Computing the Grothendieck group of affine space.

Question

For a Noetherian scheme $X$ the Grothendieck group $K(X)$ is defined as the free abelian group on coherent sheaves quotiented by the equivalence relation $\mathscr{F}=\mathscr{F}'+\mathscr{F}''$ for every short exact sequence $$0\to\mathscr{F}'\to\mathscr{F}\to\mathscr{F}''\to0$$

As part of my solution to problem III 5.4 in Hartshorne I need to show that the Grothendieck group of affine space $X=\text{Spec}( \mathbb{k}[x_1,\dots,x_n])$ is the free abelian group generated by the structure sheaf $\mathcal{O}_X$.

One can rephrase this algebraically by defining the Grothendieck group of a noetherian ring $A$ as the free abelian group on finitely generated modules quotiented by an analagous equivalence relation. Then I need to show $K(\mathbb{k}[x_1,\dots,x_n])$ is the free abelian group generated by $\mathbb{k}[x_1,\dots,x_n]$.

Unfortunately I do not really know how to approach this problem.

Answer 1

Since your scheme is affine, and since the Grothendieck group of a category obviously is invariant under equivalences, you want to compute the Grothendieck group of the category of finitely generated modules over $k[x_1,\dots,x_n]$. Since this ring is regular, it is easy to see that the Grothendieck group of its category of f.g. modules is the isomrphic to the Grothendieck group of the category of its f.g. projective modules. Now, by the Quillen-Suslin answer to Serre's question, every f.g. module over that ring is in fact free.

This tells us that the group you want is $\mathbb Z$.

Of course, using the Quillen-Suslin theorem is slightly thermonuclear. You can obtain the same conclusion simply by noting that every f.g. projective module over a polynomial ring has a finite resolution by free modules —and this is a weaker statement; you can find a proof of this here

Answer 2

That $K(k[x_1,\dotsc,x_n])\simeq\Bbb Z$ follows from a result in $K$-theory known as "homotopy invariance". Daniel Dugger has written some nice notes proving this here. For convenience, I've outlined the proof below.

For a category $\mathscr C$ let $\Bbb Z\langle\mathscr C\rangle$ be the free abelian group with one generator for every isomorphism class of objects of $\mathscr C$.

Step 1. For a ring $R$ let ${}_R\mathsf{mod}$ be the category of finitely-generated $R$-modules and let ${}_R\mathsf{proj}$ be the category of finitely-generated projective $R$-modules. Define \begin{align*} G(R) &= \frac{\Bbb Z\langle {}_R\mathsf{mod}\rangle}{S} & K(R) &= \frac{\Bbb Z\langle {}_R\mathsf{proj}\rangle}{T} \end{align*} where $S$ is the subgroup of $\Bbb Z\langle {}_R\mathsf{mod}\rangle$ generated by elements of the form $[M]-[M^{\prime}]-[M^{\prime\prime}]$ whenever there exists an exact sequence of the form $$ 0\to M^\prime\to M\to M^{\prime\prime}\to 0 $$ in ${}_R\mathsf{mod}$ and $T$ is defined similarly.

Step 2. For a regular ring $R$, the canonical map $\alpha:K(R)\to G(R)$ is an isomorphism. To prove this, define $\beta:G(R)\to K(R)$ by $\beta([M])=\sum (-1)^j\cdot[P_j]$ where $P_\bullet\to M\to 0$ is a finite projective resolution of $M$. Then some homological algebra shows that $\beta$ is well-defined and $\beta=\alpha^{-1}$. Note that $R$ needs to be regular to ensure that $M$ has a finite projective resolution.

Step 3. For a Noetherian ring $R$, the map $\phi:G(R)\to G(R[t])$ defined by $$ \phi(M)=M\otimes_R R[t] $$ is an isomorphism. This fact is known as "homotopy invariance" and the proof is somewhat involved. Again we explicitly construct the inverse $\theta:G(R[t])\to G(R)$ by the formula $$ \DeclareMathOperator{Tor}{Tor} \theta([M])=\bigl[\Tor^{R[t]}_0\bigl(M,R[t]/(t)\bigr)\bigr]-\bigl[\Tor^{R[t]}_1\bigl(M,R[t]/(t)\bigr)\bigr] $$ The details of this are written-up quite nicely on page 19 of Dugger's notes.

That $K(k[x_1,\dotsc,x_n])\simeq\Bbb Z$ then follows from $G(k)\simeq\Bbb Z$.

So, the proof of this result requires a decent amount of machinery but I don't know of a less sophisticated proof.