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 $\mathrm{sr}(\mathbf Z[X]) \not \leq 2$?

  • $\begingroup$ What is $\langle a,a_1,\dots,a_{n+1}\rangle$? $\endgroup$ Jun 5 '13 at 14:24
  • $\begingroup$ It's the ideal generated by these elements. Sorry, I thought this was transparent. $\endgroup$
    – Oblomov
    Jun 5 '13 at 14:35
  • $\begingroup$ Great question, Oblomov: +1. $\endgroup$ Jun 6 '13 at 21:19
  • 2
    $\begingroup$ 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. $\endgroup$ Jun 12 '13 at 14:21
  • $\begingroup$ 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)? $\endgroup$ Jun 16 '13 at 7:53

Sorry for putting this in a separate answer, but I think it will be cleaner this way.

The stable range of $\mathbb{Z}[x]$ is equal to 3.

I believe I now understand Vaserstein's intended argument that it's $\ge 3$:

(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 $\mathbb{Z}[x]$.

(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.

  • 4
    $\begingroup$ 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. $\endgroup$ Jun 18 '13 at 4:04
  • $\begingroup$ Wilberd: Right. I no longer understand why I thought this mattered. Thanks for making this clear. $\endgroup$ Jun 18 '13 at 13:00
  • 3
    $\begingroup$ 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 $(21+ 4x, 12, x^2 + 20)$. $\endgroup$
    – Oblomov
    Oct 24 '13 at 14:12
  • 2
    $\begingroup$ @Oblomov There is a flaw in the (very last part of) proof of Proposition 1.9 of Grunewald et al. Therefore $(21 + 4x, 12, x^2 + 20)$ is not guaranteed to be non-reducible (and I cannot show that it is reducible). I elaborate a little bit on this in my answer below. $\endgroup$
    – Luc Guyot
    Feb 17 '18 at 21:40

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 $n>1$.

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

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.

Edit This claim that the stable range of ${\mathbb Z}[X]$ is 2 is not correct according to this other answer of mine.

  • $\begingroup$ Thanks for the reference (although I don't understand the argument too). I'll try to write to the author. $\endgroup$
    – Oblomov
    Jun 6 '13 at 8:07
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    $\begingroup$ 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. $\endgroup$
    – Oblomov
    Jun 6 '13 at 8:24

The goal of my answer is to elaborate on steps (1) and (2) of the accepted answer of Steven Landsburg. In particular, I would like to make clear how a "small enough ideal" looks like and how to produce a non-reducible row like the one in Oblomov's comment. This consists essentially in revisiting the proof of [2, Proposition 1.9], with some K-theoretical shortcuts.

Addendum. I realized that the last paragraph of [2, Proof of Proposition 1.9] is flawed, so that the unimodular row $(21 + 4x, 12, x^2 + 20)$ is not guaranteed to be non-stable in $\mathbb{Z}[X]^3$. Indeed, it is assumed there that $SK_1(\mathcal{O}, 2\mathcal{O}) = \mathbb{Z}/2\mathbb{Z}$ for $\mathcal{O} = \mathbf{Z} + \mathbf{Z} \sqrt{-5}$, whereas $SK_1(\mathcal{O}, 2\mathcal{O})= 1$ by the Bass-Milnor-Serre Theorem (In other words, $f = 2$ does not have the property $(*)$ defined at the top of page 191). This glitch in the final step of the proof is harmless in the sense that changing sligthly the conductor $f$ yields a valid, but different, non-reducible unimodular row. Setting for instance $f = 4$, i.e., $B = \mathbf{Z} + 4 \mathbf{Z} \sqrt{-5} \subset \mathcal{O}$, and considering a suitable quadratic residue symbol, will do the job. My answer below is yet another way to address this problem.

The Bass-Milnor-Serre theorem [3, Theorem 11.33] is key to understand step (1): let $F$ be a totally imaginary number field and $m$ is the number of elements of finite order in $F^{\ast}$. For each ideal $I$ in the ring of integers $R$ of $F$, the relative special Whitehead group $SK_1(R, I)$ is a cyclic group of finite order $r$. For each prime $p$, $\text{ord}_p(r)$ is the nearest integer in the interval $\lbrack 0, \text{ord}_p(m) \rbrack$ to $$\min_{\mathfrak{p}} \left\lfloor \frac{\nu_{\mathfrak{p}}(I)}{\nu_{\mathfrak{p}}(pR)} - \frac{1}{p - 1} \right\rfloor$$ where $\lfloor x \rfloor$ denotes the greatest integer $\le x$ and $\mathfrak{p}$ ranges over the prime ideals of $R$ containing $p$.

