# Is SL(n,Z[x]) generated by transvections?

Is $\mathrm{SL}(n,\mathbb{Z}[x])$ equal to $E(n,\mathbb{Z}[x])$, the subgroup generated by transvections?

• The answer to the question is clearly "no", since the determinant is not trivial on $GL_n$. But the answer to the question in the title is yes, as least when $n$ is big enough, and this follows (maybe there are more simple methods) from the fact that it works for $SL(\infty,\mathbf Z)$, from the fundamental theorem of K-theory applied to SK1, and from stability results on the homology of $SL_n$. Probably you would find the exact answer in Rosenberg's Algebraic K-Theory book. Once again, there might also exist an elementary answer. – few_reps Sep 24 '15 at 8:20
• Also, this mathoverflow.net/questions/156563/… seems to answer my question in the affirmative for $n =1,3,4...$. – Yang-Mills Sep 24 '15 at 8:24
• Found ! Grunewald, Mennicke, and Vaserstein [ On the groups SL2(Z[x]) and SL2(k[x, y]), Israel Jour. Math. 86, (1994) 157–193 ] show that $SL_2(\mathbf Z[T])/E_2(\mathbf Z[T])$ surject on free groups with countable ranks ! – few_reps Sep 24 '15 at 9:54
• Cohn was probably the first to prove that $\mathrm{SL}_2(\mathbf{Z}[T]) \neq \mathrm{E}_2(\mathbf{Z}[T])$. A concrete example of a matrix not in $\mathrm{E}_2$ is $\begin{bmatrix} 1+2T & 4 \cr -T^2 & 1-2T\end{bmatrix}$. I took this example from Lam's "Serre's problem on projective modules", rk 8.11. – Oblomov Sep 24 '15 at 13:22
• Why was this downvoted and closed? It's a perfectly reasonable question with a nontrivial answer. Was this an overreaction to the OP's minor mistake of typing $GL$ instead of $SL$ in the body of the question? – Steven Landsburg Sep 24 '15 at 13:56

First, for $n=2$, the result is certainly false by the counterexample of Cohn quoted in Oblomov's comment.

For higher $n$, start with the fact that the Bass Stable Rank of a commutative ring is at most 1 more than the Krull dimension. (I'm sure you can find this hidden in Chapter V of Bass's book on K-theory, though extracting it might require some work to get familiar with his notation). Therefore ${\mathbb Z}[X]$ has stable rank at most 3. In fact, it's almost surely exactly 3. (There is some discussion here about whether this stable rank might in fact be 2, but there's something close to a proof there that this is not the case.)

Now from Bass, Chapter V, 3.3, it follows that for all $n\ge 4$, we have $$GL_n({\bf Z}[x])=GL_3({\mathbb Z}[X])E_n({\mathbb Z}[X])$$

In other words, for any matrix of size $4x4$ or greater, you can, by applying elementary operations, reduce to a matrix of the form $$\pmatrix{A&0\cr 0&I\cr}$$

where $A$ is 3 by 3 and $I$ is the $(n-3)\times (n-3)$ identity matrix.

It's not immediately clear to me how much better you can do, but I bet either that Wilberd van der Kallen (who shows up here occasionally) or Vaserstein (who I think does not) could answer this in his sleep. You might want to email one or both of them.

• Thanks Steven! I found Suslin's original paper, and indeed, the Corollary 6.6, as claimed in the first answer in mathoverflow.net/questions/156563/… is indeed applicable to $\mathbb{Z}[x]$, with $n \geq 3$, since $\mathbb{Z}$ is regular, $SK_{1}(Z) = 0$ and Krull Dimension of $\mathbb{Z}$ is $1$. – Yang-Mills Sep 25 '15 at 1:04
• You are right --- Suslin's paper supercedes the content of my answer. – Steven Landsburg Sep 25 '15 at 2:12
• Just to clarify: the conclusion of the discussion you mention about the stable range of $\mathbf Z [x]$ was that it is indeed of stable range $3$. It is proven in the paper by Grunewald, Mennicke, and Vaserstein mentionned by "few_reps" as indicated in the discussion. – Oblomov Sep 25 '15 at 7:42