课程名称︰代数导论二 课程性质︰必修 课程教师︰陈其诚 开课学院：理学院 开课系所︰数学系 考试日期（年月日）︰2012/05/31 考试时限（分钟）：170 mins 是否需发放奖励金：是 （如未明确表示，则不予发放） 试题 : Write your answer on the answer sheet. We give partial points. Let p denote a prime number. The order of a field is the cardinality. In every field, 0 ≠ 1. (1) (36 points) "yes" or "no". Either give a brief reason or give a counter example. 7 points each. (a) There is no field of order 100. (b) A field of order 343 has exactly 3 different subfield. (c) The congruent equation x^3 ≡ a (mod p) has an integer solution if and only if 3 | p - 1 and a^((p - 1)/3) ≡ 1 (mod p). (d) If K is a finite extension over Q, then K = Q(a), for some a ∈ K. (e) Every quadratic (degree 2) extension over Q is a Galois extension. (f) Every degree 3 extension over Q is a Galois extension. (2) (24 points) In the following cases, find the Galois group of K over F (8 points each). (a) F = Q and K is the splitting field of x^3 - 3. (b) F = Q and K is the splitting field of x^5 -1. (c) K is a Galois extension over Q with Galois group G and F is the fixed field of a subgroup H of G. (3) (8 points) Find a Galois extension over Q with the Galois group isomorphic to S . 7 (4) (7 poins) Find a Galois extension over Q with the Galois group isomorphic to C , the cyclic group of order 7. 7 (5) (18 points) Prove the following assertions (6 points each). (a) If K is an extension of Q, then every automorphism of K leaves Q fixed. (b) A subgroup of a solvable group is also solvable. (c) If K = F(a) is the splitting filed of the minimal polynomial f(x) ∈ F[x] of a and b is another root of f(x), then there is exactly one automorphism ψ of K leaving F fixed such that ψ(a) = b. (6) (7 points) If n is an integer, then every field contained in Q(e^(2πi/n)) is Galois extension of Q. --

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