A question on algebraic K-theory- show that $K_1(R)\cong R^\times$ if $R$ is a field.

Question

If $R$ is a ring we write $K_1(R)$ for the abelian group $K_1({\rm category\; of\; finitely\; generated\; projective\;R-modules})$. Swan's 'Algebraic K-Theory' defines $K_1$ in terms of the usual generators and relations.

If $R$ is a field, why is $K_1(R)=R^\times$? There is a map $(P,f)\mapsto {\rm det}(f)$ from $K_1(R)$ to $R^\times$ but how can we show it's an isomorphism?

Many thanks for your help.

Answer 1

Define $s_1:R^{\times}\to K_1(R)$ by $s_1(u)=(u)\oplus I_{\infty}$, an infinite diagonal matrix having $u$ and infinitely many ones on the main diagonal. Then $s_1$ is a section for $\det$, and moreover $s_1$ is surjective because every invertible matrix is equivalent with a matrix of the form $(u)\oplus I_{\infty}$, $u$ invertible.