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Let $R = \mathbb{Z}[X^{\pm1}]$ be the ring of Laurent polynomials on one indeterminate over $\mathbb{Z}$. Let $E_2(R)$ be the subgroup of $GL_2(R)$ generated by the matrices that differ from the identity by a single off-diagonal element. The question whether $SL_2(R) = E_2(R)$ is a problem of S. Bachmuth and H. Y. Mochizuki [2]. It is also an open problem in [3, MA1] and a conjecture in [4].

I am very curious about the status of this problem/conjecture or any progress on it.

A complement prompted by the first answer: P. M. Cohn proved that $SL_2(R) \neq E_2(R)$ for $R = \mathbb{Z}[X]$ by considering the matrix $\begin{pmatrix} 1 + 2X & 4 \\ -X^2 & 1 - 2X \end{pmatrix}$ after his [Lemma 8.4, 1]. This matrix is easily seen to sit in $E_2(\mathbb{Z}[X^{\pm 1}])$ by cancelling one coefficient and then using Whitehead's lemma. P. M. Cohn also proved using [Proposition 7.2, 1] that the matrix $\begin{pmatrix} 1 + XY & X^2 \\ -Y^2 & 1 - XY \end{pmatrix}$ doesn't lie into $E_2(k[X, Y])$ for any field $k$. This matrix is usually referred to as Cohn's matrix and admits the following generalization discussed by T. Y. Lam and T. Dorsey [Example VI.3.5, 5]: $C_{m, n}(X, Y) = \begin{pmatrix} 1 + XY & X^n Y^m \\ (-1)^r X^m Y^n & 1 - XY + X^2 Y^2 - \cdots + (-1)^r X^r Y^r \end{pmatrix}$ where $m, n \ge 0$ and $r = m + n - 1 \ge 0$. The first answer's author legitimately asks whether $C_{0, 3}(X - 1, p)$ lies into $E_2(\mathbb{Z}[X^{\pm 1}])$ for $p$ a prime number (it is known that $C_{m, n}(x, y) \in E_3(R)$ for any elements $x, y$ in a commutative ring $R$).


[1] "On the structure of the $GL_2$ of a ring", P. M. Cohn, 1966.
[2] "$E_2 \neq SL_2$ for most Laurent polynomial rings", S. Bachmuth, H. Y. Mochizuki, 1982.
[3] "Open problems in combinatorial group theory", G. Baumslag et al., 2000.
[4] "On finite and elementary generation of $SL_2(R)$", P. Abramenko, 2007, http://arxiv.org/abs/0808.1095.
[5] "Serre's problem on projective modules", T. Y Lam, 2006.

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This is only a long comment. I would think of $\mathbb{F}_q[x,t^{\pm 1}]$ as the function field analogue of $\mathbb{Z}[t,t^{-1}]$. For the latter, it is known that $SL_2\neq E_2$ from the results in the following paper:

  • S. Krstic and J. McCool. Free quotients of $SL_2(R[x])$. Proc. Amer. Math. Soc. 125 (1997), 1585-1588.

Applying the main result of Krstic-McCool to $R=\mathbb{F}_q[t,t^{-1}]$ implies that the matrices $$ \left(\begin{array}{cc} 1+(t-1)x^k & x^{3k}\\ (t-1)^3 & 1-(t-1)x^k+(t-1)^2x^{2k} \end{array}\right) $$ map to a basis of a free quotient of $SL_2(\mathbb{F}_q[x,t,t^{-1}])/E_2$. In particular, they witness that $SL_2$ is far from elementarily generated. It is also possible to see that the above matrices are not in the elementary subgroup using Park's realization algorithm.

The naive translation to the ring in question suggests the following candidates for matrices in $SL_2(\mathbb{Z}[t,t^{-1}])$ which are not products of elementary matrices: $$ \left(\begin{array}{cc} 1+(t-1)p & p^{3}\\ (t-1)^3 & 1-(t-1)p+(t-1)^2p^{2} \end{array}\right) $$ As I said, this is not an answer because I have no idea how to show that these matrices are in the elementary subgroup (or to factor them as product of elementary matrices).

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  • $\begingroup$ This is a very nice comment/answer! I was wondering exactly about this point; are there any concrete examples of matrices, where it is unclear at first if they are elementary. $\endgroup$ – Andreas Thom Jun 1 '16 at 9:16
  • $\begingroup$ It's a bit inconvenient to call "elementary" both the genuine ones (those with 1 on the diagonal and a single other nonzero entry) and the products (=elements of the subgroup generated by those) of such matrices... the latter is quite a poor choice of terminology... $\endgroup$ – YCor Jun 1 '16 at 15:07
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    $\begingroup$ About the use of Park's realization theorem (Theorem 2.1 in SIAM Journal on Matrix Analysis and Applications, 1999), I would be careful: T. Y. Lam claims there is a gap in the proof which necessitates a correction: the result holds only over fields, not over Euclidean domains as stated. (see Remark I.8.10 in "Serre's Problem on Projective Modules", T. Y. Lam, 2006). $\endgroup$ – Luc Guyot Jun 1 '16 at 21:04

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