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This question is related with Exercise III.5.4 (Page 230) of Hartshorne's Algebraic Geometry. Here I start with recalling the definition of Grothendieck group $K(X)$ of a noetherian scheme $X$, which is the quotient of free abelian group generated by all coherent sheaves on $X$, by the subgroup generated by all expressions $\mathscr F' -\mathscr F + \mathscr F''$ whenever there is an exact sequence $0\to \mathscr F' \to \mathscr F \to \mathscr F'' \to 0$ (cf. Exercise II.6.10, Page 148 Hartshorne's Algebraic Geometry).

Let $k$ be a field. This exercise essentially wants to prove $K(\mathbf P_k^r) \cong \mathbb Z^r$, generated by the images of the sheaves of regular functions of $\mathbf P_k^0, \mathbf P_k^1, \ldots, \mathbf P_k^{r}$ in $K(\mathbf P_k^r)$. We can prove this claim by induction according to the hints given in the book. It is trivial when $r =0$ which corresponds to a point. Suppose the claim is true for $r-1$. Since $\mathbf P_k^{r-1}$ can be considered as a closed subscheme of $\mathbf P_k^r$ and $\mathbf P_k^r - \mathbf P_k^{r-1} \cong \mathbf A_k^{r-1}$. Hence one has an exact sequence (cf. Exercise II.6.10, page 148): $$ K(\mathbf P_k^{r-1}) \xrightarrow{\alpha} K(\mathbf P_k^r) \to K(\mathbf A_k^{r-1}) \to 0.$$ I think I can prove $\alpha$ is injective by showing it splits. So it only remains to prove $K(\mathbf A_k^{r}) \cong \mathbb Z$ for any integer $r > 0$. But this is where I get stuck. $K(\mathbf A_k^1) \cong \mathbb Z$ is easy because we fully understand the module structure over PID ($k[x]$ is PID). I doubt this method can be extended to $r > 1$. One way I tried is to prove the surjective group homomorphism $ \gamma:\ K(\mathbf A_k^{r-1}) \to \mathbb Z$ is also injective, where $\gamma$ is induced by the ranks of coherent sheaves. To prove $\mathrm{ker}(\alpha)=0$, it is enough to prove any coherent torsion sheaf $\mathscr F$ is trivial in $K(\mathbf A_k^{r})$. Here a coherent sheaf $\mathscr F$ over an integral scheme $X$ is torsion if the stalk $\mathscr F_\eta = 0$ where $\eta$ is the generic point of $X$. But this is exactly where I get stuck.

I appreciate if any one could give me some hints (answers if possible) or give a new method.

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2 Answers 2

up vote 8 down vote accepted

Hailong's answer is of course exactly right, but there's also a sense in which it's overkill. That's because it invokes two facts:

Fact I: Any f.g. module over $R=k[x_1,\ldots x_r]$ has a finite resolution by projective $R$-modules.

Fact II: Any projective module over $R$ is free. (This is the Quillen/Suslin theorem).

From these it follows that

Theorem: $K_0(R)={\mathbb Z}$.

But every proof I know of Fact I shows more, namely:

Fact IA: Any f.g. module over $R$ has a finite resolution by free $R$-modules.

Given Fact IA (which is no harder than Fact I), the theorem follows immediately, with no need to invoke Fact II.

Fact IA is a theorem of Serre (or actually a theorem of Hilbert, subsequently generalized by Serre; see comments below) that long predates Quillen/Suslin. Here is a sketch of a proof:

Sketch of a proof: Let $A=k[x_1,\ldots,x_{r-1}]$, so that every f.g. $A$-module has a finite free resolution by induction. We want to show the same for $R=A[x]$.

1) If $M$ is an $A[x]$ module, we have an exact sequence $$0\rightarrow A[x]\otimes_AM\rightarrow A[x]\otimes_AM\rightarrow M\rightarrow 0$$ where the first map is $x^i\mapsto x^{i+1}\otimes m-x^i\otimes xm$ and the second is $x^i\otimes m\mapsto x^im$.

2) If $M$ is contained in a free f.g. $A[x]$-module, we can find finitely generated $A$-submodules $G,H\subset M$ so that the restriction $$0\rightarrow A[x]\otimes_AG\rightarrow A[x]\otimes_AH\rightarrow M\rightarrow 0$$ is still exact. (Proof: A free $A[x]$-module is of the form $F_0[x]$ where $F_0$ is a free $A$-module. Take $n$ large enough so that all generators of $M$ are contained in $F_0+xF_0+\ldots +x^nF_0$. Let $G=F_0+xF_0+\ldots x^nF_0$ and $H=F_0+xF_0+\ldots x^{n+1}F_0$. Check that this works.)

3) An arbitrary f.g. $A[x]$-module is a quotient of a free module. The kernel $N$ is therefore contained in a free f.g. $A[x]$-module so we have an exact sequence $$0\rightarrow A[x]\otimes_AG\rightarrow A[x]\otimes_AH\rightarrow N\rightarrow 0$$ as in 2). Use induction to construct finite free $A$-resolutions of $G$ and $H$. Tensor up to $A[x]$ and construct the mapping cone to get a finite free resolution for $N$.

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1  
I know this as Hilbert's Syzygy Theorem. –  Martin Brandenburg Jun 29 at 23:32
    
@MartinBrandenburg: Yes, for $k$ a field, this is Hilbert's syzygy theorem, which is all that's really needed for the OP's question. The argument above applies more generally to the case where $k$ is any ring over which all f.g. projectives are free. Even more generally it shows that if $k$ is regular then $k[x]$ is regular, which is Serre's theorem. –  Steven Landsburg Jun 30 at 2:39

Any module over $R=k[x_1,\dots, x_r]$ has a finite resolutions of projective modules, and projective modules are free. So, in the Grothendieck group, the class of the module is a multiple of the class of $R$.

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