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Source. Theorem 1 (Теорема 1), printed p. 66 of the Russian original (physical PDF p. 2), read on the page image. The proof begins in section 2 (p. 69), which reduces Theorems 1 and 2 to the zero estimate Theorem 3 (p. 69); it was not read here.

Statement

For every q∈Cq\in\mathbb C with 0<∣q∣<10<|q|<1, among the numbers qq, P(q)P(q), Q(q)Q(q), R(q)R(q) there are at least three algebraically independent over Q\mathbb Q. Here (p. 65)

P(z)=1−24∑n=1∞σ1(n)zn,Q(z)=1+240∑n=1∞σ3(n)zn,R(z)=1−504∑n=1∞σ5(n)zn,P(z)=1-24\sum_{n=1}^{\infty}\sigma_1(n)z^n,\quad Q(z)=1+240\sum_{n=1}^{\infty}\sigma_3(n)z^n,\quad R(z)=1-504\sum_{n=1}^{\infty}\sigma_5(n)z^n,

with σk(n)=∑d∣ndk\sigma_k(n)=\sum_{d\mid n}d^k, are Ramanujan's functions (the Eisenstein series E2,E4,E6E_2,E_4,E_6 in the variable z=e2πiτz=e^{2\pi i\tau}); they satisfy θP=(P2−Q)/12\theta P=(P^2-Q)/12, θQ=(PQ−R)/3\theta Q=(PQ-R)/3, θR=(PR−Q2)/2\theta R=(PR-Q^2)/2 with θ=z d/dz\theta=z\,d/dz (formula (1)). Equivalently, the field Q(q,P(q),Q(q),R(q))\mathbb Q(q,P(q),Q(q),R(q)) has transcendence degree at least 33, the form in which Waldschmidt's exposé restates the theorem (Théorème 4).

Proof pointer

Ingredients named in sections 1 and 2 and in the expositions: Mahler's 1969 theorem that P,Q,RP,Q,R are algebraically independent over C(z)\mathbb C(z); the differential system (1); the zero estimate Theorem 3 (p. 69), which bounds ord⁡z=0A(z,P(z),Q(z),R(z))\operatorname{ord}_{z=0}A(z,P(z),Q(z),R(z)) by 2⋅1045L1L232\cdot10^{45}L_1L_2^3 for nonzero AA with deg⁡zA≤L1\deg_zA\le L_1, deg⁡xiA≤L2\deg_{x_i}A\le L_2; an auxiliary polynomial (Lemma 2.1, p. 69); and Philippon's algebraic independence criterion (his Theorem 2.11, which gives Lemma 2.5, p. 75). Expositions: Waldschmidt, Séminaire Bourbaki exposé 824 (1997), section 2.2 (p. 118) for the statement and section 2.5 (from p. 126) for the proof; the Lecture Notes in Mathematics 1752 chapter (not read). None of the proof was checked here.

Consequence used in the corpus

Corollary 2 (pp. 66--67): for algebraic qq, P(q),Q(q),R(q)P(q),Q(q),R(q) are algebraically independent, in particular transcendental; at q=1/2q=1/2 this gives the transcendence of ∑n≥1σ(n)/2n\sum_{n\ge1}\sigma(n)/2^n.

Coverage

Statement read on the page image; proof not read. Relied on as accepted literature: refereed (Mat. Sb.), Zbl 0898.11031, expounded in the Bourbaki exposé of November 1996, and followed by the 1997 Ostrowski Prize to Nesterenko.

Bears on. #250, through Corollary 2.