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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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up vote 8 down vote accepted

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. I happen to have a scan of this which I posted 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
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@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
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