Homotopy invariance of $K_0$

Question

It is well-known that algebraic $K$-theory is $\mathbb{A}^1$-invariant for regular Noetherian schemes. The way this is proved is usually to prove that $K$-theory of coherent sheaves i.e. $G$-theory matches with the $K$-theory of vector bundles. Then $\mathbb{A}^1$-invariance is proved for $G$-theory using the localization sequence. Now let's consider the case of $K_0$. I was wondering since this proof relies on higher $K$-theory, is it possible to give a proof of $\mathbb{A}^1$-invariance of $K_0$ by just the relations coming from the definition of $K_0$

Answer 1

I will use capital letters to denote $R[t]$ modules and capital letters subscripted with $0$ to denote $R$ modules. All rings are assumed noetherian and all modules are assumed finitely generated. The key lemma is due to Swan:

Lemma: Let $M\subset N_0[t]$. Then there exists a short exact sequence of the form $$0\rightarrow X_0[t]\rightarrow Y_0[t]\rightarrow M\rightarrow 0$$.

Sketch of Proof: Let $N_k$ be the $R$-module $\sum_{i=0}^k (t^i)N_0$. Let $M_k=M\cap N_k$. Take $k$ large enough so that $M_k$ contains a generating set for $M$ over $R[t]$. Let $X_0=M_k$ and $Y_0=M_{k+1}$. Map $Y_0[t]$ to $M$ in the obvious way. Check that this works.

Corollary: If $R$ is regular and $M$ is an $R[t]$ module then $M$ has a finite projective resolution of the form $$0\rightarrow X^n_0[t]\rightarrow \ldots \rightarrow X^1_0[t]\rightarrow X^0_0[t]\rightarrow M\rightarrow 0$$

Sketch of Proof: Map a free $R[t]$-module onto $M$ and let $M'$ be the kernel. It suffices to prove the corollary for $M'$. Apply the lemma to $M'$ to get a sequence $$0\rightarrow X_0[t]\rightarrow Y_0[t]\rightarrow M'\rightarrow 0$$ Take finite projective $R$-resolutions of $X_0$ and $Y_0$, extend them to finite projective $R[t]$-resolutions of $X_0[t]$ and $Y_0[t]$, and then use the mapping cone construction to get the desired resolution of $M'$.

Theorem: If $R$ is regular, then $K_0(R)\rightarrow K_0(R[t])$ is an isomorphism.

Proof:Injectivity is clear because the map splits. For surjectivity, take $M$ projective over $R[t]$ and use the lemma to represent its $K_0$ class as an alternating sum of $[X_i[t]]$ where all the $X_i$ are projective over $R$.