[proofplan] We use the idempotent matrix model for $K_0$: finitely generated projective modules are represented by images of idempotent matrices, and equality in $K_0$ is detected after adding a common idempotent summand and conjugating by an invertible matrix. Surjectivity follows by lifting idempotent matrices across the nilpotent quotient. For injectivity, we lift the stable conjugating matrix from $R/I$ to an invertible matrix over $R$, then use the fact that idempotents congruent modulo a nilpotent ideal are conjugate. [/proofplan]

[step:Set up the idempotent model and the quotient maps] Choose an integer $N_I \ge 1$ such that $I^{N_I}=0$. For every integer $n \ge 1$, let $M_n(R)$ denote the ring of $n\times n$ matrices with entries in $R$, let $M_n(I)\trianglelefteq M_n(R)$ denote the two-sided ideal of matrices with entries in $I$, let $I_n\in M_n(R)$ denote the identity matrix, and let $GL_n(R)$ denote the group of invertible elements of $M_n(R)$. Define \begin{align*} \pi_n:M_n(R) \to M_n(R/I) \end{align*} to be the entrywise quotient homomorphism induced by $\pi$. We use the standard idempotent model of $K_0$. An idempotent matrix $p \in M_n(R)$ determines the finitely generated projective left $R$-module $\operatorname{im}(p:R^n \to R^n)$, and its class will be denoted by $[p]\in K_0(R)$. Direct sum of idempotents represents addition: \begin{align*} [p]+[q]=[p\oplus q]. \end{align*} If $u\in GL_n(R)$ and $q=upu^{-1}$, then $p$ and $q$ have isomorphic images, hence define the same class in $K_0(R)$. In this model the induced map $K_0(\pi)$ sends \begin{align*} [p]\mapsto [\pi_n(p)]. \end{align*} Since $I^{N_I}=0$, the matrix ideal $M_n(I)\trianglelefteq M_n(R)$ is nilpotent. Indeed, a product of $N_I$ matrices with entries in $I$ has every entry equal to a finite sum of products of $N_I$ elements of $I$, hence lies in $I^{N_I}=0$. Therefore \begin{align*} M_n(I)^{N_I}=0. \end{align*} [/step]

[step:Lift idempotents to prove surjectivity]Let $y\in K_0(R/I)$. By the idempotent model, write $y$ as a difference \begin{align*} y=[\bar p]-[\bar q], \end{align*} where $\bar p\in M_m(R/I)$ and $\bar q\in M_n(R/I)$ are idempotent matrices. Because $I$ is nilpotent, [citetheorem:8666] applies to the quotient homomorphisms $M_m(R)\to M_m(R/I)$ and $M_n(R)\to M_n(R/I)$. Hence there exist idempotents $p\in M_m(R)$ and $q\in M_n(R)$ such that \begin{align*} \pi_m(p)=\bar p \end{align*} and \begin{align*} \pi_n(q)=\bar q. \end{align*} Then \begin{align*} K_0(\pi)([p]-[q])=[\bar p]-[\bar q]=y. \end{align*} Thus $K_0(\pi)$ is surjective.[/step]

[guided]We want to show that every $K_0$-class over the [quotient ring](/page/Quotient%20Ring) $R/I$ comes from a $K_0$-class over $R$. In the idempotent model, an arbitrary class of $K_0(R/I)$ is represented as a formal difference of idempotents: \begin{align*} y=[\bar p]-[\bar q], \end{align*} where $\bar p\in M_m(R/I)$ and $\bar q\in M_n(R/I)$ satisfy $\bar p^2=\bar p$ and $\bar q^2=\bar q$. The key input is idempotent lifting. The theorem [citetheorem:8666] requires a nilpotent ideal. Here the kernel of $M_m(R)\to M_m(R/I)$ is $M_m(I)$, and the preceding step proved that $M_m(I)$ is nilpotent because $I$ is nilpotent. The same argument applies to $M_n(I)$. Therefore we may lift $\bar p$ and $\bar q$ to idempotents $p\in M_m(R)$ and $q\in M_n(R)$ satisfying \begin{align*} \pi_m(p)=\bar p \end{align*} and \begin{align*} \pi_n(q)=\bar q. \end{align*} The induced map $K_0(\pi)$ is defined by reducing idempotent matrices entrywise modulo $I$. Hence \begin{align*} K_0(\pi)([p]-[q])=[\pi_m(p)]-[\pi_n(q)]=[\bar p]-[\bar q]=y. \end{align*} So every element of $K_0(R/I)$ has a preimage in $K_0(R)$.[/guided]

