[proofplan] We use the Dieudonne determinant in the convention where its target is the abelianization $D^\times/[D^\times,D^\times]$. First we check that elementary matrices have determinant $1$, so the determinant descends from $GL(D)$ to $K_1(D)$. Then Gaussian elimination over the division ring reduces every stable invertible matrix, modulo elementary matrices, to a diagonal matrix, and the abelian nature of the [stable Whitehead quotient](/theorems/8654) identifies a diagonal matrix with the product of its diagonal entries modulo multiplicative commutators. This gives an inverse map from $D^\times/[D^\times,D^\times]$ to $K_1(D)$ and proves that the descended Dieudonne determinant is an isomorphism. [/proofplan]

[step:Descend the Dieudonne determinant through the elementary subgroup]Let \begin{align*} \pi:D^\times\to D^\times/[D^\times,D^\times] \end{align*} denote the quotient homomorphism from the unit group of $D$ to its abelianization. For $n\ge 1$, let $I_n$ denote the $n\times n$ identity matrix over $D$, and for $1\le i,j\le n$, let $E_{ij}\in M_n(D)$ denote the matrix whose $(i,j)$-entry is $1_D$ and whose other entries are $0_D$. We use the standard Dieudonne determinant convention: for each $n\ge 1$ there is a [group homomorphism](/page/Group%20Homomorphism) \begin{align*} \delta_n:GL_n(D)\to D^\times/[D^\times,D^\times] \end{align*} such that $\delta_n(e_{ij}(r))=1$ for every elementary transvection $e_{ij}(r)=I_n+rE_{ij}$ with $i\ne j$, and \begin{align*} \delta_n(\operatorname{diag}(a_1,\dots,a_n))=\pi(a_1\cdots a_n) \end{align*} for all $a_1,\dots,a_n\in D^\times$. The product $a_1\cdots a_n$ is interpreted in the displayed order, and its class in the abelianization is independent of changes by multiplicative commutators. The compatibility \begin{align*} \delta_{n+1}(\operatorname{diag}(A,1))=\delta_n(A) \end{align*} defines a stable homomorphism \begin{align*} \delta:GL(D)\to D^\times/[D^\times,D^\times]. \end{align*} Since every generator of $E(D)$ is an elementary transvection and each such generator maps to $1$, we have $E(D)\subseteq \ker \delta$. Therefore $\delta$ factors uniquely through the quotient map $q:GL(D)\to K_1(D)=GL(D)/E(D)$, giving a homomorphism \begin{align*} \det_D:K_1(D)\to D^\times/[D^\times,D^\times] \end{align*} satisfying $\det_D(q(A))=\delta(A)$ for every $A\in GL(D)$.[/step]

[guided]The first point is to be explicit about the determinant convention. Since $D$ need not be commutative, the ordinary product of diagonal entries is not a well-defined element of a commutative target unless we pass to the abelianization. Let \begin{align*} \pi:D^\times\to D^\times/[D^\times,D^\times] \end{align*} be the quotient map. The Dieudonne determinant is the homomorphism \begin{align*} \delta_n:GL_n(D)\to D^\times/[D^\times,D^\times] \end{align*} characterized by elementary invariance and diagonal normalization: \begin{align*} \delta_n(e_{ij}(r))=1 \end{align*} for each elementary transvection $e_{ij}(r)=I_n+rE_{ij}$, and \begin{align*} \delta_n(\operatorname{diag}(a_1,\dots,a_n))=\pi(a_1\cdots a_n) \end{align*} for diagonal matrices with $a_1,\dots,a_n\in D^\times$. The order of multiplication is harmless only after applying $\pi$, because changing the order changes the product by elements of the multiplicative commutator subgroup. The stabilization maps send $A\in GL_n(D)$ to $\operatorname{diag}(A,1)\in GL_{n+1}(D)$. The Dieudonne determinant is stable under this operation: \begin{align*} \delta_{n+1}(\operatorname{diag}(A,1))=\delta_n(A). \end{align*} Hence the maps $\delta_n$ assemble into a single homomorphism \begin{align*} \delta:GL(D)\to D^\times/[D^\times,D^\times]. \end{align*} Now we check the quotient condition. The group $E(D)$ is generated by stable elementary transvections. Each generator has Dieudonne determinant $1$, so every finite product of such generators also has Dieudonne determinant $1$. Thus \begin{align*} E(D)\subseteq \ker\delta. \end{align*} By the [universal property of quotient groups](/theorems/2701), $\delta$ factors uniquely through \begin{align*} K_1(D)=GL(D)/E(D). \end{align*} This gives the descended map \begin{align*} \det_D:K_1(D)\to D^\times/[D^\times,D^\times]. \end{align*}[/guided]

