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Hello to all,

I have been perusing Harthorne for some time, and I noticed something: it is well known that the class group on $\mathbb{P}^n_k$ is $\mathbb{Z}$. But as I look at Harthorne's proof it seems to mee that it works in much greater generality. Namely if I consider any projective scheme $X=Proj(A)$, where $A$ is a graded $UFD$ in such a way that there exists an irreducible element of degree $1$, then the exact same reasoning shows that the class group of $X$ is also $\mathbb{Z}$, and generated by the prime divisor $(a)$. Is this true ?

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1 Answer 1

up vote 3 down vote accepted

Well, if you read on to Chapter 2, exercise 6.3, then it is stated that: $$Cl(A) \cong Cl(X)/\mathbb Z[H]$$ here $[H]$ represents the hyperplane section. So the answer is yes.

There is a less well-known but very nice generalization. Suppose that $X$ is smooth. Let $R=A_m$ be the local ring of A at the irrelevant ideal. Then one has a (graded) isomorphism of $\mathbb Q$- vector spaces:

$$CH(X)_{\mathbb Q}/[H]CH(X)_{\mathbb Q} \cong A_*(R)_{\mathbb Q}$$

Here $CH(X)$ is the Chow ring of $X$ and $A_*(R)$ is the total Chow group of $R$. Details can be found in this paper by Kurano.

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Thank you for that answer, that's a relief. Although it still seems kind of weird to me that Hartshorne would impose such an unnecessary restriction like considering only polynomial rings over fields –  louis de Thanhoffer de Völcsey Mar 26 '10 at 10:03

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