Introduction to Algebraic Number Theory by Frédérique Oggier
By Frédérique Oggier
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Extra resources for Introduction to Algebraic Number Theory
We consider the projection π : O → O/pO. We have that π(pi ) = π(pO + fi (θ)O) = fi (θ)O mod pO. Consequently, pi is a prime ideal of O, since fi (θ)O is. Furthermore, since pi ⊃ pO, we have pi | pO, and the inertial degree fpi = [O/pi : Fp ] is the degree of φi , while epi denotes the ramification index of pi . Now, every prime ideal p in the factorization of pO is one of the pi , since the image of p by π is a maximal ideal of O/pO, that is e epg pO = p1p1 · · · pg and we are thus left to look at the ramification index.
Let α be a unit. Then α ∈ Z× p ⇐⇒ α ∈ Zp and 1 ∈ Zp ⇐⇒ |α|p ≤ 1 and |1/α|p ≤ 1 ⇐⇒ |α|p = 1. α 3. We are now interested in the ideals of Zp . Let I be a non-zero ideal of Zp , and let α be the element of I with minimal valuation ordp (α) = k ≥ 0. We thus have that α = pk (ak + ak+1 p + . ) where the second factor is a unit, implying that αZp = pk Zp ⊂ I. We now prove that I ⊂ pk Zp , which concludes the proof by showing that I = pk Zp . If I is not included in pk Zp , then there is an element in I out of pk Zp , but then this element must have a valuation smaller than k, which cannot be by minimality of k.
The symmetric bilinear form K ×K (x, y) → → Q TrK/Q (xy) is non-degenerate. Proof. Let us assume by contradiction that there exists 0 = α ∈ K such that TrK/Q (αβ) = 0 for all β ∈ K. By taking β = α−1 , we get TrK/Q (αβ) = TrK/Q (1) = n = 0. 2. PRIME DECOMPOSITION Now if we had that ∆K = 0, there would be a non-zero column vector (x1 , . . , xn )t , xi ∈ Q, killed by the matrix (TrK/Q (αi αj ))1≤i,j≤n . Set γ = n i=1 αi xi , then TrK/Q (αj γ) = 0 for each j, which is a contradiction by the above lemma.