Posted by Jason Polak on 03. May 2014 · Write a comment · Categories: modules · Tags:

Let \(R\) be a finite ring. The example we’ll have in mind at the end is the ring of \(2\times 2\) matrices over a finite field, and subrings. A. Kuku proved that \(K_i(R)\) for \(i\geq 1\) are finite abelian groups. Here, \(K_i(R)\) denotes Quillen’s \(i\)th \(K\)-group of the ring \(R\). In this post we will look at an example, slightly less simple than \(K_1\) of finite fields, showing that these groups can be arbitrarily large. Before we do this, let us briefly go over why this is true

But even before this, can you think of an example showing why this is false for \(i=0\)?

Of course, \(K_0(R)\) is often not finite even if \(R\) is: for instance, if \(R = \mathbb{F}_q\) then \(K_0(\mathbb{F}_q)\) is just the Grothendieck group of the monoid under direct sum of finite-dimensional vector spaces, and hence \(K_0(\mathbb{F}_q)\cong \mathbb{Z}\). Wait, is \(K_0(R)\) ever finite for any ring \(R\), besides the zero ring?

A Brief Sketch

Let us examine briefly why \(K_i(R)\) is finite for \(i \geq 1\), following Kuku’s book “Representation Theory and Higher Algebraic \(K\)-Theory”, filling in some of the quicker points in that presentation. The case \(i = 1\) is taken care of because \(K_1(R)\) is a quotient of \(R^\times\). Suppose \(i > 1\). Recall that we can define the higher \(K\)-groups as the homotopy groups \(\pi_i(\mathbf{B}\mathrm{GL}(R)^+)\) where \(\mathbf{B}(-)\) denotes the classifying space, \(\mathrm{GL}(R)\) is the infinite linear group and the \(+\) denotes Quillen’s plus construction.

Given any plus construction \(f:X\to X^+_P\) with respect to a perfect normal subgroup \(P\subseteq \pi_1(X)\), it is not hard to show that \(\mathbf{B} P^+\) has the same homotopy type as the universal cover for \(\mathbf{B} G^+\). This gives us a way to compute homotopy for \(i\geq 2\) because if \(\widetilde{X}\to X\) is a universal cover of \(X\) then \(\pi_i(\widetilde{X})\cong\pi_i(X)\) for all \(i\geq 2\).

So actually we can replace \(\mathbf{B}\mathrm{GL}(R)^+\) with \(\mathbf{B} E(R)^+\) where \(E(R)\) is the infinite elementary matrix group, defined as the direct limit \(\varinjlim E_n(R)\) where \(E_n(R)\) is the subgroup of \(\mathrm{GL}_n(R)\) generated by matrices all of whose diagonal entries are \(1\) and just one nonzero entry on some off-diagonal spot.

The bottom line is that we can compute \(K_i(R)\) via \(\pi_i(\mathbf{B} E(R)^+)\) for \(i\geq 2\) (this is true for any ring, and we have not used that \(R\) is finite yet). We will do this in a number of steps.

First, let us prove that \(H_i(\mathbf{B} E(R)^+)\) is finite. For this it suffices to prove that \(H_i(\mathbf{B} E(R))\) is finite, because the plus construction preserves homology. In order to do this, it suffices to prove that \(H_i(E(R),\mathbb{Z})\) is finite for \(i\geq 2\), where now we work with group homology, and this group in turn is isomorphic to \(H_i(E_k(R),\mathbb{Z})\) for some sufficiently large \(k\) by Suslin’s stability theorem. Of course \(H_1(E_k(R),\mathbb{Z})\) is finite, being the abelianisation of a finite group. We also claim \(H_i(E_k(R),\mathbb{Z})\) is finite. In fact, \(H_i(G,\mathbb{Z})\) is a finite abelian group for \(i\geq 1\) whenever \(G\) is any finite group because multiplication by the order \(|G|\) is the zero map, and the homology groups are finitely generated since, e.g. they can be calculated via the bar resolution (for example, if \(G\) is cyclic of order \(m\) then \(H_i(G,\mathbb{Z})\cong \mathbb{Z}/m\) for odd \(i\) and zero for nonzero even \(i\)).

