作者sckm160913 (Goat)
看板NTU-Exam
標題[試題] 100下 陳其誠 代數導論二 期末考
時間Thu May 31 20:08:21 2012
課程名稱︰代數導論二
課程性質︰必修
課程教師︰陳其誠
開課學院:理學院
開課系所︰數學系
考試日期(年月日)︰2012/05/31
考試時限(分鐘):170 mins
是否需發放獎勵金:是
(如未明確表示,則不予發放)
試題 :
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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