For which rings R is SL_n(R) generated by transvections? - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-23T20:44:16Z http://mathoverflow.net/feeds/question/59884 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/59884/for-which-rings-r-is-sl-nr-generated-by-transvections For which rings R is SL_n(R) generated by transvections? Tom De Medts 2011-03-28T19:18:01Z 2012-09-02T03:37:49Z <p>Let $R$ be a commutative ring $R$ with $1$, and $n \geq 2$ an integer.</p> <blockquote> <p>Under which conditions is the group $SL_n(R)$ generated by transvections?</p> </blockquote> <p>(A transvection is a matrix with $1$ everywhere on the diagonal and exactly one other non-zero entry.) This is certainly the case if $R$ is a field, or if $R$ is a Euclidean domain, but I'm wondering whether there is a complete answer to the question.</p> http://mathoverflow.net/questions/59884/for-which-rings-r-is-sl-nr-generated-by-transvections/60536#60536 Answer by Tom De Medts for For which rings R is SL_n(R) generated by transvections? Tom De Medts 2011-04-04T09:39:30Z 2011-04-04T09:39:30Z <p>I'm answering my own question based on the excellent reference given by Max and the additional comments of Jim Humphreys. There is nothing new in my answer, but I think it's useful to close the question in this way.</p> <p>Following Hahn-O'Meara, we write $E_n(R)$ for the subgroup of $SL_n(R)$ generated by transvections (also called elementary matrices).</p> <blockquote> <p>Theorem [H-O'M, Thm 4.3.9]. Let $R$ be a commutative ring. If $R$ is a Euclidean domain or a semilocal ring, then $SL_n(R) = E_n(R)$ for all $n$; If $R$ is a Hasse domain of a global field, then $SL_n(R) = E_n(R)$ for all $n \geq 3$ (and in many cases, but not always, also for $n=2$).</p> </blockquote> <p>There are some other more general results known based on the so-called <em>stable rank</em> of the ring $R$, but as Jim pointed out, it seems hopeless to find a complete answer to the question.</p> http://mathoverflow.net/questions/59884/for-which-rings-r-is-sl-nr-generated-by-transvections/61094#61094 Answer by Luis H Gallardo for For which rings R is SL_n(R) generated by transvections? Luis H Gallardo 2011-04-08T19:36:32Z 2011-04-08T19:42:23Z <p>A nice account of the case</p> <p>$$n =2$$</p> <p>is given by I. Reiner in his review of a paper of P.M. Cohn below</p> <p>The review is very detailed, Hope the tex may compile...</p> <p>(that NOT worked !)</p> <p>I give then just the review to try in MR:</p> <p>MR0207856 (34 #7670) Cohn, P. M. On the structure of the ${\rm GL}_{2}$ of a ring. Inst. Hautes Études Sci. Publ. Math. No. 30 1966 5–53. 20.70 (16.48)</p> <p>and the beginning of the review:</p> <p>This well-written article encompasses a wealth of information about general linear groups over certain classes of rings. The author generalizes many earlier results about such groups, and gives a number of new and striking results. We proceed to describe some of the main theorems. Assume throughout that the underlying ring $R$ has a unity element and is associative, though not necessarily commutative. Denote by $U(R)$ its groups of units. (1) Let $\text{GL}_n(R)$ be the group of $n\times n$ invertible matrices over $R$, and $D_n(R)$ its subgroup of diagonal matrices. Let $E_n(R)$ be the group generated by the set of transvections ${I+ae_{ij}\colon a\in R,1\leq i,j\leq n,i\neq j}$, where ${e_{ij}}$ is a set of matrix units. Define $\text{GE}_n(R)=D_n(R)\cdot E_n(R)$, the subgroup of $\text{GL}_n(R)$ generated by elementary matrices. Of course, $E_n(R)\Delta\text{GE}_n(R)$. The author calls $R$ a generalized Euclidean ring (GE-ring) if $\text{GL}_n(R)=\text{GE}_n(R)$ for all $n$. </p> <p>$$\dots$$</p> http://mathoverflow.net/questions/59884/for-which-rings-r-is-sl-nr-generated-by-transvections/106153#106153 Answer by Bruce A. Magurn for For which rings R is SL_n(R) generated by transvections? Bruce A. Magurn 2012-09-02T03:37:49Z 2012-09-02T03:37:49Z <p>Further results are known: L. Vaserstein's paper "SL_2 of Dedekind rings of arithmetic type" proves these rings are generalized euclidean when they have a unit of infinite order. Integral group rings of finite groups are generalized euclidean when the group has no homomorphic image among the generalized quaternion groups of order a multiple of 4, no image among the binary polyhedral groups, and the abelianization of the group has generalized euclidean integral group ring. The finite abelian G with ZG euclidean include the cyclic groups, and Z/2 x Z/2, by the 1984 paper "Generalized euclidean group rings" by Dennis, Magurn &amp; Vaserstein. But ZG is not generalized euclidean when SK_1(Z[G/[G,G]]) is non-vanishing, as it is for Z/4 x Z/2 x Z/2, for instance. So this is a delicate property!</p>