Hi, lately i started posting a lot in this part of the forum (group theory mainly), cause i had this course on faculty and im not that good at abstract maths. but im trying to tackle some basics down so i can get my self into it more.., now let's get down to the task i wanted to ask you folks here:

This task was on my exam and i partially solve it but i need to know if the rest of it im doing ok:

Task: Let Z[i] = {a+bi | a,b E Z} is set of complex numbers with the normal addition and multiplication defined for complex numbers, check the following:

1)Is Z[i] a ring?

2)Does Z[i] posses divisors of zero?

3)Is Z[i] a field?

4)Is (x-y)(x+y)=x^2-y^2 for every element of Z[i] (give explanation)

5)Does Z[i] posses finite cyclic subgroup?

My explanation:

First as the task states the set is discrete since the real and imaginary part belongs to Z.

1)For (Z[i], +, *) to be ring (+ addition, * multiplication defined standard for complex numbers) by definition

a. Z[i], + must be abelian group

b. Z[i], * must be semigroup (* posses associativity law on the elements of Z[i])

c. Distribution law for * over + must hold in the structure

Z[i] is abelian group!

-first set closure: let x,y E Z[i] such that x=a+bi and y=c+di for some a,b,c,d E Z

now x+y=(a+bi)+(c+di) = (a+c)+i(b+d) , a+b E Z, b+d E Z => x+y E Z

- associativity: let x,y,z E Z[i] x and y defined as previously and z = e+fi, e,f E Z

now x+(y+z) = (a+bi)+((c+di)+(e+fi)) = (a+bi)+((c+e)+i(d+f)) = a+(c+e) + i(b+(d+f)) = (a+c)+e + i((b+d)+f)) = ((a+c)+i(b+d))+(e+if)=((a+ib)+(c+id))+(e+if)=(x+y) +z = > associativity holds.

- identity element in Z[i], + is 0, let x E Z[i] => x+0=(a+bi)+(0+i0)=(a+0)+i(b+0)=a+ib=x.

- inverse element to x is -x that is -(a+ib)=-a-bi, indeed: x+(-x) = (a+ib)+(-a-ib)=

(a+(-a))+i(b+(-b))=(a-a)+i(b-b)=0+i0=0 (the id in Z[i] according to +).

so from all above => Z[i] is group under op. + for complex numbers, now all left over is to prove that Z[i] is closed under * and distribution law for * over + holds..

Z[i] is closed under * indeed, let x,y E Z => (a+ib)*(c+id)=ac+adi+bci-bd=ac-bd+i(ad+bc), ac-bd E Z, as well as ad+bc => x*y E Z[i], thus is closed.

now distribution laws:

let, x,y,z E Z[i], then x*(y+z) = x*((c+id)+(e+if)) = x*((c+e)+i(d+f))=(a+ib)*((c+e)+i(d+f))=(a(c+e)+ia( d+f) + ib(c+e)-b(d+f)) = (a(c+e)-b(d+f))+i(a(d+f)+b(c+e))

now the other side of the identity:

x*y+x*z = (a+ib)*(c+id)+(a+ib)*(e+if)=ac-bd+i(ad+bc)+ae-bf+i(af+be)=ac+ae-bd-bf+iad+ibc+iaf+ibe=a(c+e)-b(d+f)+i(a(d+f)+b(c+e)) and this is equal to the result bit upper so distribution law holds, thus Z[i],+,* is ring.

Now that aside, 1) is complete i think this one i did ok, i just wrote the whole work cause i think i can use this fact that Z[i] is ring to solve part 4) which most concerns me and i didnt solve it on the exam cause i didnt even look at it dont know whu :P,

the second one divisors of zero are simply x*y=0 but x and y are not equal to 0.

(a+ib)*(c+id)=0+i0 => ac-bd = 0, ad+bc = 0 => ac=bd and ad=-bc, from where a=bd/c and b=ac/d => d=-bc/(bd/c)=> -bc^2/bd and similar you express c and find the values of all a,b,c,d such that x*y=0 for x,y different then 0.. nvm this one is not that hard to do i guess..

Now task 3) about field i know how to do it, now about task 4, i tought of it this way:

Since i prove that Z[i] is ring follows that Z[i],+ is abelian group thus there's inverse element to every element in Z[i] so the cancellation laws hold in Z[i] now, i can take it down this way (im not sure if this is the right approach):

(x-y)*(x+y)=x^2-y^2, now for x^2 -x^2 is inverse according to +, so i apply it on the left side for both sides of the identity => -x^2+(x-y)(x+y)=-x^2+x^2-y^2 =>

cancellation law kicks in => -x^2+(x-y)(x+y)=-y^2, now for -y^2 inverse is y^2 now i apply y^2 on right sides on both sides of the identity => -x^2+(x-y)(x+y)+y^2=e (the identity or 0), now -x^2+(x-y)(x+y)+y^2=-x^2+x^2+xy-yx-y^2+y^2, cancellation law now again kicks in thus => +xy-yx = e, now cause for normal multiplication commutativity law holds => +xy-xy=e => e=e, so indeed everything cancelled on the long run, but i dont know if this is consistent as proof, i need some help from you guys to assure me if im doing ok, thanks!!

And for the last one, can i just say that there's no generator in Z[i] thus there cant be finite cyclic subgroup..