So this shows that \(H_i(\mathbf{B} E(R)^+)\) is a finite abelian group. Of course, the same is true of \(H_i(\mathbf{B}\mathrm{GL}(R)^+)\). But, the reason we switched to \(\mathbf{B} E(R)^+\) is because \(\mathbf{B} E(R)^+\) is homotopy equivalent to a universal cover of \(\mathbf{B}\mathrm{GL}(R)^+\) and hence is simply connected. Now we apply a form of the Hurewicz theorem in topology (see Spanier, 9.6.15 for a general statement) that says that given such a simply connected space like \(X = \mathbf{B} E(R)^+\) the Hurewicz homomorphism \(\pi_i(X)\to H_i(X)\) has both finitely genenerated kernel and finitely generated cokernel. Hence we have a sequence \(0\to A\to \pi_1(X)\to C\to 0\) where \(A\) and \(C\) are finitely generated: \(A\) is the kernel of the Hurewicz homomorphism and in fact \(C\) is finite since \(C = \pi_1(X)/A\) injects into a finitely generated (and in our case finite!) group.

Going back to our specific case, we have shown that for \(i\geq 2\) we have \(K_i(R) = H_i(\mathbf{B} E(R)^+)\) is a finitely generated group. Now we can apply a well-known theorem of Cartan and Serre, which gives us an injection

\(\pi_i(\mathbf{B} E(R)^+)\otimes_{\mathbb{Z}}\mathbb{Q}\to H_i(\mathbf{B} E(R)^+)\otimes_{\mathbb{Z}}\mathbb{Q} = 0.\)

This holds for any \(H\)-space, and in particular the classifying space of a group. Hence for \(i \geq 2\) our \(K\)-group \(K_i(R) = \pi_i(\mathbf{B} E(R)^+)\) is a finitely generated torsion group, and hence finite.


If \(R\) is a finite ring then, how big can \(K_i(R)\) be for \(i > 0\)? Here is an example I thought of in the process of writing up some notes for myself. Before this, let us look at a warmup example: \(M_n(\mathbb{F}_q)\), the ring of \(n\times n\) matrices over a finite field. Actually, by Morita invariance \(K_1(M_n(\mathbb{F}_q))\) is the same as \(K_1(\mathbb{F}_q) \cong \mathbb{F}_q^\times\cong \mathbb{Z}/(q-1)\). So the size of \(K_1(\mathbb{F}_q)\) already goes to infinity as \(q\to\infty\).

Incidentally, more generally Quillen showed that \(K_{2i-1}(\mathbb{F}_q)\cong \mathbb{Z}/(q^i-1)\) for \(i\geq 1\), so Quillen’s computation already shows that \(K_i(\mathbb{F}_q)\) for \(i\not=0\) can be arbitrarily large. However, this computation is already quite complicated and certainly too much to do in one reasonably-sized blog post!

Let’s get back to \(M_n(\mathbb{F}_q)\): by the Morita invariance we see that no matter how big \(n\) is, its \(K\)-theory remains constant. However, there are subrings of \(M_n(\mathbb{F}_q)\) whose \(K_1\) groups are much bigger! Here is one example. Consider \(R = T_2(\mathbb{F}_q)\), the subring of \(M_2(\mathbb{F}_q)\) consisting of upper triangular matrices with entries in \(\mathbb{F}_q\), and consider the two-sided ideal \(I\) of \(R\) consisting of matrices of the form \(\left(\begin{smallmatrix} 0 & a\\ 0 & 0\end{smallmatrix}\right)\) where \(a\in \mathbb{F}_q\).

For any ring \(R\) and ideal \(I\) there exists an exact sequence
\(K_1(R)\to K_1(R/I)\to K_0(R,I)\to K_0(R)\to K_0(R/I).\)
Here \(K_0(R,I)\) is a relative \(K\)-group that isn’t important for the discussion at the moment. The important thing in this case is that we have an extra special piece of information: namely, that \(R\to R/I\) is a split surjection, which implies (nontrivial!) that \(K_0(R,I)\to K_0(R)\) is injective, so \(K_1(R)\to K_1(R/I)\) is surjective. Moreover (exercise) we see that \(R/I\cong \mathbb{F}_q^2\), so that \(K_1(R)\) has cardinality at least \((q-1)^2\), which is bigger than \(K_1(M_n(\mathbb{F}_q))\). Of course, this also implies that the ring \(T_2(\mathbb{F}_q)\) of \(\mathbb{F}_q\)-upper triangular matrices is not Morita equivalent to the full matrix ring. Of course for split surjections it also turns out that \(K_0(R)\to K_0(R/I)\) is also surjective so we also could have concluded that \(T_2(\mathbb{F}_q)\) is not Morita equivalent to \(\mathbb{F}_q\) by a slightly easier \(K_0\)-calculation.

Leave a Reply

Your email address will not be published. Required fields are marked *