Let $R$ be a principal ideal domain with field of fractions $K$, then $N_{GL_n(K)}(GL_n(R)) = K^\times GL_n(R)$

Question

Let $R$ be a principal ideal domain with field of fractions $K$. Let $\mathcal{G}_n(K)$ denote the set of subgroups of $GL_n(K)$, where $GL_n(K)$ acts by conjugation on $\mathcal{G}_n(K)$:

$$GL_n(K) \times \mathcal{G}_n(K) \rightarrow \mathcal{G}_n(K)$$ $$(g, G) \mapsto gGg^{−1}$$

For $G \in \mathcal{G}_n(K)$, we denote the stabilizer of $G$ under this action as $$N_{GL_n(K)}(G) := \operatorname{Stab}_{GL_n(K)}(G)= \{g \in GL_n(K) | gGg^{−1} = G\} \subset GL_n(K)$$ In particular, it is a subgroup of $GL_n(K)$. Then, according to my Commutative Algebra lecture notes, it can be shown that

$$N_{GL_n(K)}(GL_n(R)) = K^\times GL_n(R)$$

where $K^\times \subset GL_n(K)$ is the subgroup of homotheties $cI_n$, where $c \in K^\times$.

Could anyone provide a reference for a proof? I don’t really know where to look for it and I don’t even understand intuitively why this result should hold.

Edit: the lecture notes also say that one of the inclusions (the $ K^\times GL_n(R) \subset N_{GL_n(K)}(GL_n(R))$ one) is trivial, but I don’t see it. Actually, I don’t understand what $ K^\times GL_n(R) $ is supposed to denote.

Answer 1

To prove the inclusion $\mathrm{N}_{\mathrm{GL}_n(K)}(\mathrm{GL}_n(R))\subset K^\times\mathrm{GL}_n(R)$, we need the elementary divisors theorem:

Let $R$ a PID, $F$ a free $R$-module, $M\subset F$ a finitely generated submodule $\neq 0$. Then there exists a basis $\mathcal{B}$ of $F$, $e_1,\dots, e_m \in \mathcal{B}$, non-zero elements
$a_1,\dots,a_m\in R$ such that: (i) $a_1e_1,\dots , a_me_m$ form a basis of $M$ over $R$. (ii) $a_i\mid a_{i+1}$ for $i=1,\dots, m-1$.

This is Theorem III.7.8 of Lang, Algebra. The proof can be found there.

For the proof, take an element $g=[g_{ij}]\in \mathrm{N}_{\mathrm{GL}_n(K)}(\mathrm{GL}_n(R))$. There exists $a\in R$ such that $ag_{ij}\in R$ for any $i,j$ and $(ag_{ij}\mid 1\le i,j \le n)=R$. Let $F=R^n, \phi: F\to F$ be the homomorphism defined by $ag$, $M$ the image of $\phi$. It suffices to prove that $\phi$ is an isomorphism. Since we can regard $\phi$ as the restriction of $K^n\overset{\sim}{\to}K^n$ to $R^n$, injectivity of $\phi$ follows. Take a basis $e_1,\dots, e_n$ of $F$, $a_1,\dots,a_m\in R$ as in the elementary divisors theorem. Then $a_1$ must be a unit in $R$ by the condition $(ag_{ij}\mid 1\le i,j \le n)=R$. Take any element $u\in F$. There exists $b\in R ,v\in F$ such that $u=bv=\!^t(bv_1,...,bv_n)$ and $(v_1,\dots,v_n)=R$. Since $F/Rv$ is torsion-free i.e. free, we can extend $v$ to a basis of $F$. There exists an automorphism $\psi$ of $F$ such that $\psi(e_1)=v$. There exists $\psi'$ such that $\phi\psi'=\psi \phi$ and $u=bv\in M$ so that $\phi$ is also surjective.