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Let $n$ be a positive integer and $A$ be a commutative ring. The ring $A$ is said to be of Bass stable range $\mathrm{sr}(A)\leq n$ if for $a, a_1, \dots, a_n \in A$ one has the following implication:

$$1 \in \langle a, a_1, \dots, a_{n}\rangle \implies \exists \ x_1, \dots, x_n \in A, 1 \in \langle a_1+x_1a,\ \dots, a_n + x_n a\rangle.$$ (Above, $\langle \cdot \rangle$ denotes the ideal generated by the elements inside).

Of course, one says that $A$ is of stable range $n$ if $\mathrm{sr}(A) \leq n$ and $\mathrm{sr}(A) \not \leq n-1$.

Bass proved that if $A$ is noetherian of Krull dimension $d$ then $\mathrm{sr}(A)\leq d+1$.

Examples are known; for example Vaserstein proved that $\mathrm{sr}(k[x_1,\dots,x_n]) = n+1$ when $k$ is a subfield of the real numbers.

My question is : is the stable range of the ring of integer polynomials $\mathbf Z[X]$ known?

What I wrote before shows that $\mathrm{sr}(\mathbf Z[X]) \leq 3$ and it seems very likely that $\mathrm{sr}( \mathbf Z[X])=3$.

A refinement of my previous question is : could you provide an explicit unimodular triplet of polynomials $(P_1,P_2,P_3) \in \mathbf Z[X]$ showing that $\mathrm{sr}(\mathbf Z[X]) \not \leq 2$?

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What is $\langle a,a_1,\dots,a_{n+1}\rangle$? –  Fernando Muro Jun 5 '13 at 14:24
It's the ideal generated by these elements. Sorry, I thought this was transparent. –  Oblomov Jun 5 '13 at 14:35
Great question, Oblomov: +1. –  Georges Elencwajg Jun 6 '13 at 21:19
I think Vaserstein proved that $sr(k[x1,…,xn])=n+1$ if $k$ is a subfield of the real numbers (theorem 8 in his 1971 paper "Stable rank of rings and dimensionality of topological spaces"). E.g. for $k$ a finite field, this is, in general, wrong, as pointed out in Steven Landsburg's comment to Jeremy Rickard's answer (cf. theorem 18.2 in the cited paper by Vaserstein/Suslin). In fact, for $k$ algebraic over a finite field, $sr(k[x1,…,xn]) \le n$ as soon as $n \ge 2$: see Vaserstein/Suslin, corollary 17.4. –  Torsten Schoeneberg Jun 12 '13 at 14:21
My above comment was meant to suggest a modification of the line "Examples are known ..."; I cannot edit the question myself. Or am I wrong (always quite possible)? –  Torsten Schoeneberg Jun 16 '13 at 7:53
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5 Answers

up vote 2 down vote accepted

Sorry for putting this in a separate answer, but I think it will be cleaner this way. I believe I now understand Vaserstein's intended argument:

1) There are rings of the form $A={\mathbb Z}[x]/(h)$ such that $SK_1(A)\neq 0$. One way to get such a ring is to start with the ring of integers in a quadratic field, let $I$ be a "sufficiently small" ideal (I confess to not being exactly sure what this means) and look at the subring generated by $1$ and $I$. For some definition of "sufficiently small", Bass has shown that this gives us $SK_1(A)\neq 0$.

2) Take a non-zero element of $SK_1(A)$ and represent it by a Mennicke symbol $\[\overline{f},\overline{g}\]$.

3) Then $(f,g,h)$ is a unimodular row over ${\bf Z}[x]$.

4) Clearly, the Mennicke symbol $\[\overline{f},\overline{g}\]$ does not lift to $K_1({\mathbb Z}[x])$.

5) It follows from Lemma 17.1 of the paper referenced by Jeremy Rickard that the row $(f,g,h)$ is not reducible.

I'm still just a tad unclear on why it is, in point 1), that we can take the kernel of ${\mathbb Z}[x]\rightarrow A$ to be principal. Is this obvious? I'll add a comment if I nail this down.

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If the ideal $I$ is not principal, this makes no difference. Suppose $[\bar f,\bar g]$ is unimodular modulo $I$. Choose $p$, $q\in \Bbb Z[x]$ so that $h:=pf+qg-1\in I$. Then $(f,g,h)$ is a unimodular row that is not reducible. –  Wilberd van der Kallen Jun 18 '13 at 4:04
Wilberd: Right. I no longer understand why I thought this mattered. Thanks for making this clear. –  Steven Landsburg Jun 18 '13 at 13:00
The answer is given in a paper of Grunewald, Mennicke and Vaserstein (On the groups $SL_2(\mathbf Z[x])$ and $SL_2(k[x,y])$). Israel J. Math. 86 (1994), no. 1-3, 157–193). One example of unimodular row that is not reducible is the following $(21+ 4x, 12, x^2 + 20)$. –  Oblomov Oct 24 '13 at 14:12
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After spending considerable time trying to construct a counterexample, I turned to Google and found the book "Rings Related to Stable Range Conditions" by Huanyin Chen, which claims, on page 338, that the stable range of ${\mathbb Z}[X]$ is in fact 2. This would explain my inability to find a counterexample (though not, perhaps, my willingness to put aside other urgent projects in order to look for one). However, I've not been able to understand his argument.

