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

YCor
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Chern
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    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 08:20
  • But $GL(n,Z[x])$ is just matrices with determinant $1$ or $-1$, so it should be true for both, no? – Chern Sep 24 '15 at 08:23
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    Also, this http://mathoverflow.net/questions/156563/suslins-stability-theorem-for-chevalley-groups seems to answer my question in the affirmative for $ n =1,3,4...$. – Chern Sep 24 '15 at 08:24
  • Just to be sure : do you know of transvections that have determinant $-1$ ? That would imply that we don't have the same definition. The link you give, indeed, contains an answer to the question in the title and seems to imply that $SL_n(\mathbf Z[T])=E_n(\mathbf Z[T])$ for $n\geq 3$. I don't know what happens for $n=2$, but that has certainly been studied (by Bass ?) – few_reps Sep 24 '15 at 09:28
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    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 09:54
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    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
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    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
  • Note that the term "transvection" means a more general type of elements, namely an element of the form $e+uv$, where $u$ is a column, $v$ is a row and $vu=0$. Over a (skew-)field they are exactly the conjugates of the elementary transvections $e+\xi e_{ij}$, but in general not every transvection lies in the elementary subgroup. That does not affect your question, of course, but for other rings it might. – Andrei Smolensky Sep 24 '15 at 21:48
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    Hi all! I wished to "show" that if $ A, B\in \mathbb{Z}[x]^{n\times n}$ and if $ A \sim B$, in the sense that there are matrices $ P, Q \in GL(n,\mathbb{Z}[x]) $ such that $ PAQ = B $, then, there is a sequence of row and column operations that converts $ A $ into $ B $ (which are adding a multiple of one row(column) to another row(column),switching two rows (columns) and multiplying a row(column) by a unit i.e.1,-1). It seems this is not true for $ n = 2 $, but true for $ n \geq 3 $. This is what I actually wanted. Can someone give a list of references that I can use to quote the full result? – Chern Sep 24 '15 at 22:55
  • @swaqar: Your most recent comment repeats the error that I corrected in your original post (and that led to this question's being closed for a while). Clearly if $P=-Q$ is the identity, what you're asking for is impossible. You want $SL$, not $GL$. – Steven Landsburg Sep 24 '15 at 23:02
  • Hi Steven! Can we discuss this in a private chat? – Chern Sep 24 '15 at 23:08
  • @swaqar: My mistake. I see you included "multiplying a row by a unit" as an allowable operation. Given that, your comment is fine. – Steven Landsburg Sep 24 '15 at 23:12
  • Thank you Steven! So, what reference should I give for this result in my paper? I don't know K-Theory at all, and I am trying to look into some K-Theory books, but I have not been able to find the exact result.So, first of all, is the result of $ n \geq 3 $ actually true? Second, which reference could I use for it? – Chern Sep 24 '15 at 23:18
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    @StevenLandsburg and upvoters of his comment: because not everybody was able to guess the correct meaning (GL instead of SL was one thing, "generated by the group" was confusing though not incorrect). Also I have seen in MO several questions where an apparently naive question was very close to a natural and more interesting one, and people tend to interpret it as the corrected question but the author insisted that it was really the naive one (here I mean naive at a research level). In its original formulation it was clear to me that the question had to be closed (or clarified by its author). – YCor Sep 25 '15 at 08:12

1 Answers1

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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.

Community
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Steven Landsburg
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  • Thanks Steven! I found Suslin's original paper, and indeed, the Corollary 6.6, as claimed in the first answer in http://mathoverflow.net/questions/156563/suslins-stability-theorem-for-chevalley-groups 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 $. – Chern Sep 25 '15 at 01:04
  • You are right --- Suslin's paper supercedes the content of my answer. – Steven Landsburg Sep 25 '15 at 02:12
  • 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 07:42