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(math311)[2008](f)final~65^_10501.pdf
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Solutions to Final Exam for Math 311, 2008
Problem 1(15 points). Prove that : C M2(R)given by(a + bi)=
. 1
.
a b
3
is a homomorphism of rings. Is one-to-one? Is onto?
.3ba
Answer: Check ((a1 +b1i)+(a2 +b2i)) = (a1 +b1i)+ (a2 +b2i)and
((a1 +b1i) (a2 +b2i)) = (a1 +b1i) (a2 +b2i)directly. is one-to-one,
but it is not onto.

Problem 2 (20points) Computation.
(1). Find the order of8 Z12. Answer:3.
(2). Find the order of .i C. . Answer:4.

2
(3). If a in a group G has order 3, .nd the order of a. Answer:3. (4). List all the elements in G15. Answer: G15 = {1,2,4,7,8,11,13,14}(5). Find thekernel of the homomorphism :G15 G15, where is given
2
by (a)= a. Answer: ker()= {1,4,11,14}.
Problem 3(20points). Multiplechoice(no reasons needed, thereisonly one
correct answer for eachproblem).
(1). Whichof the following rings is an integral domain?
(a). Z. (b). 2Z. (c). Z Z. (d). Z33.
(2). Whichof the following rings is a .eld?

(a) Z10. (b). Z. (c). Z Z (d). Z11.
(3). Which of the following maps from C to itself is a homomorphism of
rings?

3
(a) (z)=2z. (b). (z)= z.
(c). (z)= .z. (d). (a +bi)= a .bi .
(4). Which of the following maps from the additive group R to the mul-tiplicative group R. is a group homomorphism?

2 a
(a). (a)=2a (b). (a)= a(c). (a)= |a| (d). (a)= e.
Answer: (1) a. (2) d. (3) d. (4) d.
Problem 4(15 points) Let G be a .nite group, a G be a given element. Suppose that: G G givenby
(x)= axaxa
is a homomorphism, prove that
33
(1). (e)= a. (2). a= e.
(3). Use (ax)=(a)(x)to prove thatax = xa for all x G.
(4). G is an abelian group.

3
Proof. (1) (e)= aeaea = a.
3
(2)
Sinceisa homomorphism,(e)= e,so a= e.

(3)
Sinceisa homomorphism,wehave(ax)=(a)(x), that is,


a(ax)a(ax)a = a 5(axaxa).
Canceling a2 on the left and axa on the right, we have
4
xa = a x.
3
Using a= e proved in (2), we get xa = ax.
3
(4) Using ax = xa and a= e, wehave
32 2
(x)= axaxa = ax = x
for all x G. Since is a homomorphism, (xy)=(x)(y)for all x,y G. That is,(xy)(xy)= xxyy for all x,y G. Canceling x on the left and y on the right, we have
xy = yx
for all x,y G. This proves G is abelian group.
Problem 5(10 points) Let F be a .nite .eld with m elements. Prove that am .a =0for everya F.
Proof. The proof is similar to the proof of the Fermats little theorem.
.
Since F is a .eld, F= F .{0}is a group under multiplication. The
. m.1
multiplicative group Fhas order m .1. So a=1for alla F.. This implies am .a =0 for all a F..If a = 0, clearly am .a = 0. Therefore wehaveproved am .a =0for every a F.
Problem 6(20points) Let R be a .nite commutative ring with unity1. An element a R is calleda nilpotent elementif there existsapositiveinteger
n
n such that a=0.
(1). Prove that if a and b areboth nilpotent elements, then ab and a+b are
nilpotent el