[step:Reduce every invertible matrix to a diagonal representative modulo elementary matrices]Let $A\in GL_n(D)$. We prove that the class of $A$ in $K_1(D)$ is represented by a diagonal matrix. Because $A$ is invertible, its first column has a nonzero entry: if the first column were zero, then $A e_1=0$ for the first standard column vector $e_1\in D^n$, contradicting injectivity of left multiplication by $A$ on the right $D$-module $D^n$. If $A_{11}=0$, choose an index $i>1$ with $A_{i1}\ne 0$ and left-multiply by the elementary transvection $I_n+E_{1i}$. This replaces the first row by the sum of the first and $i$-th rows, so the new $(1,1)$-entry is $A_{i1}\ne0$. Thus, using only elementary row addition, we obtain a pivot $a\in D^\times$ in the $(1,1)$-position. For each $i>1$, left-multiply by $I_n-A_{i1}a^{-1}E_{i1}$. This replaces row $i$ by row $i$ minus $A_{i1}a^{-1}$ times row $1$ on the left, and the new $(i,1)$-entry is $A_{i1}-A_{i1}a^{-1}a=0$. For each $j>1$, right-multiply by $I_n-a^{-1}A_{1j}E_{1j}$. This replaces column $j$ by column $j$ minus column $1$ multiplied on the right by $a^{-1}A_{1j}$, and the new $(1,j)$-entry is $A_{1j}-aa^{-1}A_{1j}=0$. All these matrices are elementary transvections, so the class in $K_1(D)$ is unchanged. After these operations, $A$ is equivalent modulo $E(D)$ to a block diagonal matrix \begin{align*} \operatorname{diag}(a,A_1) \end{align*} with $a\in D^\times$ and $A_1\in M_{n-1}(D)$. The block diagonal matrix is invertible because it is obtained from $A$ by multiplying by invertible elementary transvections. Since $a\in D^\times$, its inverse block is $a^{-1}$, and the inverse of $\operatorname{diag}(a,A_1)$ has lower-right block an inverse for $A_1$. Hence $A_1\in GL_{n-1}(D)$. Repeating the argument by induction on $n$ gives elements $a_1,\dots,a_n\in D^\times$ such that \begin{align*} q(A)=q(\operatorname{diag}(a_1,\dots,a_n)) \end{align*} in $K_1(D)$.[/step]

[guided]We must be careful about the side on which coefficients multiply, because $D$ is not assumed commutative. A left multiplication by $I_n+\lambda E_{i1}$ adds $\lambda$ times row $1$ on the left to row $i$, while a right multiplication by $I_n+\mu E_{1j}$ adds column $1$ multiplied on the right by $\mu$ to column $j$. Since $A\in GL_n(D)$, the first column cannot be zero. Indeed, view $D^n$ as a right $D$-module of column vectors and let $A$ act by left multiplication. If the first column were zero, then $A e_1=0$ for the first standard column vector $e_1\in D^n$, contradicting injectivity of this right $D$-linear automorphism. Choose an index with a nonzero first-column entry. If needed, adding that row to the first row by an elementary transvection produces a nonzero pivot $a\in D^\times$ in the $(1,1)$-position. For $i>1$, left multiplication by $I_n-A_{i1}a^{-1}E_{i1}$ changes the $(i,1)$-entry to \begin{align*} A_{i1}-A_{i1}a^{-1}a=0. \end{align*} The coefficient is on the left of row $1$, so the order $A_{i1}a^{-1}$ is the correct one. For $j>1$, right multiplication by $I_n-a^{-1}A_{1j}E_{1j}$ changes the $(1,j)$-entry to \begin{align*} A_{1j}-aa^{-1}A_{1j}=0. \end{align*} Here the coefficient multiplies column $1$ on the right, so the order $a^{-1}A_{1j}$ is the correct one. All matrices used in these row and column operations are elementary transvections. Therefore multiplication by them does not change the image under the quotient map $q:GL(D)\to K_1(D)$. The resulting matrix has the form $\operatorname{diag}(a,A_1)$ with $A_1\in M_{n-1}(D)$. It is invertible because it is obtained from the invertible matrix $A$ by multiplying by invertible elementary transvections. Since $a$ is a unit, the lower-right block of the inverse matrix gives an inverse for $A_1$, so $A_1\in GL_{n-1}(D)$. Induction on $n$ now reduces $A_1$ to a diagonal matrix modulo elementary transvections, and hence $A$ has the same $K_1(D)$-class as some $\operatorname{diag}(a_1,\dots,a_n)$ with every $a_i\in D^\times$.[/guided]

