By "fundamental copies" of $SL_2(R)$ in $SL_n(R)$, I mean those embedded along the diagonal (for instance, if $n=3$, those are the upper left and lower right corner copies of $SL_2(R)$ embedded in $SL_3(R)$).

When $R$ is a field or a Euclidean ring, $SL_n(R)$ is generated by elementary matrices. However, this need not be the case in general. My question is, whether one can nevertheless generate $SL_n(R)$ by its fundamental copies of $SL_2(R)$, even if $SL_n(R)$ is bigger than its elementary subgroup. I have the impression that the answer is at least "yes" when $R$ is a principal ideal domain or, more generally, a Bezout domain.

A similar question holds for the affine Kac-Moody group $SL_2(R[t,t^{-1}])$: when is it generated by its two fundamental copies of $SL_2(R)$, where the two embeddings of $SL_2(R)$ are given as follows:

$\begin{pmatrix}a &b\\ c &d \end{pmatrix}\mapsto \begin{pmatrix}a &b\\ c &d \end{pmatrix}$ and $\begin{pmatrix}a &b\\ c &d \end{pmatrix}\mapsto \begin{pmatrix}d &ct^{-1}\\ bt &a \end{pmatrix}$

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    $\begingroup$ Just a remark: if you have the $n-1$ "fundamental" copies of $SL_2(R)$, then you have enough permutation matrices to have all the $\binom{n}{2}$ copies of $SL_2$ (in particular you have all the elementary matrices). $\endgroup$ – Aurel Mar 30 '17 at 13:48
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    $\begingroup$ The "similar question" deserves distinct answers and would be welcome as a separate question. $\endgroup$ – YCor Feb 15 '18 at 20:49

$SL(n, R)$ is indeed generated by the fundamental copies of $SL(2, R)$ when $R$ is a Bezout domain (or even a Hermitian ring), this is called Smith normal form. In fact, $SL(n, R)$ is not only generated by such subgroups, but is a product of $(n-1)^2$ fundamental copies of $SL(2, R)$. See $SL_2$-factorizations of Chevalley groups for the proof of a slightly weaker bound of $n^2-n$ factors in case $R$ is a Bezout domain.

On the other hand, for any ring $R$ of stable rank at most $2$, one can express $SL(n, R)$ as a product of $\frac{1}{2}(3n^2-5n)$ copies of $SL(2, R)$. Here is a link.

Even more, $SL(n, R)$ being generated by $SL(2, R)$'s is equivalent to the map $SL(2, R)/E(2, R)\to SL(n, R)/E(n, R)$ being surjective. The question whether it is so for a particular ring $R$ is known as the "surjective stability for $K_1$", and there are lots of papers on the subject. To put it short, in terms of dimension of $R$ the best one can get is "stable rank at most 2", but there are plenty of examples of geometric, arithmetic or analytic nature having stable rank greater than 2 (or not known).

As for the last question, concerning the Kac—Moody groups of type $\widetilde{A}_1$, there is a paper by S. Sidki, but it deals with groups over fields.

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    $\begingroup$ About "this is called the Smith Normal Form". I am afraid that there isn't such a thing for Bézout domain: it has been conjectured that there exists Bézout domain which is not a Elementary Divisor domain, see introduction of alpha.math.uga.edu/~lorenz/Bezout.pdf. Would you have a reference for your claim about Hermitian rings? $\endgroup$ – Luc Guyot Mar 31 '17 at 9:46
  • $\begingroup$ Thanks, it is good to know there is no simple answer for general rings. Note that over Bezout domains, the generation of Chevalley groups by fundamental copies of $SL_2$ is in fact in Steinberg's "Lectures on Chevalley groups" (Theorem 15 and Lemma 43 pages 99-100). Actually, his argument seems to work for arbitrary Kac-Moody groups, so this should also answer my second question over Bezout domains. $\endgroup$ – Timothée Marquis Mar 31 '17 at 10:39
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    $\begingroup$ @LucGuyot This is a certain abuse of language, of course. What is meant here is that when one tries to compute the Smith normal form of a particular matrix, the algorithm one uses does not require the ring to be Euclidean. One only has to be able to reduce rows and columns to obtain more and more zeroes. This is what Hermitian rings do. And when one reduces the matrix to a single 2x2 block — well, this in one more copy of $SL_2$. $\endgroup$ – Andrei Smolensky Mar 31 '17 at 10:56

L. Vaserstein obtain a generation result in a very large generality, which encompasses all finitely generated commutative algebras over fields or PIDs:

Theorem (Vaserstein, 2006, unpublished): for every commutative noetherian ring that is finitely generated (by $k$ elements) over a subring of Krull dimension $\le 1$ and every $r\ge 3$, every matrix decomposes as a product of one matrix in the upper $\mathrm{SL}_2$ and a bounded (lazy bound: $\le 28k^2r^2$) number of elementary matrices.

(Note that obviously every elementary matrix is a product of a bounded (bound depending only on $r$) number of (elementary) matrices in one of the diagonal $\mathrm{SL}_2$.)

For somewhat absurd reasons, this paper is unpublished. I did in 2007 a detailed check of the proof, including a careful reading of his long 1976 paper with Suslin on which the paper strongly relies. Vaserstein's paper can be found on his web page: http://www.math.psu.edu/vstein/pm2.pdf; the above theorem follows from his Corollary 2.

Note that in case of finitely generated algebra over a field/PID, this is considerably more general than what is cited in Andrei's answer: the case of Bézout rings is strictly contained in the case of Krull dimension 1, and the case of stable $\le 2$ does not much go beyond Krull dimension 2: it covers the case of Krull dimension 1 and also maybe a few rings of Krull dimension 2.

