K1 of a field

Example

Let $F$ be a field and $F^\times$ its multiplicative group. Then the determinant induces an isomorphism $$\det \colon K_1(F) \xrightarrow{\ \cong\ } F^\times.$$ Check of surjectivity: for $u \in F^\times$, the matrix $\mathrm{diag}(u, 1, \dots, 1)$ has determinant $u$. Check of injectivity: let $A \in GL_n(F)$ have determinant $1$. Gaussian elimination using only the operations "add a multiple of one row to another" brings $A$ to $\mathrm{diag}(1, \dots, 1, d)$. A zero pivot is avoided by first adding another row with a nonzero entry in that column. Then $d = \det A = 1$, so $A$ is a product of elementary matrices and lies in the elementary subgroup $E_n(F)$. Hence the kernel of $\det$ on $K_1(F) = GL(F)/E(F)$ is trivial, that is, SK1 of a field is $0$. For example, $K_1(\Q) \cong \Q^\times$ and $K_1(\F_q) \cong \F_q^\times$, cyclic of order $q - 1$.