Algebraic results using lower K-theory as a blackbox

There is an algebraic K-theory seminar at my school and we are struggling to find applications from areas other than topology. We'd like a nice statement like "If X then Y" whose proof makes unexpected use of $K_0(R)$ or $K_1(R)$, where X and Y are purely algebraic statements about rings (or something like that).

An example from group cohomology (not K-theory) is the following Sylow-style theorem, which is elementary to state.

Problem A. Suppose $G$ is a finite group of order $mn$ where $\gcd(m,n)=1$. Suppose $A$ is a normal abelian subgroup of order $m$. Prove that $G$ has a subgroup of order $n$, and show that any two such subgroups are conjugate.

Surprisingly a quick proof can be given with group cohomology. We're looking for something similar, except maybe for rings/fields, and instead of group cohomology it must use K-theory.

I've also seen the following result as a bonus problem in commutative algebra, and my prof told me the only solution he knew used K-theory. I don't know the solution though.

Problem B. Suppose $R\subseteq S$ is an integral extension of commutative domains. Suppose that $S$ is a PID. Prove that $R$ is a PID.

Not sure if the above statement is true as it stands, but I'm definitely pretty sure it's true if you assume that $R$ is a finitely-generated free $R$-module.

Anything else like Problem B above?

• For problem B, $\mathbb{R}[x,y]/(x^2+y^2-1)\subset\mathbb{C}[x,y]/(x^2+y^2-1)$ is a counterexample. I asked JB about this problem, and he said he wasn't sure what the right statement was supposed to be. – Julian Rosen Nov 1 '14 at 21:29
• My research gives some combinatorial results using algebraic K-theory of Grassmannians, but it doesn't sound like that's what you want. – Matt Samuel Nov 1 '14 at 21:55
• @Julian: in your example, is $S$ is f.g. free over $R$? – Ehsaan Nov 1 '14 at 22:10
• Yes, a basis for $S$ as an $R$-module is $1,i$. – Julian Rosen Nov 1 '14 at 22:57

Let $R$ be the ring $\mathbb{Z}[T]$ after inverting $T$ and $T^m-1$ for all $m\geq 1$. One can prove that $R$ is a principal ideal domain [Gray2]. Is it a Euclidean domain? It's hard to prove that Euclidean domains are principal ideal domains, so maybe this a job for $K$-theory after all.

Recall that for a commutative ring $SK_1(R)$ is the quotient $SL(R)/E(R)$ or equivalently, the kernel of map $K_1(R)\to R^\times$ induced by the determinant. It is easy to prove [Ros, Prop. 2.3.2] that if $R$ is a Euclidean domain, then $SK_1(R) = 0$.

However, for this particular $R$, Grayson in [Gray2] notes a remark of Stein and Franks that the results in [Gray1] imply that $SK_1(R) = SSR$ where $SSR$ is a certain abelian group that functions as the receptacle for obstructions to a diffeomorphism (of a certain type, see [Gray2] for details) of a compact smooth manifold to be isotopic to a Morse-Small diffeomorphism. Apparently, these are important in the theory of dynamical systems. However, $SSR$ can be defined completely algebraically.

Anyways, H.W. Lenstra Jr. proved that $SSF\cong \oplus_{n\geq 1}{\rm Cl}(\mathbb{Z}[\zeta_n,1/n])$ where $\zeta_n$ is a primitive $n$-th root of unity and ${\rm Cl}(-)$ denotes the class group. One can see this is nonzero by looking at $n$ a prime power, so that the class group of $\mathbb{Z}[\zeta_n,1/n]$ is the same as that of $\mathbb{Z}[\zeta_n]$. There are infinitely many of these that are nontrivial. For example ${\rm Cl}(\mathbb{Z}[\zeta_{23}]) \cong \mathbb{Z}/3$. You could also consider this another aspect of $K$-theory since for a Dedekind domain the class group is the reduced $0$-th $K$-group of $R$.

Thus, $SK_1(R)$ is nonzero and hence $R$ cannot be a Euclidean domain. So we get an example of a principal ideal domain that is not Euclidean domain.

• [Gray1] - Grayson, '$K$-Theory of Endomorphisms'
• [Gray2] - Grayson, '$SK_1$ of an Interesting Principal Ideal Domain'
• [Ros] - Rosenberg, 'Higher Algebraic $K$-Theory and Its Applications'

If you are prepared to consider $K_{0}$ of varieties (schemes) rather than rings, then the paper Algebraic $K$-theory and sums-of-squares formulas by Dugger and Isaksen fits the bill.

Let $F$ be a field. The sums-of-squares problems asks for which $r,s$ and $n$ there exist identities $$(x_{1}^{2} + \cdots + x_{r})(y_{1}^{2} + \cdots + y_{s}^{2}) = z_{1}^{2} + \cdots + z_{n}^{2}$$

where the $z_{i}$ are bilinear in $X=(x_{1},\dots,x_{r})$ and $Y=(y_{1},\dots,y_{s})$.

Dugger and Isaksen establish bounds on possible triples $[r,s,n]$ by considering algebraic vector bundles on the space $\mathbb{P}^{s-1} - V_{q}$, where $V_{q}$ is the subvariety of $\mathbb{P}^{s-1}$ given by $\sum_{i} x_{i}^{2} = 0$. They show that the existence of an $[r,s,n]$ formula implies the existence of a certain vector bundle over $\mathbb{P}^{s-1} - V_{q}$, which then places some constraints on the structure of $K_{0}(\mathbb{P}^{s-1} - V_{q})$.

In section 3 of their paper, which is essentially independent of the rest, they calculate $K_{0}(\mathbb{P}^{s-1} - V_{q})$, using fairly classical methods. This result is the black box they use to obtain bounds on the possible triples $[r,s,n]$.