By Ehud de Shalit

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N be n distinct embeddings of K in another extension L/F, which are the identity on F . Let ω1 , . . , ωn be a basis of K over F. 4) det(σi (ωj )) = 0. Proof. 5) ci σi (ωj ) = 0. i=1 Since the σi are F -linear, and the ωj form a basis of K over F, it follows that ci σi = 0 identically on K. This contradicts Artin’s theorem. 2. Norm and Trace. Let L/K be a finite separable field extension and embed it in a Galois extension M/K. Let Γ = EmbK (L, M ) be the set of n = [L : K] embeddings of L into M over K.

However, if i = j then (ζ i − ζ j ) = p is relatively prime to l, and we get that χ(σl ) = l. Since χ is an isomorphism from G to (Z/pZ)× , the order of σl is the (multiplicative) order of lmodp. We have obtained the following theorem. 42 4. 1. Let l be a prime different than p. Then under the identification of G with (Z/pZ)× , σl maps to l and the order of Gl is the minimal f such that lf ≡ 1modp. 1. Let p = 7, so that G is cyclic of order 6. It has four subgroups, {1} , H2 = {±1}√ , H3 = {1, 2, 4} and G.

N−1 ). 10) e OK /pOK OK /pi i . 11) OK /pOK Z[X]/(f, p) ¯ Fp [X]/(f) Fp [X]/(hi )ei . Both decompositions are as direct sums of rings that can not be further decomposed as direct sums (because they are local - a local ring is not a product of two subrings). 12) 1= εi of 1 as a sum of mutually orthogonal minimal idempotents (ε2i = εi , εi εj = 0 if i = j, and εi is not the sum of two mutually orthogonal idempotents). [Given a decomposition R = Ri let εi be the unit of Ri . ] It is an easy exercise to show that such a decomposition of 1 is unique.