Bass' stable range condition for principal ideal domains

Question

In his algebraic K-Theory book Bass gives the following property on a ring $R$ and a number $n$:

For every $n$ elements $v_1, \ldots, v_n$ that generate the unit ideal there are numbers $r_1, \ldots r_{n-1}$ such that $v_1 + r_1 v_n, v_2 + r_2 v_n, \ldots, v_{n-1} + r_{n-1} v_n$ also generate the unit ideal.

He then goes on to show that a noetherian, d-dimensional ring has this property for all $n \geq d+2$, but the proof is long and nontrivial.

My question now is: Is there an easier way to see this for a principal ideal domain and say $n=3$? Or even more concretely, given three numbers $a,b,c \in \mathbb Z$ with $gcd(a,b,c) = 1$ why are there numbers $n,m \in \mathbb Z$, such that also $gcd(a+nc,b+mc) = 1$?

P.S. I have a more technical question along the same lines waiting for the lucky answerer! All of this comes from my trying to understand van-der-Kallen homology stability of general linear groups.

Answer 1

EDIT 3 : Sorry for editing this old answer one more time, but I want to also point out for future readers that there is a proof that Dedekind domains have stable range $2$ which is very similar to my proof for PID's in Satz K.13 of the book Algebra by Jantzen and Schwermer. The whole Appendix K of that book is a lovely introduction to the whole notion of stable range.

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EDIT 2 : I just learned of a super-short proof of the special case of the Bass Stable Range theorem alluded to in the question (the one giving the stable range for Noetherian $d$-dimensional rings). It's a little more abstract than what I did below for PID's, but not much harder. See Section 2 of

MR0217052 (36 #147) Estes, Dennis; Ohm, Jack Stable range in commutative rings. J. Algebra 7 1967 343–362.

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EDIT : Here's a proof that works for $R$ a a PID, which implies that the condition of generating the unit ideal is the same as having gcd equal to $1$.

For some $n \geq 2$ consider a tuple $(a_1,\ldots,a_{n+1})$ of elements of $R$ whose gcd is $1$. We want to find $r_1,\ldots,r_n \in R$ such that $\text{gcd}(a_1+r_1 a_{n+1},\ldots,a_n + r_n a_{n+1}) = 1$.

There are three cases. If $a_{n+1}=0$, then there is nothing to do. If $a_i=0$ for some $1 \leq i \leq n$, then we can take $r_i=1$ and $r_j=0$ for $j \neq i$.

The most interesting case is when none of the $a_i$ equal $0$. In this case, we will only need $r_1$ (the rest of the $r_i$ can be taken to be $0$). Set $b = \text{gcd}(a_2,\ldots,a_n)$, and let $p_1,\ldots,p_k$ be the distinct primes dividing $b$. For each $i$, we know that $p_i$ cannot divide both $a_1$ and $a_{n+1}$. This implies that there exists some $c_i \in \{0,1\}$ such that $$a_1 + c_i a_{n+1} \neq 0 \quad (\text{mod } p_i).$$ By the Chinese remainder theorem, there exists some $r_1 \in R$ such that $$r_1 = c_i \quad (\text{mod } p_i)$$ for $1 \leq i \leq k$, which implies that $$a_1 + r_1 a_{n+1} \neq 0 \quad (\text{mod } p_i)$$ for all $1 \leq i \leq k$. We conclude that the gcd of $a_1+r_1 a_{n+1}$ and $b$ equals $1$, and thus that the gcd of $a_1+r_1 a_{n+1},a_2,\ldots,a_n$ is $1$.

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Here is what was my original answer:

This does not exactly answer your question, but it is much easier to prove that the complexes that van der Kallen needs are highly connected for $\mathbb{Z}$ than for general rings. This was originally done by Maazen in his unpublished thesis, which can be downloaded here. There is also a different proof of this connectivity in Step 2 of the proof of Theorem B in my paper "The complex of partial bases for $F_n$ and finite generation of the Torelli subgroup of $\text{Aut}(F_n)$" with Matt Day, available on my webpage.

Answer 2

To illustrate a few different techniques, I will give a couple more proofs that PID's have stable rank $\le 2$, along with some generalizations.

Theorem: An $n$-dimensional domain has stable rank $\le n+1$, and an $n$-dimensional ring has stable rank $\le n+2$.

This is Theorem 3.4 in "Generating Ideals in Prufer Domains" by Heitmann. I do not know of a simple proof, but I am listing it for completeness.

The case for one-dimensional domains has a different, much easier proof, which I will sketch. Two facts that are easy to check from the definitions: (1) a ring has stable rank $\le n+1$ if each proper homomorphic image has stable rank $\le n$, and (2) $R$ and $R/J(R)$ have the same stable rank. Using these we reduce to showing that von Neumann regular rings have stable rank 1. Given elements $a$ and $b$, write $a = ue$, where $u$ is a unit and $e$ is an idempotent. Then $(a + (1-e)b)(eu^{-1}b + 1-e) = eb + (1-e)b = b$, so $(a,b) = (a + (1-e)b)$. (Coincidentally, this also shows that VNR rings are Bezout.)

