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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?2012-08-27

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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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    Dear Matt, You're welcome, and I'm glad that you found the comments in the thread helpful. Best wishes,2012-08-28