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What is the zeta function of $\mathbb{P}^1_{\mathbb{Q}}$? Thanks

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What part of the definition are you having trouble applying? –  Qiaochu Yuan Aug 27 '12 at 18:10

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I'm going to assume this means: what is the zeta function of $\mathbb{P}^1_{\mathbb{Z}}$?

There is the standard formula that $\zeta(X_1\coprod X_2, s)=\zeta(X_1, s)\zeta(X_2, s)$. Since $\mathbb{P}^1=\mathbb{A}^1\coprod \mathbb{A}^0$ and $\zeta(\mathbb{A}^0, s)=\zeta(Spec(\mathbb{Z}), s)=\zeta(s)$ the Riemann zeta function, we only need to figure out the zeta function of $\mathbb{A}^1$.

This is a pretty standard argument. Since $\mathbb{A}^1=Spec(\mathbb{Z}[x])$ the closed points correspond to the ideals $(p, f(x))$ where $f(x)$ is irreducible when reduced over $\mathbb{F}_p[x]$. Thus $$\zeta(\mathbb{A}^1, s):=\prod_{x \ closed} (1-|\kappa(x)|^{-s})^{-1}=\prod_p \prod_{f \ irred}(1-p^{-s \text{deg}(f)})^{-1}$$

After some re-writing you find that $\zeta(\mathbb{A}^1, s)=\zeta(s-1)$. Thus $\zeta(\mathbb{P}^1, s)=\zeta(s-1)\zeta(s)$.

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What is the difference with $\zeta(\mathbb{P}^1_{\mathbb{Q}})$? –  user17090 Aug 27 '12 at 18:34
    
I'm pretty sure if you have a variety $X$ over $\mathbb{Q}$, then the zeta function of $X$ means find a model over $\mathbb{Z}$ and take the zeta function of that model (of course, depending on the variety, this could mean the zeta function depends on the choice of model). I could be wrong and maybe there is some intrinsic definition though ... –  Matt Aug 27 '12 at 18:39
    
@Hollowdead: Dear Hollow and Matt, The intrinsic definition is in terms of the Galois action on the etale cohomology. In this particular case it will give the same answer as Matt's. Regards, –  Matt E Aug 27 '12 at 19:37
    
@MattE: Thanks. Where can I find the precise intrinsic definition? –  user17090 Aug 27 '12 at 21:27
    
@Hollowdead: Dear Hollow, This thread at the n-category cafe is pretty helpful, as are the references it contains. Regards, –  Matt E Aug 28 '12 at 1:36

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