Bass' stable range condition for principal ideal domains - MathOverflow most recent 30 from http://mathoverflow.net 2013-06-20T01:40:46Z http://mathoverflow.net/feeds/question/110457 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/110457/bass-stable-range-condition-for-principal-ideal-domains Bass' stable range condition for principal ideal domains FJH 2012-10-23T18:39:06Z 2012-10-24T16:28:21Z <p>In his algebraic K-Theory book Bass gives the following property on a ring $R$ and a number $n$:</p> <p>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. </p> <p>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.</p> <p>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> <p>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.</p> http://mathoverflow.net/questions/110457/bass-stable-range-condition-for-principal-ideal-domains/110472#110472 Answer by Andy Putman for Bass' stable range condition for principal ideal domains Andy Putman 2012-10-23T20:47:22Z 2012-10-24T16:28:21Z <p>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 $1$.</p> <p>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 $\text{gcd}(a_1+r_1 a_{n+1},\ldots,a_n + r_n a_{n+1}) = 1$. </p> <p>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 $j \neq i$.</p> <p>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 <code>$b = \text{gcd}(a_2,\ldots,a_n)$</code>, 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 <code>$c_i \in \{0,1\}$</code> 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 $1$.</p> <hr> <p>Here is what was my original answer:</p> <p>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 <a href="http://www.math.rice.edu/~andyp/MaazenThesis.djvu" rel="nofollow">here</a>. 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 <a href="http://www.math.rice.edu/~andyp/papers/" rel="nofollow">webpage</a>.</p>