# Difference between zeta sum and Euler product?

The fact that $$\sum_{n=1}^\infty \frac{1}{n^s} = \prod_{p}\frac{1}{1-p^{-s}}$$ is a consequence of unique factorization of primes.

We could form a similar sum and a similar product of irreducibles in a number field which does not have unique factorization. Since the unique factorization does not hold the vales would be probably different.

• Are the always different?

• Can we get any information about the class number from their difference?

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"unique prime factorization", or "unique factorization into primes", but not "unique factorization of primes" – joriki May 16 '11 at 14:42

There are a few issues with what you are proposing: what it means for an element of $\mathcal{O}_K$ to be "positive" is unclear (so which elements are we summing over on the left?), which value of $\alpha^s$ to choose when $\alpha$ is not a positive real number is unclear, and that because not all prime ideals of $\mathcal{O}_K$ are principal, taking a product over irreducible elements (assuming we already picked which ones are "positive") will not even be capturing all the information about the number field, much less the failure of unique factorization. The fix (or at least the one I know of, perhaps there are others) is to consider ideals in $\mathcal{O}_K$ instead of positive integers, and prime ideals instead of irreducible elements, and take norms so that we only ever take positive integers to complex powers. Then because there is unique factorization of ideals into prime ideals, we have the Euler product for the Dedekind zeta function $$\sum_{I\subseteq\mathcal{O}_K}\frac{1}{N_{\mathbb{Q}}^K(I)^s}=\prod_{P\subseteq\mathcal{O}_K}\left(1-\frac{1}{N_{\mathbb{Q}}^K(P)^s}\right)^{-1}.$$