What is the zeta function of $\mathbb{P}^1_{\mathbb{Q}}$? Thanks
What is the zeta function of the projective line over $\mathbb{Q}$
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1What part of the definition are you having trouble applying? – 2012-08-27
1 Answers
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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0What is the difference with $\zeta(\mathbb{P}^1_{\mathbb{Q}})$? – 2012-08-27
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0I'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 ... – 2012-08-27
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0@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, – 2012-08-27
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0@MattE: Thanks. Where can I find the precise intrinsic definition? – 2012-08-27
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0@Hollowdead: Dear Hollow, [This thread](http://golem.ph.utexas.edu/category/2010/07/zeta_functions_dedekind_versus.html) at the n-category cafe is pretty helpful, as are the references it contains. Regards, – 2012-08-28
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0@MattE Thanks for pointing that out. The comments on that post are amazing. It is quite a useful read. – 2012-08-28
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0Dear Matt, You're welcome, and I'm glad that you found the comments in the thread helpful. Best wishes, – 2012-08-28