[step:Lift invertible matrices across the nilpotent quotient] Let $\bar u\in GL_n(R/I)$. We prove that $\bar u$ has an invertible lift in $GL_n(R)$. Choose matrices $a,b\in M_n(R)$ such that \begin{align*} \pi_n(a)=\bar u \end{align*} and \begin{align*} \pi_n(b)=\bar u^{-1}. \end{align*} Then \begin{align*} \pi_n(ab)=I_n \end{align*} and \begin{align*} \pi_n(ba)=I_n. \end{align*} Thus there exist $x,y\in M_n(I)$ such that \begin{align*} ab=I_n+x \end{align*} and \begin{align*} ba=I_n+y. \end{align*} Since $M_n(I)^{N_I}=0$, the matrices $x$ and $y$ are nilpotent with $x^{N_I}=0$ and $y^{N_I}=0$. Define \begin{align*} c_x=\sum_{k=0}^{N_I-1}(-x)^k \in M_n(R) \end{align*} and \begin{align*} c_y=\sum_{k=0}^{N_I-1}(-y)^k \in M_n(R). \end{align*} The finite geometric identity gives \begin{align*} (I_n+x)c_x=c_x(I_n+x)=I_n \end{align*} and \begin{align*} (I_n+y)c_y=c_y(I_n+y)=I_n. \end{align*} Therefore $ab$ and $ba$ are invertible. Since $ab$ is invertible, $a$ has a right inverse $bc_x$. Since $ba$ is invertible, $a$ has a left inverse $c_yb$. The left and right inverses are equal: \begin{align*} c_yb=c_ybabc_x=bc_x. \end{align*} Hence $a$ is invertible. It is an invertible lift of $\bar u$, because $\pi_n(a)=\bar u$. [/step]

[step:Conjugate idempotents that are congruent modulo a nilpotent ideal]Let $p,q\in M_n(R)$ be idempotents such that \begin{align*} p-q\in M_n(I). \end{align*} We show that $p$ and $q$ are conjugate by an element of $GL_n(R)$. Define \begin{align*} v=qp+(I_n-q)(I_n-p)\in M_n(R). \end{align*} Modulo $M_n(I)$, the congruence $p\equiv q$ gives \begin{align*} \pi_n(v)=\pi_n(p)\pi_n(p)+(I_n-\pi_n(p))(I_n-\pi_n(p))=I_n. \end{align*} Thus $v-I_n\in M_n(I)$. Since $M_n(I)$ is nilpotent, the same finite geometric series argument used above shows that $v$ is invertible. We now compute the intertwining relation. Using $p^2=p$ and $q^2=q$, \begin{align*} vp=qp p+(I_n-q)(I_n-p)p=qp \end{align*} because $(I_n-p)p=0$. Similarly, \begin{align*} qv=q qp+q(I_n-q)(I_n-p)=qp \end{align*} because $q(I_n-q)=0$. Hence \begin{align*} vp=qv. \end{align*} Multiplying on the right by $v^{-1}$ gives \begin{align*} v p v^{-1}=q. \end{align*} Therefore $p$ and $q$ are conjugate in $M_n(R)$.[/step]

