9
$\begingroup$

Is Euler product formula equivalent to fundamental theorem of arithmetic (unique factorization theorem) ?

$$\sum_{n=1}^\infty\frac{1}{n^s} = \prod_{p \text{ prime}} \frac{1}{1-p^{-s}}$$

We know that we can derive Euler product formula from fundamental theorem of arithmetic. Is the opposite also true ? Can we use Euler product formula to derive fundamental theorem of arithmetic.

The reason we are interested in this question is, if we can derive fundamental theorem of arithmetic from Euler product formula, does does this mean that Riemann Zeta function is a "complex version" of fundamental theorem of arithmetic ?

$\endgroup$
5
$\begingroup$

The answer is affirmative. Since $$\frac{1}{1-p^{-s}}=1+\frac{1}{p^s}+\frac{1}{p^{2s}}+\ldots, $$ we have the identity: $$\prod_{p}\frac{1}{1-p^{-s}}=\sum_{n\geq 1}\frac{h(n)}{n^s}$$ where $h(n)\in\mathbb{N}_{\geq 1}$ counts the number of ways of writing $n$ as a product of powers of different primes. Assuming $h(n)>1$ for some $n$, the identity $$\zeta(s)=\sum_{n\geq 1}\frac{h(n)}{n^s}$$ cannot hold for every $s\in\mathbb{R}_{>1}$. This proves the fundamental theorem of arithmetic.

$\endgroup$
  • $\begingroup$ Just trying to understand the last part. Could you please explain why the identity cannot hold for every $s>1$ ? E.g. how do we know $h(n)$ is not bounded above? $\endgroup$ – Antinous Jan 21 at 21:50
  • $\begingroup$ @Antinous: $h(n)$ is bounded above, it is constantly one. The point is that the Riemann $\zeta$ function cannot have two different representations as a Dirichlet series: the coefficients $h(n)$ depend on the inverse Mellin transform. $\endgroup$ – Jack D'Aurizio Jan 21 at 21:57

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.