课程名称︰代数导论一 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）︰2011/12/29 考试时限（分钟）：190 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. In this examina- tion, G denotes a group and e always is the identity of G. Also, R denote a ring, "I" denote an ideal of R, and if R is commutative and a∈R, (a) = { ar | r∈R }. If ψ is a homomorphism, then ker(ψ) denote the kernel. Let Z denote the ring of integers and let Z[i] = { a + bi | a,b ∈ Z }, where i^2 = -1. (1) (5 points each) "yes" or "no". Either give a brief reason or give a coun- ter example. _ (a) If there exists a surjective group homomorphism ψ:G → G with _ o(ker(ψ)) = 13 and o(G) = 3, then G is a cyclic group. (b) The number 29 is an irreducible element in Z[i]. (c) The number 2011 is an irreducible element in Z[i]. (d) G is cyclic, if o(G) = 343 and the order of G[7] := { g∈G | g^7 = e } equals 7. (e) If o(G) = 343 and the order of G[7] equals 49, then x^49 = e for all x∈G. (f) If R is an integral domain with 1∈R and I = (a) is maximal ideal, then a must be irreducible. (g) If R is commutative and the order (cardinality) of R is finite, then R is a field. (h) If R is an integral domain such that 6a = 0 for all a∈R, then either 2a = 0, for all a or 3a = 0, for all a. (2) (15 points) Let R be the ring of all continuous functions on the closed interval [-1,1]. Show that R is not an integral domain. Show that if ψ:R → R is a surjective ring homomorphism, than there wxist some (↑实数域的R) a∈[-1,1] such that ψ(f) = f(a), for every f∈R (5 points). (3) (15 points) Suppose 1∈R and the only right-ideals of R are {0} and R. Show that for every nonzero a∈R, there exists some a'∈R such that aa' = 1 (10 points. Hint: consider aR = { ar | r∈R }). Show that R must be a division ring (5 points). ＿ ＿ (4) (10 points) Show that Z[√-6] = { a + b√-6 | a,b ∈ Z} is not a Euclidean ＿ ＿ (Hint: -2 * 3 = √-6 * √-6 ). (5) (10 points) Suppose R is an integral domain and a,b ∈ R. Show that |n| |n| M := { n∈Z | a = b } is an ideal of Z (5 points). Show that if there exist some n∈M so that n + 1∈M, then a = b (5 points). (6) (10 points) Let R = Z[i]. Show that R/(3) is a field (5 points) whose order (cardinality) equals 9 (5 points). --

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