Independently of the above theorem, it follows from [1, Corollary 5.2] that $SK_1(R, I)$ surjects onto $SK_1(R, J)$ if $I \subset J$ and $R/I$ is semi-local. For $R$ and $I$ as in the theorem, the group $SK_1(R, I)$ increases in size as $I$ decreases with respect to inclusion. This is reflected in the above formula and it is immediate to see that for every decreasing sequence of ideals $I$, the corresponding sequence $(SK_1(R, I))_I$ stabilizes. Moreover, we can find $I$ such that $SK_1(R, I)$ has the maximal cardinality $m$ and $I$ is then small enough in the sense that any ideal contained in $I$ yields the same maximal $SK_1$ group.

We are now (almost) in position to provide an order $A$ in some $R$ as above such that $\mathbf{Z}[X]$ surjects onto $A$ and $SK_1(A)$ is not trivial. Indeed, consider $R = \mathbf{Z}[i]$, the Gaussian integers, set $A = \mathbb{Z} + 4i \mathbb{Z}$ and $I = 4 \mathbf{Z}[i]$ (this $I$ is sufficiently small for our purpose, but doesn't maximize $SK_1$, unlike $8 \mathbf{Z}[i]$). We will establish that $SK_1(A)$ surjects onto $SK_1(R, I) \simeq \mathbf{Z}/2\mathbf{Z}$, the latter isomorphism being given by Bass-Milnor-Serre's theorem. To do so, let us prove the following lemma:

Let $S$ be a one-dimensional Noetherian domain and let $J$ be an ideal of $S$ such that $S/J \simeq \mathbf{Z}/n\mathbf{Z}$ for some $n \ge 0$. Then $SK_1(S) = SK_1(S, J)$.

Proof. In the exact sequence [3, Theorem 13.20 and Example 13.22] $$K_2(S/J) \rightarrow SK_1(S, J) \rightarrow SK_1(S) \rightarrow SK_1(S/J)$$ the last term, namely $SK_1(S/J)$, is trivial since $S/J$ is Artinian and hence of stable range $1$. In addition, the image of $K_2(S/J)$ in $SK_1(S, J)$ is also trivial. Indeed $K_2(S/J)$ is generated by the Steinberg symbol $\{-1 + J, -1 + J\}_{S/J}$ because $S/J\simeq \mathbf{Z}/n \mathbf{Z}$ [3, Exercises 13A.10 and 15C.10]. As $\{-1, -1\}_S$ is a lift of the previous symbol in $K_2(S)$ our claim follows and the proof is now complete.

By the previous lemma we have $SK_1(A) = SK_1(A, I)$. Since the inclusion $A \subset R$ induces an epimorphism from $SK_1(A, I)$ onto $SK_1(R, I)$, we deduce that $SK_1(A) \neq 1$.

It is actually possible to find a non-trivial element in $SK_1(A)$ by considering the power residue symbol $\binom{12}{1 + 4i}_2 = -1$ [2, Section 11C]. This yields an non-trivial Mennicke symbol $[12, 1 + 4i]$ and eventually the non-reducible row $(12, 1 + X, X^2 + 16)$ using the ring epimorphism induced by $X \mapsto 4i$.

Erratum. I claimed erroneously in former versions of this answer that the same technique easily applies to $\mathbb{Z}[X^{\pm 1}]$. My conclusions were wrong and I fail to answer the following

Question. Is the stable rank of $\mathbb{Z}[X^{\pm 1}]$ equal to $3$?

[1] H. Bass, "K-theory and stable algebra", 1964.
[2] F. Grunewald et al., "On the groups $SL_2(\mathbf{Z}[x])$ and $SL_2(k[x,y])$", 1994.
[3] B. Magurn, "An algebraic introduction to K-theory", 2002.


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 $A=1+J$.

Edit Vaserstein's claim that the stable range of ${\mathbb Z}[X]$ is $\ge 3$ (and hence 3) is correct. See this other answer of mine for more details.

  • $\begingroup$ 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?" $\endgroup$ Jun 6 '13 at 17:42
  • $\begingroup$ 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.) $\endgroup$ Jun 6 '13 at 17:51

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 $\mathrm{SL}_2(A)= E_2(A)$?)

  • $\begingroup$ 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". $\endgroup$ Jun 7 '13 at 17:59

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