[step:Compress diagonal representatives to one unit]Let $a_1,\dots,a_n\in D^\times$. Define \begin{align*} d=\pi(a_1\cdots a_n)\in D^\times/[D^\times,D^\times]. \end{align*} Choose any representative $a\in D^\times$ with $\pi(a)=d$. The stable Whitehead quotient theorem [citetheorem:8654] applies because $D$ is a unital ring, $GL(D)$ is the stable general linear group under the stabilization maps $A\mapsto\operatorname{diag}(A,1)$, and $E(D)$ is the stable elementary subgroup generated by the elementary transvections. Hence $K_1(D)=GL(D)/E(D)$ is abelian. Therefore conjugate elements of $GL(D)$ have the same image in $K_1(D)$. For a permutation $\sigma\in S_n$, let $P_\sigma\in GL_n(D)$ denote the permutation matrix with $(P_\sigma)_{i,\sigma(i)}=1_D$ and all other entries equal to $0_D$. Conjugating a diagonal matrix by $P_\sigma$ permutes its diagonal entries, so diagonal entries may be reordered without changing their $K_1$-class. Also, \begin{align*} \operatorname{diag}(a,1)\operatorname{diag}(b,1)=\operatorname{diag}(ab,1) \end{align*} inside $GL_2(D)$ for every $a,b\in D^\times$. Hence the class of a diagonal matrix depends only on the product of its diagonal entries modulo $[D^\times,D^\times]$: \begin{align*} q(\operatorname{diag}(a_1,\dots,a_n))=q(\operatorname{diag}(a,1,\dots,1)). \end{align*} Indeed, if $a_1\cdots a_n=a c$ with $c\in [D^\times,D^\times]$, then $q(\operatorname{diag}(c,1))=1$ because every multiplicative commutator $xyx^{-1}y^{-1}$ appears as the commutator of $\operatorname{diag}(x,1)$ and $\operatorname{diag}(y,1)$ in $GL_2(D)$, and commutators vanish in the abelian quotient $K_1(D)$.[/step]

[guided]The goal is to show that a diagonal matrix carries only one invariant: the product of its diagonal entries in the abelianized unit group. The noncommutative issue is that $a_1a_2$ and $a_2a_1$ need not be equal in $D^\times$, so we must always interpret products after applying \begin{align*} \pi:D^\times\to D^\times/[D^\times,D^\times]. \end{align*} First, the stable Whitehead quotient theorem [citetheorem:8654] says that the stable quotient $GL(R)/E(R)$ is an abelian group for every unital ring $R$. We apply it with $R=D$. The present $GL(D)$ is defined using the same stabilization maps $A\mapsto\operatorname{diag}(A,1)$, and the present $E(D)$ is generated by the same elementary transvections, so the theorem gives that $K_1(D)=GL(D)/E(D)$ is abelian. Thus if $g,h\in GL(D)$, then the images of $g h g^{-1}$ and $h$ in $K_1(D)$ are equal, because \begin{align*} q(g h g^{-1})=q(g)q(h)q(g)^{-1}=q(h) \end{align*} in an abelian group. For a permutation $\sigma\in S_n$, let $P_\sigma\in GL_n(D)$ be the permutation matrix with $(P_\sigma)_{i,\sigma(i)}=1_D$ and all other entries equal to $0_D$. Conjugating a diagonal matrix by $P_\sigma$ only reorders its diagonal entries. Since conjugate elements have the same image in $K_1(D)$, diagonal entries may be permuted without changing the $K_1$-class of the diagonal matrix. Second, two diagonal entries can be multiplied into one slot. For $a,b\in D^\times$, we have the literal matrix identity \begin{align*} \operatorname{diag}(a,1)\operatorname{diag}(b,1)=\operatorname{diag}(ab,1). \end{align*} Since $\operatorname{diag}(1,b)$ is conjugate to $\operatorname{diag}(b,1)$ by the transposition matrix, the previous paragraph gives \begin{align*} q(\operatorname{diag}(1,b))=q(\operatorname{diag}(b,1)). \end{align*} Therefore \begin{align*} q(\operatorname{diag}(a,b))=q(\operatorname{diag}(a,1))q(\operatorname{diag}(1,b))=q(\operatorname{diag}(ab,1)). \end{align*} Repeating this compression finitely many times gives \begin{align*} q(\operatorname{diag}(a_1,\dots,a_n))=q(\operatorname{diag}(a_1\cdots a_n,1,\dots,1)). \end{align*} Finally, replacing the product $a_1\cdots a_n$ by another representative of the same class in $D^\times/[D^\times,D^\times]$ does not change the $K_1$-class. If $c=xyx^{-1}y^{-1}$ is a multiplicative commutator in $D^\times$, then \begin{align*} \operatorname{diag}(c,1)=[\operatorname{diag}(x,1),\operatorname{diag}(y,1)] \end{align*} as an element of $GL_2(D)$. Its image in the abelian group $K_1(D)$ is therefore the identity. Since every element of $[D^\times,D^\times]$ is a finite product of such commutators, $\operatorname{diag}(c,1)$ has class equal to the identity element of $K_1(D)$ for every $c\in [D^\times,D^\times]$.[/guided]

