# Bass' stable range condition for principal ideal domains

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.

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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.

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.

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 $1$.

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.

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Just what I was hoping for! As for the more technical question: van der Kallen claims 'a standard application of this' is that for an arbitrary unimodular $v \in R^n$, $n\geq 5$ there is an automorphism of $R^n$ that sends $v$ to a vector with first component $0$, and that fixes (not necessarily pointwise) the set of vectors with last component $0$ or $1$ und second to last component $0$. As you seem knowledgeable in these things, do you know a proof? Or should I make this a new question, or even a private question or not ask you at all? – FJH Oct 24 '12 at 10:26
Regarding your original answer: Maazen's thesis and the published paper use different filtrations though, and the thesis' one seems a whole lot more complicated. And by now I really am interested in this general proof. (A copy of the thesis is also available on van der Kallen's homepage by the way.) – FJH Oct 24 '12 at 10:32
@Andy One should be a little careful about zeroes. When the starting vector is $(8,0,0,0,5)$ it does not suffice to use $r_1$. – Wilberd van der Kallen Oct 24 '12 at 14:48
@FJH This was not my claim. We wanted to send $v$ to a vector with first component 1, not 0. Zero would also be possible, but that was not the claim. First one adds multiples of later entries to $v_1$ and $v_2$ to arrange that $(v_1,v_2,v_3)$ becomes unimodular. Then one transforms $(v_1,v_2,v_3)$ to $(1,0,0)$. All this leaves the last two coordinates alone and can be achieved by the action of $GL_n$. – Wilberd van der Kallen Oct 24 '12 at 14:53
Yes, sorry, it must be $1$ not $0$. and regarding andy's answer: it still suffices to use one $r$ though, right? The $a_1,\ldots,a_n$ can just be permuted to make his argument apply (unless they are all $0$ in which case there's nothing to do). And thanks a lot! – FJH Oct 24 '12 at 16:16

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 $(a,b,c) = (a + tc, b - sc)$.

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