[guided]The point of this step is to convert the weaker relation "equal modulo $I$" into the stronger relation "conjugate over $R$." Suppose $p,q\in M_n(R)$ are idempotents and $p-q\in M_n(I)$. We introduce the matrix \begin{align*} v=qp+(I_n-q)(I_n-p). \end{align*} This expression is designed so that it acts like $q$ on the image of $p$ and like $I_n-q$ on the complement selected by $I_n-p$. First we prove that $v$ is invertible. Since $p-q\in M_n(I)$, the reductions of $p$ and $q$ modulo $I$ agree. Write $\bar p=\pi_n(p)=\pi_n(q)\in M_n(R/I)$. Reducing $v$ entrywise gives \begin{align*} \pi_n(v)=\bar p\bar p+(I_n-\bar p)(I_n-\bar p). \end{align*} Because $\bar p$ is idempotent, $\bar p^2=\bar p$ and $(I_n-\bar p)^2=I_n-\bar p$. Hence \begin{align*} \pi_n(v)=\bar p+(I_n-\bar p)=I_n. \end{align*} Thus $v-I_n\in M_n(I)$. The ideal $M_n(I)$ is nilpotent, so $(v-I_n)^{N_I}=0$. Therefore the finite geometric series \begin{align*} \sum_{k=0}^{N_I-1}(-(v-I_n))^k \end{align*} is a two-sided inverse for $v=I_n+(v-I_n)$. Hence $v\in GL_n(R)$. Now we verify that $v$ conjugates $p$ to $q$. Multiplying $v$ on the right by $p$ gives \begin{align*} vp=qp p+(I_n-q)(I_n-p)p. \end{align*} Since $p^2=p$ and $(I_n-p)p=p-p^2=0$, this becomes \begin{align*} vp=qp. \end{align*} Multiplying $v$ on the left by $q$ gives \begin{align*} qv=q qp+q(I_n-q)(I_n-p). \end{align*} Since $q^2=q$ and $q(I_n-q)=q-q^2=0$, this becomes \begin{align*} qv=qp. \end{align*} Therefore $vp=qv$. Since $v$ is invertible, multiplying on the right by $v^{-1}$ yields \begin{align*} v p v^{-1}=q. \end{align*} So idempotents that are congruent modulo the nilpotent ideal $M_n(I)$ are conjugate over $R$.[/guided]

[step:Use stable conjugacy over the quotient to prove injectivity]Let $x\in K_0(R)$ satisfy \begin{align*} K_0(\pi)(x)=0. \end{align*} Write \begin{align*} x=[p]-[q], \end{align*} where $p\in M_m(R)$ and $q\in M_n(R)$ are idempotents. The equality $K_0(\pi)(x)=0$ says that \begin{align*} [\pi_m(p)]=[\pi_n(q)] \end{align*} in $K_0(R/I)$. By the standard stable idempotent interpretation of equality in the idempotent-matrix model for $K_0$ stated in the setup step, there exist an integer $r\ge 1$, an idempotent $\bar h\in M_r(R/I)$, and an invertible matrix \begin{align*} \bar u\in GL_{m+n+r}(R/I) \end{align*} such that, after replacing idempotents by block sums of the same size, \begin{align*} \bar u\bigl(\pi_m(p)\oplus 0_n\oplus \bar h\bigr)\bar u^{-1}=0_m\oplus \pi_n(q)\oplus \bar h. \end{align*} By [citetheorem:8666], choose an idempotent $h\in M_r(R)$ such that \begin{align*} \pi_r(h)=\bar h. \end{align*} Define idempotents $P,Q\in M_{m+n+r}(R)$ by \begin{align*} P=p\oplus 0_n\oplus h \end{align*} and \begin{align*} Q=0_m\oplus q\oplus h. \end{align*} By the invertible lifting step, choose $u\in GL_{m+n+r}(R)$ such that \begin{align*} \pi_{m+n+r}(u)=\bar u. \end{align*} Set \begin{align*} Q_1=uPu^{-1}\in M_{m+n+r}(R). \end{align*} Then $Q_1$ is idempotent, and reducing the displayed conjugacy modulo $I$ gives \begin{align*} \pi_{m+n+r}(Q_1)=\pi_{m+n+r}(Q). \end{align*} Hence \begin{align*} Q_1-Q\in M_{m+n+r}(I). \end{align*} By the preceding step, $Q_1$ and $Q$ are conjugate in $M_{m+n+r}(R)$. Since $P$ is conjugate to $Q_1$ by $u$, it follows that $P$ is conjugate to $Q$ over $R$. Therefore $P$ and $Q$ define the same class in $K_0(R)$: \begin{align*} [P]=[Q]. \end{align*} Using additivity under block sum, \begin{align*} [p]+[h]=[q]+[h]. \end{align*} Cancelling $[h]$ in the abelian group $K_0(R)$ gives \begin{align*} [p]=[q]. \end{align*} Thus $x=[p]-[q]=0$, so $K_0(\pi)$ is injective.[/step]