If one is not familiar with VNR rings, then the one-dimensional Noetherian case is a little easier, since it is a special case of the following fact.

Theorem: Any ring with only finitely many maximal ideals has stable rank 1, and thus any ring where every nonzero element is contained in only finitely many maximal ideals has stable rank $\le 2$.

The proof proceeds as before, but is even simpler since, if $R$ has only finitely many maximal ideals, then $R/J(R)$ is a finite direct product of fields, and obviously a direct product of rings of stable rank $\le n$ has stable rank $\le n$.

For another generalization, one has:

Theorem: Bezout domains have stable rank $\le 2$.

Proof: Given elements $a,b,c$, write $(a,b) = (d)$, say $a = dx$, $b = dy$, and $d = as + bt = d(xs + yt)$. The case $d = 0$ is trivial. Otherwise, we get $xs + yt = 1$. Then $(a + tc)y - (b - sc)x = (ay - bx) + (xs+ty)c = 0 + 1\cdot c = c$. Therefore $(a,b,c) = (a + tc, b - sc)$.

Answer 3

See the paper

Anderson, D.D.; Juett, J.R., Stable range and almost stable range, J. Pure Appl. Algebra 216, No. 10, 2094-2097 (2012). ZBL1261.19002.

for a simple proof of Bass's Theorem

Answer 4

The following theorem, due to Raymond Heitmann (Generating non-Noetherian modules efficiently., Doi:10.1307/mmj/1029003021, Corollary 2.4.(ii) to Theorem 2.1), takes away the Noetherianity restriction.

An elementary proof, paraphrased below, was found by Thierry Coquand, Sur un théorème de Kronecker concernant les variétés algébriques., Doi: 10.1016/j.crma.2003.12.008. I reproduce it here because the argument is so simple and deserves to be better known.

Theorem: If $A$ is a commutative ring of Krull dimension $\dim A\leq d$, and $a, b_0,\cdots,b_d\in A$, then there exist $x_0,\cdots,x_d\in A$ such that $\sqrt{aA+b_0A+\cdots+b_dA}$ $=$ $\sqrt{(b_0+ax_0)A+\cdots+(b_d+ax_d)A}$.

Proof: Induction on $\dim A$. When $\dim A=-1$, $A=0$ is the trivial ring. It has a unique ideal, which is generated by every subset of $A$, including the empty set.

Now let $\dim A\geq0$. It clearly suffices to find $x_0,\cdots,x_d\in A$ such that $a\in\sqrt{I_d}$, where $I_d$ is the ideal $(b_0+ax_0)A+\cdots+(b_d+ax_d)A$. Let $J$ $=$ $b_dA+\sqrt{0}:b_d$, the ideal generated by $b_d$ and the $y\in A$ for which $yb_d$ is nilpotent.

If $\mathfrak{p}$ is a minimal prime ideal of $A$ and $b_d\in\mathfrak{p}$, then $b_d\in\mathfrak{p}A_{\mathfrak{p}}$. But $\mathfrak{p}A_{\mathfrak{p}}$ is the unique prime ideal of the ring $A_{\mathfrak{p}}$, so that $b_d$ must be nilpotent in $A_{\mathfrak{p}}$. Hence $sb_d$ is nilpotent in $A$ for some $s\in A-\mathfrak{p}$. Then $s\in J-\mathfrak{p}$, and thus $J$ is not contained in any minimal prime of $A$.

It follows that $\dim A/J$ $<$ $\dim A$. By the induction hypothesis, there are $n\in\mathbb{N}$ and $x_0,\cdots,x_{d-1}\in A$ such that $(a\text{ mod }J)^n$ $\in$ $I_{d-1}\cdot A/J$, where $I_{d-1}$ is the ideal $(b_0+ax_0)A+\cdots+(b_{d-1}+ax_{d-1})A$. So $a^n-b_dz-x_d$ $\in$ $I_{d-1}$ for suitable $z$ and $x_d$ in $A$ with $x_db_d$ nilpotent.

Then $a^{n+1}$ $\in$ $I_{d-1}+b_dA+ax_dA$, so that $a$ $\in$ $\sqrt{I_{d-1}+b_dA+ax_dA}$ $=$ $\sqrt{I_d}$, where $I_d$ denotes $(b_0+ax_0)A+\cdots+(b_d+ax_d)A$ as before, with all elements $x_i$ now determined. The latter equality follows from the identities $\sqrt{I+K}$ $=$ $\sqrt{I+\sqrt{K}}$, for any pair of ideals $I,K$ of $A$, and $\sqrt{bA+cA}$ $=$ $\sqrt{(b+c)A}$, for any $b,c\in A$ with $bc$ nilpotent - indeed, in that case $b^2$ $=$ $b(b+c)-bc$ implies that some power of $b$ is in $(b+c)A$, and the same goes for $c$. $\,\square$

Bass' Stable Range Theorem (valid for all commutative rings having finite Krull dimension) follows at once, as for any ideal $I$ of $A$ one has $1\in I$ if and only if $1\in\sqrt{I}$.