We saw previously that $K_2(F) = 0$ for a finite field $F$, where $K_2$ is the second $K$-group of $F$. It may be helpful to refer to that post for the definitions of this functor.

I thought that it might be disappointing because we did all that work to compute the second $K$-group of a finite field and it turned out to be zero! But now we will see something a little more interesting: a ring with nonzero $K_2$.

## Matsumoto’s Theorem and the MSK Theorem

Let us recall Matsumoto’s theorem:

**Theorem (Matsumoto)**. Let $F$ be a field. The group $K_2(F)$ is the abelian group generated by the symbols $(x,y)$ for $x,y\in F^\times$ with the relations

- $(x,y)(x’,y) = (xx’,y)$ and $(x,y)(x,y’) = (x,yy’)$ for all $x,x’,y,y’\in F$
- $(x,1-x) = 1$ whenever $x\not=1$

Matsumoto’s theorem gave us a way to compute the second $K$-group of a field. But there is a similar theorem that allows us to compute the $K$-theory of a commutative local ring.

**Theorem (Maazen, Stienstra, and van der Kallen).**Let $R$ be a commutative local ring. Then $K_2(R)$ is the multiplicative abelian group generated by the symbols $(r,s)$ such that $1-rs\in R^\times$, and such that these relations are satisfied:

- $(r,s)(s,r) = 1$
- $(r,s)(r,t) = (r,s+t-rst)$
- $(r,st) = (rs,t)(tr,s)$

Let’s call this the MSK theorem. I want to stress here that even though I am using the notation $(r,s)$ for the generators in Matsumoto’s theorem and MSK, these two are *different presentations* and *one is not a special case of the other*.

By using relation #2, we see that $(r,0) = 1$ for any $r$. Note that $(r,0)$ is in fact a valid generator for any $r\in R$. By using relation #3, we also see that $(r,1) = 1$, when $(r,1)$ is a generatorâ€”that is, when $1-r$ is a unit. I’ll leave these as quick exercises.

Now, we can prove for example that $K_2(\Z/3) = 0$. Indeed, any generator of the form $(r,0)$ or $(r,1)$ is trivial. But $(2,2)$ is not even a generator because $1-2\cdot 2= 0$.

## What about $\Z/4$?

Now we come to the interesting part. The ring $\Z/4$ is a commutative local ring with maximal ideal $(2)$. Unlike Matsumoto’s theorem, which is limited to fields, the MSK theorem gives us a presentation of $K_2(\Z/4)$. Again, let us figure out the possible nontrivial generators. The only possibilities are

$$(2,2),(2,3),(3,2).$$ Now, compute with relation schema #2 that $(2,3)(2,3) = 1$. Therefore, $(2,3) = (3,2)$ by #1. Also, $(2,3)(2,2) = 1$ so again, $(2,3) = (2,2)$. Hence there is only a single generator we need to consider, $(2,2)$ and $(2,2)^2 = 1$. Therefore the only possibilities for $K_2(\Z/4)$ are $\Z/2$ or the trivial group. Let’s write $\Z/2$ multiplicatively as $\{\pm 1\}$.

To decide which one we have, we observe (little exercise) that if we map $(2,2)\mapsto -1$, we get a well-defined group homomorphism $K_2(\Z/4)\to \Z/2$ because all the relations in the MSK theorem are satisfied by $-1\in\Z/2$: indeed, they just say $(-1)(-1) = 1$. Hence, $K_2(\Z/4) = \Z/2$.