[step:Construct the inverse from the abelianized unit group]Define \begin{align*} s:D^\times/[D^\times,D^\times]\to K_1(D) \end{align*} by \begin{align*} s(\pi(a))=q(\operatorname{diag}(a)) \end{align*} for $a\in D^\times$, where $\operatorname{diag}(a)$ is viewed as an element of $GL_1(D)\subset GL(D)$. This map is well-defined. If $\pi(a)=\pi(b)$, then $b^{-1}a\in [D^\times,D^\times]$. By the commutator argument in the previous step, \begin{align*} q(\operatorname{diag}(b^{-1}a))=1. \end{align*} Since multiplication in $GL_1(D)$ gives $\operatorname{diag}(b)\operatorname{diag}(b^{-1}a)=\operatorname{diag}(a)$, applying $q$ gives \begin{align*} q(\operatorname{diag}(a))=q(\operatorname{diag}(b)). \end{align*} The same matrix identity \begin{align*} \operatorname{diag}(a)\operatorname{diag}(b)=\operatorname{diag}(ab) \end{align*} shows that $s$ is a group homomorphism.[/step]

[guided]The definition of $s$ uses a representative $a\in D^\times$ of the coset $\pi(a)$, so the only point to check is independence of that representative. Suppose $a,b\in D^\times$ satisfy $\pi(a)=\pi(b)$. Then $b^{-1}a\in [D^\times,D^\times]$, because $\pi(b^{-1}a)=\pi(b)^{-1}\pi(a)=1$. The previous step proved that every element of the multiplicative commutator subgroup gives the identity class after embedding into $GL_1(D)\subset GL(D)$, so \begin{align*} q(\operatorname{diag}(b^{-1}a))=1. \end{align*} Multiplying in $GL_1(D)$ in the displayed order gives $\operatorname{diag}(b)\operatorname{diag}(b^{-1}a)=\operatorname{diag}(a)$. Applying the quotient homomorphism $q$ therefore gives \begin{align*} q(\operatorname{diag}(a))=q(\operatorname{diag}(b)). \end{align*} Thus $s$ is well-defined. To check the homomorphism property, take $a,b\in D^\times$. In $GL_1(D)$ we have \begin{align*} \operatorname{diag}(a)\operatorname{diag}(b)=\operatorname{diag}(ab). \end{align*} Applying the quotient homomorphism $q$ gives $s(\pi(a))s(\pi(b))=s(\pi(ab))$, so $s$ is a group homomorphism.[/guided]

[step:Verify the two maps are inverse isomorphisms]For $a\in D^\times$, \begin{align*} (\det_D\circ s)(\pi(a))=\det_D(q(\operatorname{diag}(a)))=\pi(a). \end{align*} Thus $\det_D\circ s=\operatorname{id}_{D^\times/[D^\times,D^\times]}$. Conversely, let $\xi\in K_1(D)$. Choose $A\in GL_n(D)$ with $q(A)=\xi$. By the diagonal reduction and compression steps, there is $a\in D^\times$ such that \begin{align*} q(A)=q(\operatorname{diag}(a)). \end{align*} Then \begin{align*} (s\circ\det_D)(\xi)=s(\det_D(q(A)))=s(\pi(a))=q(\operatorname{diag}(a))=\xi. \end{align*} Therefore $s\circ\det_D=\operatorname{id}_{K_1(D)}$. The descended Dieudonne determinant is a group isomorphism \begin{align*} K_1(D)\cong D^\times/[D^\times,D^\times]. \end{align*}[/step]

[guided]We verify both compositions. For $a\in D^\times$, the map $s$ sends the coset $\pi(a)$ to the $K_1(D)$-class of the one-by-one matrix $\operatorname{diag}(a)$. The Dieudonne determinant is normalized on diagonal matrices, so \begin{align*} (\det_D\circ s)(\pi(a))=\det_D(q(\operatorname{diag}(a)))=\pi(a). \end{align*} Hence $\det_D\circ s$ is the identity on $D^\times/[D^\times,D^\times]$. For the other composition, let $\xi\in K_1(D)$. Choose $A\in GL_n(D)$ such that $q(A)=\xi$. The diagonal reduction step gives a diagonal representative for $q(A)$, and the compression step replaces that diagonal representative by a one-by-one diagonal matrix without changing its $K_1(D)$-class. Thus there exists $a\in D^\times$ such that \begin{align*} q(A)=q(\operatorname{diag}(a)). \end{align*} Using this representative, we compute \begin{align*} (s\circ\det_D)(\xi)=s(\det_D(q(A)))=s(\pi(a))=q(\operatorname{diag}(a))=\xi. \end{align*} Therefore $s\circ\det_D$ is the identity on $K_1(D)$, and the descended Dieudonne determinant is an isomorphism.[/guided]