(Input welcome! I do not know if there's a commutative algebra that is finitely generated over a PID, has stable rank $\le 2$ and Krull dimension $\ge 2$. There is a MO post resulting in the conclusion that $\mathbf{Z}[x]$ has stable rank 3. According to this paper by Gabel (Pacific J. Math. 1975), polynomial rings over fields with $s\ge 4$ indeterminates have stable rank $\ge 3$, and for subfields of the real field this already holds when $s\ge 2$.)

Added: Vaserstein proved his result soon after Shalom proved and gave talks on a Kazhdan Property T result relying on the fact that for a ring $R$ of stable rank $d$, $\mathrm{SL}_r(R)$ for $r\ge d+1$ is boundedly generated by elementary matrices along with its upper left copy of $\mathrm{SL}_{d+1}(R)$. For $d=2$ this is of course the easy result on stable rank 2 retrieved later (link in Andrei's post). This is briefly mentioned in Shalom's ICM 2006 paper, with a reference to a paper which never appeared, and explained in detail in 2006/07 pdf notes of mine based on his talks. Combining Shalom and Vaserstein's result lead to the result that the subgroup generated by elementary matrices in $\mathrm{SL}_r(R)$ has Property T for every finitely generated commutative ring and $r\ge 3$. What happened a bit later is that the Property T result was obtained by other means, not using bounded generation, by Ershov and Jaikin-Zapirain (Inv. Math. 2010), in a wider generality ($R$ not necessarily commutative). This superseded the Property T result, but not Vaserstein's beautiful bounded generation result, which obviously is of interest independently of Kazhdan's Property T.

  • $\begingroup$ @LucGuyot thanks very much; I edited accordingly all three answers by Steve Landsburg to clarify what's going on. $\endgroup$ – YCor Feb 16 '18 at 17:30

The goal of my answer is only to append further definitions and references to Andrei Smolensky’s accepted answer.

The OP’s first question asks whether $SL_n(R)$, for $R$ a principal ideal domain (PID), is generated by the images of the fundamental embeddings of $SL_2(R)$. Andrei Smolensky suggests that it can be verified by inspecting the proof of the Smith Normal Form theorem for PIDs. Indeed, every matrix used in the reduction is either an elementary matrix (i.e., a matrix that differs from the identity by an off-diagonal element) or a matrix coming from one of the fundamental embeddings. Hence the result follows from Aurel’s comment.

As indicated by Andrei Smolensky, this result can be generalized both to rings of stable range at most $2$ and to Hermit rings in the sense of Kaplansky, using the spirit of the Smith Normal Form theorem.

The stable rank of a ring $R$ is, if it exists, the least integer $n$ such that for every $(n + 1)$-row $(r_1, \dots, r_{n + 1}) \in R^{n + 1}$ verifying $r_1 R + \dots + r_{n + 1}R = R$ we can find $(\lambda_1, \dots, \lambda_n) \in R^n$ such that $(r_1 + \lambda_1 r_{n + 1})R + \dots + (r_n + \lambda_n r_{n + 1})R =R$.

Assertion. If the stable rank of $R$ is at most $2$, then $SL_n(R)$ is generated by the images of the fundamental embeddings of $SL_2(R)$.

Proof. Let $A \in SL_n(R)$. Since the first column of $A$ generates $R$, we can find a product $E$ of elementary matrices such that the coefficients of index $(1, 1)$ and $(2, 1)$ of $A’ = EA$ generates $R$. Using the top left-hand corner embedding, we can turn these two coefficients respectively into $1$ and $0$. Using the coefficient of index $(1,1)$ as a pivot we can reduce to a matrix in $SL_{n - 1}(R)$ and conclude by induction.

As Dedekind domainsKrull dimension is at most $1$, it follows from the Bass Cancellation Theorem that Dedekind domains have stable rank at most $2$. This provides us with examples of rings which fail simultaneously to be Bézout rings and generalized Euclidean rings (a generalized Euclidean ring is a ring $R$ such that $SL_n(R)$ is generated by elementary matrices for every $n \ge 2$.)

A Bézout domain has stable rank at most $2$, see this MO post. Therefore, the above assertion answers also OP’s second question.

A Bézout domain is also a commutative Hermit ring in the sense of Kaplansky [Theorem 3.2, 1], i.e., a ring such that, for every $(r,s) \in R^2$ there is $A \in SL_2(R)$ and $d \in R$ such that $(r, s)A = (d, 0)$.

Hence the same answer can be inferred from

Assertion. If $R$ is a Hermit ring in the sense of Kaplansky, then $SL_n(R)$ is generated by the images of the fundamental embeddings of $SL_2(R)$.

Proof. Let $A \in SL_n(R)$. Using the top left-hand corner embedding, we can cancel the coefficient of index $(1, 2)$. Using a row permutation followed by a matrix from the latter embedding, we can cancel all coefficients in the first column except the coefficient of index $(1, 1)$. Iterating over columns, we can thus turn $A$ into a trigonal matrix $T$ whose diagonal coefficients are necessarily invertible. Further elementary row transformations can then be used to turn $T$ into a diagonal matrix $D$. By Whitehead's lemma, $D$ can be turned into the identity matrix using elementary row transformations as well.

The class of commutative Hermit rings encompasses the class of commutative principal ideal rings [2], Bézout rings with finitely many minimal prime ideals [3, Theorem 2.2] and Bézout rings whose zero divisors lie in the nil radical [1, Theorem 3.2].

[1] "Elementary divisors and modules", I. Kaplansky, 1949 (MR0031470).
[2] "On the structure of principal ideal rings", T. W. Hungerford, 1968 (MR0227159).
[3] "Elementary divisors and finitely presented modules", M. Larsen, J. Lewis and T. Shores, 1974 (MR0335499).


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