[proofplan] We use the standard presentation $K_1(F)=GL(F)/E(F)$, where $GL(F)$ is the stable general linear group and $E(F)$ is the stable elementary subgroup. For each $n$, let $GL_n(F)$ denote the group of invertible $n\times n$ matrices over $F$, write $SL_n(F)=\ker(\det_n:GL_n(F)\to F^\times)$, and let $E_n(F)\le GL_n(F)$ be the subgroup generated by elementary transvections in size $n$. The stable determinant is well-defined on $K_1(F)$ because elementary matrices have determinant $1$, and the inclusion $F^\times=GL_1(F)\subset GL(F)$ gives a section. The only point is injectivity: an element with determinant $1$ is represented by a matrix in $SL_n(F)$, and over a field every determinant-one matrix is a product of elementary matrices. [/proofplan]

[step:Define the stable determinant on $K_1(F)$] Let $GL(F)=\varinjlim_n GL_n(F)$ under the stabilization maps $A\mapsto \operatorname{diag}(A,1)$, and let $E(F)\trianglelefteq GL(F)$ be the stable elementary subgroup generated by all elementary transvections $e_{ij}(a)$ with $a\in F$ and $i\neq j$. By the standard definition, \begin{align*} K_1(F)=GL(F)/E(F). \end{align*} For each $n\geq 1$, let \begin{align*} \det_n:GL_n(F)&\to F^\times \end{align*} be the determinant homomorphism. Since \begin{align*} \det_{n+1}(\operatorname{diag}(A,1))=\det_n(A) \end{align*} for every $A\in GL_n(F)$, the maps $\det_n$ induce a homomorphism \begin{align*} \det:GL(F)&\to F^\times. \end{align*} Every elementary transvection has determinant $1$, so $E(F)\subseteq \ker(\det)$. Hence $\det$ descends to a homomorphism \begin{align*} \overline{\det}:K_1(F)&\to F^\times. \end{align*} This is the stable determinant map in the special case of the field $F$. [/step]

[step:Construct the inverse candidate from one-dimensional matrices] Define \begin{align*} s:F^\times&\to K_1(F) \end{align*} by sending $u\in F^\times$ to the class in $K_1(F)$ of the matrix $(u)\in GL_1(F)\subset GL(F)$. This map is a [group homomorphism](/page/Group%20Homomorphism) because multiplication in $GL_1(F)$ is multiplication in $F^\times$. Moreover, \begin{align*} (\overline{\det}\circ s)(u)=u \end{align*} for every $u\in F^\times$. Thus $\overline{\det}$ is surjective, and $s$ is its section. [/step]

[step:Show that determinant-one matrices over a field are elementary]We prove that every matrix in $SL_n(F)$ maps to the identity element of $K_1(F)$. For $n=1$, the group $SL_1(F)$ consists only of the $1\times 1$ identity matrix. Assume $n\geq 2$, and let $A\in SL_n(F)$. Gaussian elimination over the field $F$ uses elementary row additions and signed row interchanges to reduce $A$ to a diagonal matrix. The signed interchange of two adjacent rows is elementary because the corresponding $2\times 2$ block, with first row $(0,1)$ and second row $(-1,0)$, factors as \begin{align*} e_{12}(1)e_{21}(-1)e_{12}(1). \end{align*} Thus all pivoting operations used in Gaussian elimination are products of elementary matrices, after replacing a row interchange by the signed interchange above and absorbing the sign into the resulting diagonal entries. Consequently there exists a product $U\in E_n(F)$ such that \begin{align*} UA=\operatorname{diag}(d_1,\dots,d_n) \end{align*} with each $d_i\in F^\times$. Since $\det(U)=1$ and $\det(A)=1$, we have \begin{align*} d_1\cdots d_n=1. \end{align*} For each $u\in F^\times$, the determinant-one diagonal block is elementary: \begin{align*} \operatorname{diag}(u,u^{-1}) = e_{12}(u)e_{21}(-u^{-1})e_{12}(u)e_{12}(-1)e_{21}(1)e_{12}(-1). \end{align*} Placing these $2\times 2$ factorizations into adjacent diagonal positions shows that \begin{align*} \operatorname{diag}(d_1,\dots,d_n)\in E_n(F), \end{align*} because \begin{align*} \operatorname{diag}(d_1,\dots,d_n)=\prod_{i=1}^{n-1}\operatorname{diag}(1,\dots,1,d_1\cdots d_i,(d_1\cdots d_i)^{-1},1,\dots,1) \end{align*} as a product of elementary determinant-one diagonal blocks in the stable elementary group. Since $UA\in E(F)$ and $U\in E(F)$, it follows that $A\in E(F)$.[/step]

[guided]The kernel of the determinant consists of stable classes represented by determinant-one matrices, so we need a concrete reason why such matrices disappear in $K_1(F)=GL(F)/E(F)$. The reason is the classical elementary generation result over a field: determinant-one matrices are generated by elementary transvections. Let $A\in SL_n(F)$. If $n=1$, then $A=(1)$, so there is nothing to prove. Suppose $n\geq 2$. Because $F$ is a field, Gaussian elimination can choose a nonzero pivot in each nonzero column and use elementary row additions to clear entries below and above the pivot. If the nonzero pivot is not already in the desired row, we move it there by a signed row interchange. This signed interchange is allowed inside the elementary group because the $2\times 2$ block with first row $(0,1)$ and second row $(-1,0)$ is \begin{align*} e_{12}(1)e_{21}(-1)e_{12}(1), \end{align*} a product of elementary matrices. Thus each row operation used in the elimination process is multiplication on the left by an elementary matrix. After finitely many such operations, there is a product $U\in E_n(F)$ such that \begin{align*} UA=\operatorname{diag}(d_1,\dots,d_n) \end{align*} for some $d_1,\dots,d_n\in F^\times$. Taking determinants gives \begin{align*} d_1\cdots d_n=\det(U)\det(A)=1. \end{align*} Here $\det(U)=1$ because $U$ is a product of elementary matrices, and $\det(A)=1$ because $A\in SL_n(F)$. It remains to see that every diagonal matrix with determinant $1$ is also elementary. For $u\in F^\times$, the explicit factorization \begin{align*} \operatorname{diag}(u,u^{-1}) = e_{12}(u)e_{21}(-u^{-1})e_{12}(u)e_{12}(-1)e_{21}(1)e_{12}(-1) \end{align*} expresses the basic determinant-one diagonal block as a product of elementary matrices. Since $d_1\cdots d_n=1$, the diagonal matrix $\operatorname{diag}(d_1,\dots,d_n)$ can be written as a product of such adjacent blocks, placed along the diagonal in the stable group. Hence $\operatorname{diag}(d_1,\dots,d_n)\in E(F)$. Now $UA\in E(F)$ and $U\in E(F)$. Since $E(F)$ is a subgroup of $GL(F)$, multiplying by $U^{-1}$ gives $A\in E(F)$. Therefore every determinant-one representative is trivial in $K_1(F)$.[/guided]

[step:Identify the determinant map as an isomorphism] Let $\alpha\in K_1(F)$ satisfy $\overline{\det}(\alpha)=1$. Choose a representative $A\in GL_n(F)$ for $\alpha$. Then $\det_n(A)=1$, so $A\in SL_n(F)$. By the previous step, $A\in E(F)$, and therefore $\alpha$ is the identity element of $K_1(F)$. Hence $\overline{\det}$ is injective. Since $\overline{\det}$ is both surjective and injective, it is an isomorphism: \begin{align*} K_1(F)\cong F^\times. \end{align*} [/step]