$K_1(\mathbb{Z}_4)$ and $K_1(\mathbb{Z_4}[t])$

Question

I am an amateur in $K$-theory, I have just started reading from "The K-book" by Charles Weibel. I have only read the definition of $K_1$ which is stated as a quotient of $GL(R)$. The union of the sequence $R^{ \times} = GL_1(R) \subset GL_2(R) \subset ...\subset GL_n(R) \subset GL_{n+1}(R) \subset..$ is called $GL(R)$ and $K_1(R)$ is defined to be $GL(R)/[GL(R),GL(R)].$

In one of the examples he has proved that if $R$ is any commutative ring then $K_1(R) \cong R^{\times} \oplus SK_1(R)$. Where $SK_1(R)$ is denoted as the kernel of a surjective map from $K_1(R) \rightarrow R^{\times}$ which is induced from the determinant map $GL(R) \rightarrow R^{\times}$. I understand the splitting. But when I am trying to compute $SK_1(R)$ for $R = \mathbb{Z}_4$ or $R = \mathbb{Z}_4[t]$ explicitly I am at a loss.

If you could please guide me to the right direction I would be grateful.

Answer 1

As mentioned in Example 1.3.5, the group $SK_1$ vanishes over euclidean domains. The reason is that then you can transform any invertible matrix into diagonal form using elementary row and column operations, by the Smith normal form algorithm. (Note that we are using more than just the abstract elementary divisor theorem for PIDs, we really need that the transformation matrices are built from elementary matrices. In fact, there are examples of PIDs with nonzero SK_1.)

Example 1.3.5 is stated only for euclidean domains, but you don't need to be without zero-divisors, you just need a euclidean algorithm. For example, the same argument also applies to local rings such as $\mathbb{Z}/4$, and you can set up a similar argument for $\mathbb{Z}/4[t]$.