課程名稱︰代數導論一
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）︰2011/12/29
考試時限（分鐘）：190 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. 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
(↑實數域的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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