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The following proposition is from Algebraic Number Theory by Neukirch (Proposition 1.11, Chapter 3, p.191), but I doubt that exat sequence. Let $K$ be a number field, $O$ be the ring of integers. And $\mathfrak{p}\mid \infty$ means $\mathfrak{p}$ is a infinite place.

Let $\Gamma=\lambda(O^{\star})$ denote the complete lattice of units in trace-zero space, $$H=\{(v_{\mathfrak{p}}) \in \prod_{\mathfrak{p}|\infty} \mathbb{R} \mid \sum_{\mathfrak{p}|\infty}{ v_{\mathfrak{p}}}=0 \}.$$ There is an exact sequence $$ 0\rightarrow H/ {\Gamma} \rightarrow CH^{1}(\bar{O})^{0} \rightarrow CH^{1}(O) \rightarrow 0\quad.$$

Let $Div(\bar{O})$ be a replete divisor (Arakelov divisor), $CH^{1}(\bar{O})=Div(\bar{O})/P(\bar{O})$ be a replete divisor class group. Let $CH^{1}(\bar{O})^{0}$ be the kernel of $\deg: CH^{1}(\bar{O}) \rightarrow \mathbb{R}$, which is $$\deg(\sum_{\mathfrak{p}}{v_{\mathfrak{p}} \mathfrak{p} })=\log(\prod_{\mathfrak{p}}\mathfrak{N}(\mathfrak{p})^{v_{\mathfrak{p}}}).$$

My question comes from the exactness of $$0\rightarrow H\stackrel{\alpha}{\longrightarrow} Div(\bar{O})^{0} \stackrel{\beta}{\longrightarrow} Div(O)\rightarrow 0,$$ which is crucial in the proof. $\beta$ is the projection map, mapping all the finite place part of $Div(\bar{O})^{0}$ to $Div(O)$. It is surjective. $\alpha$ is defined to be $$\quad \alpha((v_{\mathfrak{p}}))=\sum_{v_{\mathfrak{p}}\mid\infty}{\frac{v_{\mathfrak{p}}}{f_{\mathfrak{p}}} \mathfrak{p}}\quad$$ (where $f_{\mathfrak{p}}=[K_{\mathfrak{p}}: \mathbb{R}]$, $K_{\mathfrak{p}}$ is the completion of $K$ with respect to place $\mathfrak{p}$, actually one can show: for $\mathfrak{p}\mid \infty$, $f_{\mathfrak{p}}=1$ if $\mathfrak{p}$ is a real embedding, and $f_{\mathfrak{p}}=2$ if $ \mathfrak{p}$ is a complex embedding.)

The point is that $Ker(\beta)$ is larger than $Im(\alpha)$. Say, one can pick $\mathfrak{p}_{1}$ as a finite place with $v_{\mathfrak{p}_{1}}=1$, and some infinite place $\mathfrak{p}_{2}$, such that $v_{\mathfrak{p}_{2}}=\log(p^{f_{\mathfrak{p}_{1}}})$. And all the other $v_{\mathfrak{p}}=0$. This is in the $Ker(\beta)$, but not in $Im(\alpha)$!

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    $\begingroup$ I'm confused with your counterexample. Doesn't your divisor $1\mathfrak p_1+v_{\mathfrak p_2}\mathfrak p_2$ map down by the projection map $\beta$ to $1\mathfrak p_1$, rather than zero? $\endgroup$ – John M Jul 6 '11 at 4:45
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I think Neukirch is correct while your counter example does not work as John M said. We have ${\rm Ker} \beta=\{D=\sum_{p} v_pp\in {\rm Div}(\bar{o})^0|\forall p\not| \infty: v_p=0\} =\{D=\sum_{p|\infty} v_pp| 0=\deg D=\sum_{p|\infty}v_pf_p\} =\{D=\sum_{p|\infty} (x_p/f_p)p | \sum_{p|\infty}x_p=0\} ={\rm Im}\alpha.$

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