Intuition behind the Euler product formula for the Riemann zeta function

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I’ve written about the Riemann zeta function in other posts and touched briefly on the product formula representation. Here, I wanted to go through a derivation of the product formula. Like most maths related subjects I write about it won’t be particularly mathematically rigorous, but will aim more to convey the intuition behind the product formula representation.

The Riemann zeta function is defined as follows for complex numbers $s$ with real part greater than $1$:

\[\zeta(s) = \sum_{n=1}^{\infty} \frac{1}{n^{s}} = 1 + \frac{1}{2^{s}} + \frac{1}{3^{s}} + \frac{1}{4^{s}} + \frac{1}{5^{s}} + \cdots\]

The infinite series converges when the real part of $s$ exceeds $1$. The Euler product formula expresses $\zeta(s)$ as a product of terms involving prime numbers:

\[\zeta(s) = \prod_{p \text{ prime}} \frac{1}{1 - p^{-s}}\]

To derive this we start with the infinite series representation and writing each term as its prime factorisation:

\[\begin{aligned} \frac{1}{2^{s}} &= \left( \frac{1}{2^{s}} \right) \cdot 1 \\ \frac{1}{3^{s}} &= \left( \frac{1}{3^{s}} \right) \cdot 1 \\ \frac{1}{4^{s}} &= \left( \frac{1}{2^{s}} \right) \cdot \left( \frac{1}{2^{s}} \right) \\ \frac{1}{5^{s}} &= \left( \frac{1}{5^{s}} \right) \cdot 1 \\ \frac{1}{6^{s}} &= \left( \frac{1}{3^{s}} \right) \cdot \left( \frac{1}{2^{s}} \right), \\ \end{aligned}\]

and so on. This is just using the fundamental theorem of arithmetic. Rewriting the series with these terms gives us

\[\zeta(s) = (1 \cdot 1) + \left( \frac{1}{2^{s}} \cdot 1 \right) + \left( \frac{1}{3^{s}} \cdot 1 \right) + \left( \frac{1}{2^{s}} \cdot \frac{1}{2^{s}} \right) + \left( \frac{1}{5^{s}} \cdot 1 \right)+ \left( \frac{1}{3^{s}} \cdot \frac{1}{2^{s}} \right) + \cdots\]

Next, we can group the terms by the prime numbers involved:

\[\zeta(s) = \left( 1 + \frac{1}{2^{s}} + \frac{1}{2^{2s}} + \frac{1}{2^{3s}} + \cdots \right) \cdot \left( 1 + \frac{1}{3^{s}} + \frac{1}{3^{2s}} + \frac{1}{3^{3s}} + \cdots \right) \cdot \left( 1 + \frac{1}{5^{s}} + \frac{1}{5^{2s}} + \frac{1}{5^{3s}} + \cdots \right) + \cdots\]

Here we see that each term being multiplied is a geometric series. Recall that in general a geometric series can be written as

\[\sum_{n=0}^{\infty} a r^{n} = \frac{a}{1 - r}\]

And so, each factor of the above product can be simplified:

\[\begin{aligned} 1 + \frac{1}{2^{s}} + \frac{1}{2^{2s}} + \frac{1}{2^{3s}} + \cdots &= \frac{1}{1 - \frac{1}{2^{s}}} \\ 1 + \frac{1}{3^{s}} + \frac{1}{3^{2s}} + \frac{1}{3^{3s}} + \cdots &= \frac{1}{1 - \frac{1}{3^{s}}} \\ 1 + \frac{1}{5^{s}} + \frac{1}{5^{2s}} + \frac{1}{5^{3s}} + \cdots &= \frac{1}{1 - \frac{1}{5^{s}}} \\ \end{aligned}\]

Substituting the simplified terms back into the product we had above gets us the product formula.

\[\begin{aligned} \zeta(s) = \left( \frac{1}{1 - \frac{1}{2^{s}}} \right) \cdot \left( \frac{1}{1 - \frac{1}{3^{s}}} \right) \cdot \left( \frac{1}{1 - \frac{1}{4^{s}}} \right) \cdot \left( \frac{1}{1 - \frac{1}{5^{s}}} \right) \cdots \end{aligned}\] \[\zeta(s) = \prod_{p \text{ prime}} \frac{1}{1 - p^{-s}}\]

This is just such a cool result. And again, while this derivation is by no means mathematically rigorous I think it provides a good high-level overview for the intuition behind the Euler product for the Riemann zeta function.