The Euler Product: Factoring the Zeta Function over Primes

The precise statement

Theorem (Euler, 1737). For every complex number s with Re(s) > 1,

ζ(s) = ∑ₙ₌₁^∞ 1/nˢ = ∏ₚ prime (1 − p⁻ˢ)⁻¹,

where the product runs over all primes p = 2, 3, 5, 7, …. Both sides converge absolutely in this half-plane, and the identity is an equality of holomorphic functions there.

• The hypothesis Re(s) > 1 is essential: it is exactly the condition making ∑ 1/nˢ converge absolutely, since |n⁻ˢ| = n⁻Re(s).
• Each factor is the sum of a geometric series: (1 − p⁻ˢ)⁻¹ = 1 + p⁻ˢ + p⁻²ˢ + p⁻³ˢ + ⋯, valid because |p⁻ˢ| = p⁻Re(s) < 1.
• Convergence of the infinite product means the partial products over the first k primes tend to a nonzero limit; equivalently ∑ₚ p⁻ˢ converges.

The formula packages an infinite additive object (a Dirichlet series) as an infinite multiplicative one, and that translation is the whole point.

The picture: multiplication mirrors unique factorization

The intuition is that the primes are the multiplicative atoms of ℤ, and the Euler product is unique factorization written in the language of infinite series. Expand the product informally:

(1 + 2⁻ˢ + 2⁻²ˢ + ⋯)(1 + 3⁻ˢ + 3⁻²ˢ + ⋯)(1 + 5⁻ˢ + ⋯)⋯

To form one term you pick a power p⁻ᵃᵖˢ from each factor. Multiplying them gives (2^{a₂} 3^{a₃} 5^{a₅} ⋯)⁻ˢ = n⁻ˢ, where n = 2^{a₂}3^{a₃}5^{a₅}⋯. Because every integer n ≥ 1 has one and only one such prime factorization, each n⁻ˢ appears exactly once when you expand the product — no term missing, none duplicated.

So the product, fully multiplied out, reassembles precisely ∑ 1/nˢ. The Fundamental Theorem of Arithmetic is not a background fact here; it is the identity, viewed generating-function style. Absolute convergence is what licenses rearranging this infinite double-distribution without changing the value.

The key idea of the proof

The mechanism is a sieve realized analytically. Start with S = ∑ₙ 1/nˢ. Multiply by the 2-factor 2⁻ˢ and subtract:

S − 2⁻ˢS = (1 − 2⁻ˢ)S = ∑_{n odd} 1/nˢ,

because every even term 1/(2m)ˢ has been cancelled — this is the sieve of Eratosthenes in disguise. Repeat with 3: (1 − 3⁻ˢ)(1 − 2⁻ˢ)S removes all multiples of 3. Inductively, multiplying by (1 − p⁻ˢ) over the first k primes p₁ < ⋯ < p_k leaves

∏_{i≤k}(1 − pᵢ⁻ˢ) · ζ(s) = 1 + ∑′ 1/nˢ,

where the remaining sum runs only over n > 1 with all prime factors exceeding p_k. That tail is bounded by ∑_{n > p_k} n⁻Re(s) → 0 as k → ∞. Hence ∏ₚ(1 − p⁻ˢ)·ζ(s) = 1. Dividing gives the formula. Absolute convergence (Re(s) > 1) is what makes each subtraction legitimate and forces the tail to vanish.

Worked special case: s = 2 and the value ζ(2)

Take s = 2. The Euler product reads

∑ₙ 1/n² = (1 − 1/4)⁻¹(1 − 1/9)⁻¹(1 − 1/25)⁻¹(1 − 1/49)⁻¹⋯ = (4/3)(9/8)(25/24)(49/48)⋯

Euler famously evaluated the left side (the Basel problem, 1735) as π²/6 ≈ 1.6449. So the primes multiply up to π²/6:

• (4/3) = 1.3333, ×(9/8) = 1.5000, ×(25/24) = 1.5625, ×(49/48) = 1.5951, …
• Continue over primes 2,3,5,7,11,13,… and the partial products creep toward 1.6449.

Reading the identity backwards gives a startling corollary: the probability that two random integers are coprime is 6/π² ≈ 0.6079. Indeed that probability equals ∏ₚ(1 − 1/p²) = 1/ζ(2), because coprimality means no prime divides both, and the events "p divides both" are independent with probability 1/p². The Euler product is exactly the bookkeeping that makes this independence rigorous.

Why the hypothesis matters — and what breaks at Re(s) ≤ 1

Drop Re(s) > 1 and the identity collapses in an instructive way. At s = 1 the sum is the harmonic series ∑ 1/n, which diverges; correspondingly the product ∏ₚ(1 − 1/p)⁻¹ also diverges to +∞. Euler read this both ways: since the product blows up, there must be infinitely many primes (a finite product is finite). Taking logarithms, log∏ₚ(1−1/p)⁻¹ ≈ ∑ₚ 1/p, giving the sharper ∑ₚ 1/p = ∞ — primes are "denser" than the squares, whose reciprocals converge.

• For 0 < Re(s) ≤ 1 the Dirichlet series no longer converges absolutely and the term-by-term sieve is illegal; the product identity is simply false as stated.
• ζ(s) still exists there — but only via analytic continuation (Riemann, 1859), which extends ζ to all of ℂ ∖ {1} with a simple pole at s = 1.

Crucially, the Euler product proves ζ(s) ≠ 0 for Re(s) > 1 (a convergent product of nonzero factors is nonzero) — so every nontrivial zero must lie in the critical strip 0 ≤ Re(s) ≤ 1, the arena of the Riemann Hypothesis.

Significance: the template for all of analytic number theory

The Euler product is the prototype of every L-function. Its structural lesson — arithmetic (multiplicativity) becomes an Euler factorization — recurs throughout:

• Dirichlet L-functions L(s,χ) = ∏ₚ(1 − χ(p)p⁻ˢ)⁻¹, whose non-vanishing at s = 1 gives Dirichlet's theorem: infinitely many primes in every arithmetic progression a, a+q, a+2q, … with gcd(a,q) = 1.
• Dedekind zeta functions of number fields, Hecke and automorphic L-functions — each attaches an Euler factor to every prime, encoding local-global principles.
• The Prime Number Theorem (Hadamard & de la Vallée Poussin, 1896): controlling ζ's zeros near Re(s) = 1 — via the log-derivative −ζ′/ζ = ∑ Λ(n)n⁻ˢ, whose Euler product exposes the von Mangoldt weights — yields π(x) ∼ x/log x.

More generally, taking a logarithm of the Euler product turns products into sums and links ζ to prime-counting functions; the whole edifice of using complex analysis to count primes rests on this one factorization. It is, quite literally, where analytic number theory begins.

The Euler product across the s-plane: what holds where, and why the boundary Re(s) = 1 is decisive.

Region	Does ∑ 1/nˢ converge?	Does the product ∏ₚ(1−p⁻ˢ)⁻¹ converge?	What it tells you
Re(s) > 1	Yes, absolutely	Yes, absolutely; equals the sum	ζ(s) ≠ 0; Dirichlet series has an Euler factorization
s = 1	No (harmonic series diverges)	No (∏ (1−1/p)⁻¹ = ∞)	Divergence ⇒ infinitely many primes; ∑ 1/p diverges
0 < Re(s) ≤ 1	Conditionally / not at all	Does not converge	Need analytic continuation; product identity fails here
Re(s) ≤ 0	Diverges	Diverges	ζ defined only by continuation; trivial zeros at −2, −4, …