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I would like to show that

The product $\prod\limits_{\mathfrak p} (1 - N(\mathfrak p)^{-s})$ converges absolutely iff the series $\sum\limits_{\mathfrak p} N(\mathfrak p)^{-s}$ converges absolutely, where $\mathfrak p$ runs through the prime ideals of a number field $K$ and $s>1$.

I know that the product $\prod_{n \geq 1} (1 + x_n)$ with $|x_n| < 1$ converges absolutely iff the series $\sum_{n \geq 1} x_n$ converges absolutely. We have $$-\sum\limits_{\mathfrak p} N(\mathfrak p)^{-s} = -\sum\limits_{p} \sum\limits_{\mathfrak p \mid p} N(\mathfrak p)^{-s} = \sum_{n \geq 1} a_n, \qquad a_p := \sum\limits_{\mathfrak p \mid p} -N(\mathfrak p)^{-s}, a_n=0 \text{ if $n$ is not prime}$$ so applying the above theorem yields the product $$\prod_{n \geq 1} (1 + a_n) = \prod_{p} \left(1 - \sum\limits_{\mathfrak p \mid p} N(\mathfrak p)^{-s}\right)$$

I don't see how this can be equal to $$\prod\limits_{\mathfrak p} (1 - N(\mathfrak p)^{-s})$$ because my previous product will "miss" the "double" factors like $N(\mathfrak p)N(\mathfrak p')$.

Thank you!

Watson
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    Any reference of the formula? This is different from the Euler's product formula I know. Thanks! – S. Y Jan 06 '17 at 18:17
  • @S. Y : We have $\prod_{\mathfrak p} (1-N(\mathfrak p)^{-s})^{-1} = \sum_I N(I)^{-s}$ (Neukirch ANT p. 457), but the results above equivalence of convergences come from here. – Watson Jan 06 '17 at 18:46
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    Make a bijection $n \leftrightarrow \mathfrak{p}$ (as the set of prime ideals is countably infinite). It follows that the product $\prod_{\mathfrak{p}} (1 + x_{\mathfrak{p}})$ with $|x_{\mathfrak{p}}| < 1$ converges absolutely iff the series $\sum_{\mathfrak{p}} x_{\mathfrak{p}}$ converges absolutely. Now set $x_{\mathfrak{p}} = -N(\mathfrak{p})^{-s}$. – Peter Humphries Jan 06 '17 at 19:45
  • Related: https://math.stackexchange.com/questions/1319431/ – Watson Jan 07 '17 at 12:39
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    @PeterHumphries : thank you for your comment! So if I understand well, absolute convergence is important here, because otherwise the choice of a bijection $n \leftrightarrow \mathfrak p$ would have an influence on the (conditional) convergence. Am I right? – Watson Jan 07 '17 at 14:59
  • @PeterHumphries : By the way, you could provide this as an answer. I would enjoy accepting it, then :-) – Watson Jan 07 '17 at 15:24

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