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The Zeta Euler Product Encodes Unique Factorization by Vicfred

Expanding one geometric series for each prime produces every positive integer exactly once. The point is to make the formal expression readable enough to audit line by line.

Statement

Dirichlet series \(\sum a_nn^{-s}\) turn multiplicativity into Euler products. The complex variable \(s=\sigma+it\) lets analytic continuation and zero-free regions control arithmetic sums.

$$ \zeta(s)=\sum_{n\ge1}\frac1{n^s} $$

A reliable calculation names domain and codomain. The notation \(\mathsf{data}\mapsto\mathsf{claim}\) is harmless only after both \(\operatorname{dom}\) and \(\operatorname{cod}\) have been fixed.

$$ \zeta(s)=\prod_p\left(1-p^{-s}\right)^{-1}\qquad(\Re s>1) $$

Worked algebra

A worked instance is useful here because it exposes every index that the compressed statement hides.

$$ \prod_p(1+p^{-s}+p^{-2s}+\cdots)=\sum_{n=\prod_pp^{a_p}}\prod_pp^{-a_ps} $$

Conceptual compression

The compact alignment is a local map of the argument: assumptions on the first row, consequence on the second. Any generalisation must preserve that dependency.

$$ \begin{aligned} \mathsf{D}\;&:\quad \zeta(s)=\sum_{n\ge1}\frac1{n^s},\\[5pt] \mathsf{C}\;&:\quad \zeta(s)=\prod_p\left(1-p^{-s}\right)^{-1}\qquad(\Re s>1). \end{aligned} $$

Caveat

An Euler product converges absolutely only in a right half-plane. Formal rearrangement outside that region can destroy the argument.

$$ \boxed{\begin{gathered} \text{compact conclusion}\\[-2pt] \zeta(s)=\prod_p\left(1-p^{-s}\right)^{-1}\qquad(\Re s>1) \end{gathered}} $$

The notation is dense, but it is doing honest work: every delimiter records scope and every index records dependence. Removing one should require a mathematical reason.

This article was posted on Fri 10 February 2017. Facts and circumstances may have changed since publication.
Please contact me before jumping to conclusions if something seems wrong or unclear.