(G) \cdot \mathrm{GL}(n,\mathbf Z) $$ where $Z(G)$ is the centre of $G.$ Is there a simple proof/disproof of this fact? More generally, for which integral domains $R$ it is known that $\mathrm{GL}(n,R)$ "almost" coincides with its normalizer in the group $\mathrm{GL}(n,Q(R))$ where $Q(R)$ is the quotient field of $R?$

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You need to add the scalar matrices, at least. – Plop Nov 10 '11 at 0:00
@Plop: I will, thank you. – Olod Nov 10 '11 at 7:34
Perhaps you should try asking this on MathOverflow. – Derek Holt Nov 11 '11 at 8:58
Right. $\phantom{abc}$ – Olod Nov 11 '11 at 9:44
The interested readers may find Emerton's answer to the question at… – Olod Nov 11 '11 at 16:16

This is true for $GL_n(\mathbb{Z}_p)$ inside $GL_n(\mathbb{Q}_p)$ for every prime. Use $\prod\limits_p GL_n(\mathbb{Z}_p) \subset \prod\limits_p GL_n(\mathbb{Q}_p)$ and intersect with the diagonally embedded $GL_n(\mathbb{Q})$. This gives you the proof.

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For an integral domain $R$ with quotient field $k$, $GL_n(R)$ is the stabilizer in $GL_n(k)$ of the $R$-module $R^n$ in $k^n$. Anything normalizing $GL_n(R)$ sends $R^n$ to another subset of $k^n$ stabilized by $GL_n(R)$. For a PID $R$, a given $0\not=v\in k^n$ can have its "denominator" collected-up so as to write $v=t\cdot v'$ with $v'$ primitive, $t\in k^\times$. Then $GL_n(R)\cdot v = t\cdot GL_n(R)\cdot w=t\cdot R^n$. That is, for a PID, the non-zero $GL_n(R)$ orbits are scalar multiples of $R^n$. Thus, anything $h$ in the normalizer of $GL_n(R)$ maps $R^n$ to $t\cdot R^n$ for some $t\in k^\times$, so, up to $GL_n(R)$, is a scalar.

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