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Real infinite primes ramify? #216

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user202729 opened this issue Nov 17, 2023 · 1 comment
Open

Real infinite primes ramify? #216

user202729 opened this issue Nov 17, 2023 · 1 comment
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@user202729
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user202729 commented Nov 17, 2023

image

I think the real infinite primes are not supposed to ramify, they should be inert.

(also that the two conjugate complex embeddings actually correspond to the same complex infinite prime...?)

image

This according to Neukirch.

(in the context of Artin reciprocity: If less primes ramifies, the theorem would apply to more cases i.e. the theorem is stronger)


Maybe I work out an example, but I'm not sure if I understood everything correctly.

Consider the Galois extension L/K where L = ℚ(∛2, ω) and K = ℚ(∛2).

In K, the real primes are σ (the only real embedding) and τ (the pair of complex embeddings).

L is totally complex and has 6 complex embeddings, thus 3 distinct primes.

σ is inert in L, with inertia degree 2 (the residue field becomes ℂ from ℝ).

τ splits into the product of two complex primes, the inertia degree is 1 (the residue field remains ℂ).

Note that whenever the extension L/K is Galois, then the inertia degrees are all equal -- if any infinite prime τ above σ is complex, then all infinite primes τ above σ are complex (because of normality of L over K, the image of L is the same for all embedding into the algebraic closure).

@vEnhance vEnhance added bug Something isn't working math Heavy math required labels Nov 23, 2023
@user202729
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Looks like the convention can be done both ways.

image

The former way feels a bit more natural though (let f be the field extension's degree in both cases). That said, the residue field λ(𝔓)/κ(𝔭) (in the finite prime case) and the field completion L_𝔓/K_𝔭 (in the infinite prime case) is not the same thing...

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