[guided]We prove that no nonzero class of $K_0(R)$ becomes zero after reducing modulo $I$. Let $x\in K_0(R)$ satisfy \begin{align*} K_0(\pi)(x)=0. \end{align*} Using the idempotent-matrix model from the setup step, choose idempotent matrices $p\in M_m(R)$ and $q\in M_n(R)$ such that \begin{align*} x=[p]-[q]. \end{align*} The equality $K_0(\pi)(x)=0$ means that the reduced idempotents represent the same class: \begin{align*} [\pi_m(p)]=[\pi_n(q)] \end{align*} in $K_0(R/I)$. The stable idempotent interpretation of equality in this model says that equality of these two classes is witnessed after adding the same idempotent summand and conjugating. Thus there exist an integer $r\ge 1$, an idempotent $\bar h\in M_r(R/I)$, and a matrix \begin{align*} \bar u\in GL_{m+n+r}(R/I) \end{align*} such that \begin{align*} \bar u\bigl(\pi_m(p)\oplus 0_n\oplus \bar h\bigr)\bar u^{-1}=0_m\oplus \pi_n(q)\oplus \bar h. \end{align*} The theorem [citetheorem:8666] applies because $I\trianglelefteq R$ is nilpotent, so the idempotent $\bar h$ lifts to an idempotent $h\in M_r(R)$ satisfying \begin{align*} \pi_r(h)=\bar h. \end{align*} Define idempotents $P,Q\in M_{m+n+r}(R)$ by \begin{align*} P=p\oplus 0_n\oplus h \end{align*} and \begin{align*} Q=0_m\oplus q\oplus h. \end{align*} The invertible-lifting step applies to $\bar u\in GL_{m+n+r}(R/I)$, so there exists $u\in GL_{m+n+r}(R)$ with \begin{align*} \pi_{m+n+r}(u)=\bar u. \end{align*} Set \begin{align*} Q_1=uPu^{-1}\in M_{m+n+r}(R). \end{align*} Then $Q_1$ is idempotent. Reducing the equality defining $Q_1$ modulo $I$ and using the displayed conjugacy over $R/I$ gives \begin{align*} \pi_{m+n+r}(Q_1)=\pi_{m+n+r}(Q). \end{align*} Therefore \begin{align*} Q_1-Q\in M_{m+n+r}(I). \end{align*} The congruent-idempotent conjugacy step now applies to the idempotents $Q_1$ and $Q$, so they are conjugate in $M_{m+n+r}(R)$. Since $P$ is conjugate to $Q_1$ by $u$, the idempotents $P$ and $Q$ are conjugate over $R$. Conjugate idempotents define isomorphic image modules, hence the same $K_0(R)$-class: \begin{align*} [P]=[Q]. \end{align*} By the definition of block sum addition in the idempotent model, this equality is \begin{align*} [p]+[h]=[q]+[h]. \end{align*} Because $K_0(R)$ is an abelian group, we may cancel $[h]$ from both sides and obtain \begin{align*} [p]=[q]. \end{align*} Thus \begin{align*} x=[p]-[q]=0. \end{align*} So the kernel of $K_0(\pi)$ is zero, and $K_0(\pi)$ is injective.[/guided]

[step:Conclude that the induced map is an isomorphism] The quotient homomorphism $\pi:R\to R/I$ induces a [group homomorphism](/page/Group%20Homomorphism) \begin{align*} K_0(\pi):K_0(R)\to K_0(R/I). \end{align*} The surjectivity step proves that every element of $K_0(R/I)$ lies in its image, and the injectivity step proves that its kernel is zero. Hence $K_0(\pi)$ is both surjective and injective. Therefore \begin{align*} K_0(R)\cong K_0(R/I). \end{align*} [/step]