[proofplan] We identify the commutative monoid of isomorphism classes of finitely generated projective left $D$-modules with the additive monoid $\mathbb N_0=\{0\}\cup\mathbb N$ by sending a module to its left dimension over $D$. The only module-theoretic input is the basis argument for finite-dimensional left vector spaces over a division ring, which gives $P\cong D^m$ for a unique integer $m\ge 0$. Since $K_0(D)$ is the group completion of this monoid, and the group completion of $\mathbb N_0$ is $\mathbb Z$, the desired isomorphism follows. [/proofplan]

[step:Show that every finitely generated left $D$-module is finite free]Let $P$ be a finitely generated left $D$-module. Since $D$ is a division ring, $P$ is a left [vector space](/page/Vector%20Space) over $D$. Choose a finite generating set $S=\{p_1,\dots,p_n\}$ for $P$ as a left $D$-module, where $n\in\mathbb N$ if $P\ne 0$ and $S=\varnothing$ if $P=0$. Among the linearly independent subsets of $S$, choose one of maximal cardinality, and denote it by $B=\{b_1,\dots,b_m\}$, with $m\in\mathbb N_0$. We claim that $B$ spans $P$. If $s\in S\setminus B$, then maximality of $B$ implies that $B\cup\{s\}$ is linearly dependent. Therefore there exist elements $a_1,\dots,a_m,c\in D$, not all zero, such that \begin{align*} a_1b_1+\cdots+a_mb_m+cs=0. \end{align*} The coefficient $c$ is nonzero, because otherwise the displayed relation would be a nontrivial linear relation among the elements of $B$. Since $D$ is a division ring, $c^{-1}$ exists, and hence \begin{align*} s=-c^{-1}a_1b_1-\cdots-c^{-1}a_mb_m. \end{align*} Thus every element of $S$ lies in the left $D$-span of $B$, and since $S$ generates $P$, the set $B$ is a basis of $P$. Define the left $D$-[linear map](/page/Linear%20Map) \begin{align*} \theta:D^m\to P \end{align*} by \begin{align*} \theta(d_1,\dots,d_m)=d_1b_1+\cdots+d_mb_m. \end{align*} Because $B$ spans $P$, the map $\theta$ is surjective. Because $B$ is linearly independent, the kernel of $\theta$ is zero. Hence $\theta$ is an isomorphism of left $D$-modules, so $P\cong D^m$.[/step]

[guided]We need to prove freeness without using commutativity of $D$. The correct viewpoint is that a left $D$-module is a left vector space over the division ring $D$, so scalar coefficients always multiply basis vectors on the left. Let $P$ be a finitely generated left $D$-module. Choose a finite generating set $S=\{p_1,\dots,p_n\}$ if $P\ne 0$, and choose $S=\varnothing$ if $P=0$. Since $S$ is finite, there is a linearly independent subset of $S$ of maximal cardinality. Denote such a subset by $B=\{b_1,\dots,b_m\}$, where $m\in\mathbb N_0$. The key point is to show that maximal [linear independence](/page/Linear%20Independence) inside the finite generating set forces spanning. Let $s\in S\setminus B$. Since $B$ was chosen maximal among linearly independent subsets of $S$, the set $B\cup\{s\}$ is linearly dependent. Thus there are elements $a_1,\dots,a_m,c\in D$, not all zero, such that \begin{align*} a_1b_1+\cdots+a_mb_m+cs=0. \end{align*} The coefficient $c$ cannot be zero. If $c=0$, then the same displayed relation would be a nontrivial linear relation among $b_1,\dots,b_m$, contradicting the linear independence of $B$. Hence $c\ne 0$. Since $D$ is a division ring, the element $c$ has an inverse $c^{-1}\in D$. Multiplying the displayed relation on the left by $-c^{-1}$ gives \begin{align*} s=-c^{-1}a_1b_1-\cdots-c^{-1}a_mb_m. \end{align*} So every element of $S$ lies in the left $D$-span of $B$. Since $S$ generates $P$, the set $B$ spans all of $P$. Now define the left $D$-linear map \begin{align*} \theta:D^m\to P \end{align*} by \begin{align*} \theta(d_1,\dots,d_m)=d_1b_1+\cdots+d_mb_m. \end{align*} The map is surjective because $B$ spans $P$. Its kernel is zero because $B$ is linearly independent. Therefore $\theta$ is an isomorphism of left $D$-modules, and so $P\cong D^m$.[/guided]

[step:Identify isomorphism classes by dimension] Let $P$ be a finitely generated projective left $D$-module. By the previous step, there exists $m\in\mathbb N_0$ such that $P\cong D^m$. Define $\dim_D P=m$, the left dimension of $P$ over $D$. The integer $m$ is unique. Indeed, if $D^m\cong D^n$ as left $D$-modules, then a left $D$-module isomorphism sends a basis of $D^m$ to a basis of $D^n$, so the standard basis of $D^m$ has $m$ elements and its image is a basis of $D^n$ with $m$ elements. By the exchange argument for bases over a division ring, any two bases of the same left $D$-vector space have the same cardinality. Hence $m=n$. Therefore two finitely generated projective left $D$-modules are isomorphic if and only if they have the same left dimension. [/step]

[step:Identify the projective-class monoid with $\mathbb N_0$] Let $V(D)$ denote the commutative monoid of isomorphism classes of finitely generated projective left $D$-modules under direct sum. Define \begin{align*} \Phi:V(D)\to \mathbb N_0 \end{align*} by \begin{align*} \Phi([P])=\dim_D P. \end{align*} This map is well-defined by the uniqueness of dimension from the previous step. It is bijective because every class is represented by exactly one free module $D^m$, and $D^m$ is finitely generated projective for every $m\in\mathbb N_0$. For finitely generated projective left $D$-modules $P$ and $Q$, choose integers $m,n\in\mathbb N_0$ with $P\cong D^m$ and $Q\cong D^n$. Then \begin{align*} P\oplus Q\cong D^m\oplus D^n\cong D^{m+n}. \end{align*} Thus \begin{align*} \Phi([P]+[Q])=\Phi([P\oplus Q])=m+n=\Phi([P])+\Phi([Q]). \end{align*} Also $\Phi([0])=0$. Hence $\Phi$ is an isomorphism of commutative monoids \begin{align*} V(D)\cong \mathbb N_0. \end{align*} [/step]

[step:Pass from the monoid isomorphism to the Grothendieck group] By definition, $K_0(D)$ is the group completion of the commutative monoid $V(D)$. By [citetheorem:8633], group completion is characterized by its universal property and is unique up to unique isomorphism. The additive group $\mathbb Z$ is the group completion of the additive monoid $\mathbb N_0$. Indeed, the inclusion \begin{align*} \iota:\mathbb N_0\to \mathbb Z \end{align*} is a monoid homomorphism, and for every abelian group $A$ and every monoid homomorphism \begin{align*} f:\mathbb N_0\to A, \end{align*} there is a unique [group homomorphism](/page/Group%20Homomorphism) \begin{align*} \overline f:\mathbb Z\to A \end{align*} such that $\overline f\circ\iota=f$, namely the homomorphism determined by $\overline f(1)=f(1)$. Composing the monoid isomorphism $\Phi:V(D)\to\mathbb N_0$ with the group completion isomorphism gives a group isomorphism \begin{align*} K_0(D)\cong \mathbb Z. \end{align*} Under this isomorphism, the class $[P]\in K_0(D)$ maps to $\Phi([P])=\dim_D P$. This is the claimed identification. [/step]