Ramification index and inertia degree #
Given P : Ideal S lying over p : Ideal R for the ring extension f : R ā+* S
(assuming P and p are prime or maximal where needed),
the ramification index Ideal.ramificationIdx f p P is the multiplicity of P in map f p,
and the inertia degree Ideal.inertiaDeg f p P is the degree of the field extension
(S / P) : (R / p).
Main results #
The main theorem Ideal.sum_ramification_inertia states that for all coprime P lying over p,
Ī£ P, ramification_idx f p P * inertia_deg f p P equals the degree of the field extension
Frac(S) : Frac(R).
Implementation notes #
Often the above theory is set up in the case where:
Ris the ring of integers of a number fieldK,Lis a finite separable extension ofK,Sis the integral closure ofRinL,pandPare maximal ideals,Pis an ideal lying overpWe will try to relax the above hypotheses as much as possible.
Notation #
In this file, e stands for the ramification index and f for the inertia degree of P over p,
leaving p and P implicit.
If b mod p spans S/p as R/p-space, then b itself spans Frac(S) as K-space.
Here,
pis an ideal ofRsuch thatR / pis nontrivialKis a field that has an embedding ofR(in particular we can takeK = Frac(R))Lis a field extension ofKSis the integral closure ofRinL
More precisely, we avoid quotients in this statement and instead require that b āŖ pS spans S.
Let V be a vector space over K = Frac(R), S / R a ring extension
and V' a module over S. If b, in the intersection V'' of V and V',
is linear independent over S in V', then it is linear independent over R in V.
The statement we prove is actually slightly more general:
- it suffices that the inclusion
algebraMap R S : R ā Sis nontrivial - the function
f' : V'' ā V'doesn't need to be injective
If p is a maximal ideal of R, and S is the integral closure of R in L,
then the dimension [S/pS : R/p] is equal to [Frac(S) : Frac(R)].
R / p has a canonical map to S / (P ^ e), where e is the ramification index
of P over p.
If P lies over p, then R / p has a canonical map to S / P.
This can't be an instance since the map f : R ā S is generally not inferable.
Instances For
The inclusion (P^(i + 1) / P^e) ā (P^i / P^e).
Instances For
S ā§ø P embeds into the quotient by P^(i+1) ā§ø P^e as a subspace of P^i ā§ø P^e.
See quotientToQuotientRangePowQuotSucc for this as a linear map,
and quotientRangePowQuotSuccInclusionEquiv for this as a linear equivalence.
Instances For
S ā§ø P embeds into the quotient by P^(i+1) ā§ø P^e as a subspace of P^i ā§ø P^e.
Instances For
Quotienting P^i / P^e by its subspace P^(i+1) ā§ø P^e is
R ā§ø p-linearly isomorphic to S ā§ø P.
Instances For
Since the inclusion (P^(i + 1) / P^e) ā (P^i / P^e) has a kernel isomorphic to P / S,
[P^i / P^e : R / p] = [P^(i+1) / P^e : R / p] + [P / S : R / p]
If p is a maximal ideal of R, S extends R and P^e lies over p,
then the dimension [S/(P^e) : R/p] is equal to e * [S/P : R/p].
If p is a maximal ideal of R, S extends R and P^e lies over p,
then the dimension [S/(P^e) : R/p], as a natural number, is equal to e * [S/P : R/p].
Properties of the factors of p.map (algebraMap R S) #
Chinese remainder theorem for a ring of integers: if the prime ideal p : Ideal R
factors in S as ā i, P i ^ e i, then S ā§ø I factors as Ī i, R ā§ø (P i ^ e i).
Instances For
Chinese remainder theorem for a ring of integers: if the prime ideal p : Ideal R
factors in S as ā i, P i ^ e i,
then S ā§ø I factors R ā§ø I-linearly as Ī i, R ā§ø (P i ^ e i).
Instances For
The fundamental identity of ramification index e and inertia degree f:
for P ranging over the primes lying over p, ā P, e P * f P = [Frac(S) : Frac(R)];
here S is a finite R-module (and thus Frac(S) : Frac(R) is a finite extension) and p
is maximal.
Ideal.sum_ramification_inertia, in the local (DVR) case.