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Hua and Reiner in their paper titled "Automorphisms of the unimodular group" have established what will be the commutator subgroup of $\mathrm{GL}_n(\mathbb{Z})$, $\forall n\geq 0$. I have understood their proof and tried to get a result when $\mathbb{Z}$ is replaced by a PID with unity $R$. Even for the $n \leq 3$ I could not find the commutator subgroups (of course for $n =1$, they are the units in the PID). The difficulty arises because of the existence of units which are not $1$ or $-1$.

I would be obliged if I get any help in this direction or some references, where my question already has an answer. Thank you.

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    $\begingroup$ For $n\ge 3$, every elementary matrix is a commutator ($[e_{ij}(1),e_{jk}(s)]=e_{ik}(s)$) and $\mathrm{SL}_n$ of a PID is generated by elementary matrices, so the determinant yields the abelianization. So it remains only $n=2$ to consider. For $n=2$, when the PID is Euclidean then generation by elementary matrices also holds, but I'm not sure in general, and in any case the commutator formula does not apply. $\endgroup$
    – YCor
    Commented Apr 2 at 9:22
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    $\begingroup$ Related: mathoverflow.net/questions/59884/… $\endgroup$
    – Tom De Medts
    Commented Apr 2 at 9:42
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    $\begingroup$ Yes, I made no claim for $n=2$. For a Euclidean PID, $\mathrm{SL}_2(R)$ is generated by elementary matrices, but as you mention with $R=\mathbf{Z}$, unipotents need not be in the derived subgroup. $\endgroup$
    – YCor
    Commented Apr 2 at 11:19
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    $\begingroup$ So for $n\ge 1$ and a PID $R$, the determinant gives a surjective homomorphism from the abelianization of $\mathrm{GL}_n(R)$ to $R^\times$, whose kernel is trivial for $n\neq 2$. For $n=2$, when $R$ is also Euclidean, it still holds that the kernel is an elementary abelian 2-group (namely a quotient of $R/2R$). $\endgroup$
    – YCor
    Commented Apr 2 at 11:30
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    $\begingroup$ If $R$ is a Euclidean domain and has an invertible element $s$ such that $1-s$ is invertible, then all elementary matrices are commutators. This applies to $R=k[t]$ whenever $k$ is a field with $|k|>2$. As regards $R=\mathbf{F}_2[t]$, the abelianization of $\mathrm{GL}_2(R)$ is a free $\mathbf{F}_2$-module of infinite rank (as seen from the usual amalgam decomposition). $\endgroup$
    – YCor
    Commented Apr 2 at 11:43

1 Answer 1

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This answer provides references for the facts mentioned in the comments of YCor. More on this topic can be found in [2, Section 9].

For $R$ a unital ring, we denote by $\text{E}_n(R)$ the subgroup of $\text{GL}_n(R)$ generated by all the transvections $$e_{ij}(r) = I_n + r \epsilon_{ij}$$ (a.k.a the elementary matrices over $R$) with $r \in R, 1\le i \neq j \le n$ and where $I_n$ is identity matrix and $\epsilon_{ij}$ is the matrix whose $(i,j)$-entry is $1$ while all its other entries are zero.

Claim. Let $R$ be a unital ring of finite stable rank $\text{sr(R)}$. Then we have $$[\text{GL}_n(R), \text{GL}_n(R)] = \text{E}_n(R)$$ for every $n > \text{sr}(R)$.

Proof. This follows for instance from Vaserstein's Injective Stability Theorem [4, Theorem 10.15].

Corollary. Let $R$ be a commutative and unital ring of Krull dimension at most $1$, e.g., a principal ideal domain. Then $[\text{GL}_n(R), \text{GL}_n(R)] = \text{E}_n(R)$ for every $n > 2$.

Proof. As the stable rank of a commutative and unital ring of Krull dimension at most $d$, is at most $d + 1$ [3, Corollary 2.3], the result follows from the previous claim.

For $n = 2$, Section 9 of P. M. Cohn's [2] contains a wealth of valuable results.

For instance, it follows from [2, Proposition 9.2] that $$[\text{GL}_2(R), \text{GL}_2(R)] = \text{E}_2(R)$$ if $R$ is Euclidean and $1$ is the sum of two units of $R$, e.g., if $R$ is any of the following rings:

  • $\mathbb{Z}[\frac{1}{2n}]$ where $n \in \mathbb{N}_{> 0}$,
  • $\mathbb{Z}_{(p)} = \{ \frac{m}{n} \, \vert \, m,n \in \mathbb{Z}, \text{gcd}(n, p) = 1 \}$ for $p$ an odd prime number,
  • $\mathbb{Z}_p$, the ring of $p$-adic integers for $p$ an odd prime number,
  • the ring $k[X]$ of univariate polynomials over $k$ where $k$ is a field with at least $3$ elements; see how this result changes when $k$ is $\mathbb{F}_2$, the field with two elements, in this related post.
  • $\mathbb{Z}[e^{\frac{2i \pi}{3}}]$, the ring of Eisenstein integers.

Among other interesting results, Cohn's generalization [2, Theorem 9.4 and subsequent remark] of the result of Hua and Reiner [1] is particularly relevant: we have $$\text{E}_2(R)/[\text{GL}_2(R), \text{GL}_2(R)] \simeq R/N$$ where $N$ is the ideal of $R$ generated by all the elements of the form $1 - \alpha$ with $\alpha \in \text{GL}_1(R)$, provided $R$ is quasi-free. Quasi-free rings in the sense of Cohn encompass the class of discretely normed rings which contains in particular $\mathbb{Z}$, the rings of rational integers and $\mathbb{Z}[i]$, the rings of Gaussian integers. If $R$ is any of these last two rings then the quotient group $$\text{SL}_2(R)/[\text{GL}_2(R), \text{GL}_2(R)] = \text{E}_2(R)/[\text{GL}_2(R), \text{GL}_2(R)]$$ has two elements, which is Hua and Reiner's result [1] when specializing $R$ to $\mathbb{Z}$.

  • [1] L. Hua and I. Reiner, "Automorphisms of the unimodular group", 1951.
  • [2] P. M. Cohn, "On the structure of the $GL_2$ of a ring", 1966.
  • [3] R. Heitmann, "Generating non-Noetherian modules efficiently", 1984.
  • [4] B. Magurn, "An algebraic introduction to $K$-theory", 2002.
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