There is clearly a typo where he says "Clearly,${\mathbb Z}[x]$ is a euclidean integral domain; hence it is a Dedekind domain". Presumably he means ${\mathbb Z}$ instead of ${\mathbb Z}[x]$? But then I cannot fully follow the rest of the argument, possibly because it references Example 12.1.14, which is not part of the preview available on either Amazon or Google books.

In any event, though you probably already know this, the simplest class of nontrivial unimodular rows over ${\mathbb Z}[X]$ consists of those of the form $(1+aX,bX^m,cX^m)$ where $b,c$ and $m$ are arbitrary and some power of $a$ lives in the ideal $(b,c)$. I tried to find $a,b,c$ for which this row was provably not reducible, but I was insufficiently clever to pull this off.

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Thanks for the reference (although I don't understand the argument too). I'll try to write to the author. –  Oblomov Jun 6 '13 at 8:07
I found Example 12.1.14 in Chen's book (it's on page 373). It states that a Dedekind domain is of stable range $2$. The typo you found seems to be rather a mistake. –  Oblomov Jun 6 '13 at 8:24
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There's a comment at the top of page 993 of

L N Vaseršteĭn, A A Suslin, "Serre's problem on projective modules over polynomial rings, and algebraic K-theory", Math. USSR Izv., 1976, 10 (5), 937–1001,

to the effect that one of the authors had proved that the stable rank of $\mathbb{Z}[X]$ is 3, but unfortunately without any hint as to which or where or how, although they do show there that the stable rank of $\mathbb{Z}[X_1,\dots,X_n]$ is equal to $n+1$ if $n>1$.

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Is $\mathbf Z[x_1, \dots, x_n]$ really of stable range $n+1$ (and not $n+2$)? –  Oblomov Jun 6 '13 at 15:01
This is right. For a finitely generated ring of dimension at least 3, we always have stable range $\le$ the dimension of the ring. –  Steven Landsburg Jun 6 '13 at 16:15
I've written to both Suslin and Vaserstein about this and will report back if I learn anything. –  Steven Landsburg Jun 6 '13 at 16:25
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I have this from Vaserstein:

To show that ${\mathbb Z}[X]$ has stable range greater than 2, it suffices to find a quotient $A$ of ${\mathbb Z}[X]$ with $E_2(A)\neq SL_2(A)$.

For this, let $B$ be the ring of integers in an imaginary quadratic field, let $J$ be a small ideal in $B$ so that $SK_1(B,J)\neq 1$, and let $A=1+J$.

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Thanks for that, Steven. I haven't digested it yet, but presumably I'm not the only one who's wondering "What's the unimodular row?" –  Jeremy Rickard Jun 6 '13 at 17:42
Jeremy: I'm wondering that too. Presumably one can work backward from this argument to construct the row explicitly, but I'm trying to resist the temptation to start a calculation that threatens to kill my afternoon. (On another front, I quoted Vaserstein when I said "small ideal", but I presume this must mean "principal ideal" to make the argument work.) –  Steven Landsburg Jun 6 '13 at 17:51
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This is rather a comment to Steven Landsburg/ Vaserstein answer, but too long for a comment.

Finding a quotient of $\mathbf Z[X]$ with $E_2 \neq \mathrm{SL}_2$ is easy. Actually, $\mathbf Z[X]$ will do. For example, Cohn proved that the matrix : $$\begin{bmatrix} 1+2x & 4 \cr -x^2 & 1-2x \end{bmatrix} $$ is in $\mathrm{SL}_2$ but not in $E_2$.

If one prefers an example in the spirit of the one suggested by Vaserstein, then Cohn (again!) shjowed that in the ring of integers of $\mathbf Q[\sqrt{-19}]$, with $\theta= \frac{1+\sqrt{-19}}{2}$, the matrix $$ \begin{bmatrix} 3- \theta & 2+ \theta \cr -3-2\theta & 5-2\theta \end{bmatrix} $$ is in $\mathrm{SL}_2$ but not in $E_2$.

(I took these two examples of matrices from the book of T.Y. Lam on "Serre's problem on projective module, Chapter I.9.)

With these examples, it should be straightforward to make explicit a unimodular row, but I have to admit that I don't understand Vaserstein's argument. (Is it well known that $\mathrm{sr}(A)= 2 $ implies $\mathrm{SL}_2(A)= E_2(A)$?)

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Sorry, I was not able to find how to type matrices correctly –  Oblomov Jun 7 '13 at 12:42
I fixed it for you. For some reason you need to use the TeX \cr rather than the LaTeX `\`. –  Ryan Reich Jun 7 '13 at 15:23
I'm confused too, but here's my best approximate guess of what he means: First, let $A$ be a ring constructed per V's suggestion. It suffices to show that $sr(A)\ge 3$. For this, it suffices to show that $SL_3(A)\neq E_3(A)$, and for this it suffices to show that $SK_1(A)\neq 0$. And indeed, examples of such $A$ are known for which $SK_1\neq 0$. So, modulo filling in details, this works if Vaserstein's "2" was a typo for "3". –  Steven Landsburg Jun 7 '13 at 17:59
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