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Exercise 2176Number theory › IrrationalityM1Added 3 Nov 2026

Schanuel's conjecture (statement)

#Schanuel#conjecture#transcendence

Statement

State Schanuel's conjecture and explain its consequences.

Hint

If α1,…,αn∈C\alpha_1,\ldots,\alpha_n\in\mathbb{C} are Q\mathbb{Q}-linearly independent, then Q(α1,…,αn,eα1,…,eαn)\mathbb{Q}(\alpha_1,\ldots,\alpha_n,e^{\alpha_1},\ldots,e^{\alpha_n}) has transcendence degree ≥n\ge n.

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Solution

Conjecture. If α1,…,αn\alpha_1,\ldots,\alpha_n are Q\mathbb{Q}-linearly independent complex numbers, then the field Q(α1,…,αn,eα1,…,eαn)\mathbb{Q}(\alpha_1,\ldots,\alpha_n,e^{\alpha_1},\ldots,e^{\alpha_n}) has transcendence degree at least nn over Q\mathbb{Q}.

Consequences.
- Lindemann–Weierstrass theorem (special case).
- Algebraic independence of ee and π\pi (take n=2n=2, α1=1\alpha_1=1, α2=iπ\alpha_2=i\pi).
- Gelfond–Schneider theorem.

Conclusion. Schanuel's conjecture is a vast generalization; it remains open and is one of the deepest conjectures in transcendence theory.