Susan's question When can number rings be spanned (as $\mathbb{Z}$-modules) by units? smells like an algebraic K-theory question in disguise. I'll reformulate the question first:

Given an integral domain (for safety) $R$, let $A$ be the smallest subring containing the units $R^\times$. When does $A = R$? Examples: any field, the integers, the Gaussian or Eisenstein integers, an easily described class of real quadratic rings.

Consider the related high-brow property of an integral domain $R$:

(P): For all (sufficiently large?) $N$, and all partitions $n_1 + \cdots + n_t = N$, the group $GL_N(R)$ is generated by $GL_N(Z)$ and $\prod_{i=1}^t GL_{n_i}(R)$.

There are definitely better ways of writing property (P), but my point is that property (P) seems very much like an asymptotic statement about the structure theory of $GL_N(R)$ as a group containing $GL_N(Z)$.

If $A = R$, then I think $R$ has property (P). So two questions:

  1. Is property (P) equivalent to $A = R$? I'd guess yes. Edit: maybe if $K_1(R) = R^\times$?

  2. Can a K-theory expert reformulate property (P) in terms of the algebraic K-theory of $R$?

At a fundamental level, does the property $A = R$ depend only on the algebraic K-theory of $R$? Does Milnor K-theory come into play (e.g. since $R^\times$ might not equal $K_1(R)$ if $R$ is not a Euclidean domain)?

  • $\begingroup$ (P) implies $A=R$ by taking all the $n_i=1$ for $N \geq 2$, using the fact that each element of $R$ shows up as an element of a unipotent matrix in $GL_N(R)$. Have you checked (P) for $\mathbb Z[i]$? $\endgroup$ – Will Sawin Oct 24 '13 at 6:26
  • $\begingroup$ I think that A=R implies (P) for Euclidean domains. Basically, once you have $GL_N(Z)$ and the diagonal matrices in $GL_N(R)$, you get all the elementary matrices -- the generators of the Chevalley group $SL_N(R)$. Diagonal matrices in $GL_N(R)$ give you all possible determinants in $R^\times$, so that does it. $\endgroup$ – Marty Oct 24 '13 at 6:31
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    $\begingroup$ Isn't Property P equivalent to the much simpler statement obtained by taking all $n_i$'s to be $1$, since that group is contained in all other groups obtained from partitions of $N$? $\endgroup$ – James Cranch Oct 24 '13 at 11:08
  • $\begingroup$ Yes -- it is equivalent ("there are definitely better ways of writing property (P)"). But I wanted a statement with an asymptotic feel to it, and one could imagine variations on property (P) by allowing only some partitions. Simplicity is probably better though... $\endgroup$ – Marty Oct 24 '13 at 23:38

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