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TheoremStatement: Literature-sourcedProof: AI-generatedprecheck passjudge pass (gpt-5.6-terra)audited 2026-09-04
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Weak approximation for rational places

Statement

Let v1,…,vr be distinct places of Q. For each i, let ai∈Q, and let εi>0. Then there exists x∈Q such that

∣x−ai∣vi<εi(1≤i≤r).

Consequently, after completing Q at these places, the diagonal copy of Q is dense in the finite product of the local fields.

Facts & Assumptions

Given: Distinct places v1,…,vr, rational targets a1,…,ar, and positive reals ε1,…,εr.

[L1]

The places of Q are the archimedean place and the prime places (Places of the rationals).

Proof

technique · constructive
1.1L1givenconstruct

Reorder the places so that v1,…,vs are the finite places p1,…,ps and, if the archimedean place occurs, it is vr=∞. Choose integers Ni≥1 with pi−Ni<εi for 1≤i≤s. Let D0 be a common positive denominator of the finite targets a1,…,as, and replace it by D:=TD0, where T is a large integer coprime to every pi; this keeps every Dai integral and lets us later make M/D as small as we wish.

2.1L2step 1.1construct

Put M:=∏i=1spiNi+vpi(D). By [L2], there is an integer y such that y≡Dai(modpiNi+vpi(D))(1≤i≤s). Then for x0:=y/D one has vpi(x0−ai)≥Ni, hence ∣x0−ai∣pi≤pi−Ni<εi for every finite place in the list.

3.1step 1.1step 2.1cases

If ∞ is not among the chosen places, then x:=x0 works. Otherwise every number of the form xk:=x0+kM/D has the same finite-place congruence conditions as x0, because vpi(M/D)=Ni(1≤i≤s). By taking T in step 1.1 so large that M/D<εr, the arithmetic progression x0+Z(M/D) has mesh smaller than εr, so some integer k satisfies ∣xk−ar∣∞<εr.

4.1step 2.1step 3.1discharge-construct∎

The chosen x satisfies all requested inequalities. The density formulation is the same statement with the local targets first approximated by rational elements in each factor.

Depends on

Used by

Dependency tree · two levels

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Sources