A rational which is a p-adic integer for all p is an integer.

[kbuzzard]

Sounds easy, but when I say "p-adic integer for all p" I unfortunately actually mean

 ∀ (v : IsDedekindDomain.HeightOneSpectrum (𝓞 ℚ)),
  ↑((algebraMap ℚ (FiniteAdeleRing (𝓞 ℚ) ℚ)) x) v ∈ IsDedekindDomain.HeightOneSpectrum.adicCompletionIntegers ℚ v

so that adds a bit of a twist. Probably one should start by proving that for all such v, there's a prime p : Nat such that v=(p), and then show that ↑((algebraMap ℚ (FiniteAdeleRing (𝓞 ℚ) ℚ)) x) v ∈ IsDedekindDomain.HeightOneSpectrum.adicCompletionIntegers ℚ v is an interesting way of saying that p doesn't divide the denominator of v.

[Ruben-VandeVelde]

claim

at least for a little bit

[Ruben-VandeVelde]

claim