Question

Please, explain me
is Z2 a subring of Z6?

Answer 1

The book says: \(Z_2\) is not even a subset of \(Z_6\). I don't get it.

Answer 2

@xapproachesinfinity

Answer 3

Well, what defines a ring?

Answer 4

a set equipped 2 operation, namely + and * in which
for +: associativity , abelian
for *: associativity, left/right distribution
distribute * over +, that is a(b+c) = ab +ac from both sides (left and right)

Answer 5

Does it have to be closed as well under these operations?

Answer 6

sure

Answer 7

I am asking because I am partly not sure and partly think that this is the way to the answer you're looking for. It seems to me that in Z2 you will have 1+1=0 and in Z6 1+1=2, but I am going to get out my abstract algebra book because I am curious to figure this out with you now! =D

Answer 8

My question on it is: if we check the axiom of subring, that is \(Z_2\) has no additive opposite element in it, I agree that \(Z_2\) is not a subring of \(Z_6\)
but the reason they give me is \(Z_2\) is not a subset of \(Z_6\) why?

Answer 9

\(Z_6=\{[0],[1],[2],....,[5]\}\)
and \(Z_2=\{[0],[1],[2]\}\)
obviously, \(Z_2\subset Z_6\)

Answer 10

I believe that 0, 2, 4 forms a subset of Z6 since every operation on this set is closed.

Answer 11

For example, 0+0=0, 0+2=2, 0+4=4, 2+2=4, 2+4=0, 0*0=0, 0*2=0, 0*4=0, 2*2=4, 2*4=2

Answer 12

You can't do that with Z2 since 0 and 1 won't be closed and form a subset, does that make sense? It's all because 1+1=2 in Z6, otherwise it would work since 0*0=0, 0*1=0, 0+0=0, 0+1=1 all work otherwise.

Answer 13

eh didn't the ring theory, number theory or whatever this is related to
Z2 means class with mod 2? right?

Answer 14

oh, I confused!! what is the definition of a subset of a set?

Answer 15

@xapproachesinfinity engruent class, not just class

Answer 16

https://www.google.com/?gws_rd=ssl#q=subset+of+a+set
hence, trivial,

Answer 17

http://www.math.niu.edu/~beachy/abstract_algebra/guide/section/51soln.pdf
problem 24a

Answer 18

Yeah I don't know the terminology very well, is what I was saying right or are we talking about different problems now?

Answer 19

Oh! Yes look at 24b it is exactly what I am describing, just a different example! =D

Answer 20

I understand 24b also, :) but 24a hehehe... how???

Answer 21

Z2 = {0, 1} is a subset of Z6

Answer 22

Oh you're refering to that link.. idk why it says Z2 is not a subset of Z6

Answer 23

oh, I put [2] on it :) my bad
but it is trivial that Z2 is a subset of Z6, right? why they said so?

Answer 24

I think they meant to write subring? Yeah seems weird, I thought maybe I just didn't know the terminology but now I'm just completely confused since I see what you guys are saying. Good luck!

Answer 25

Thank you. It is not my assignment. I read and wonder how, just want to clarify it. My problem is on the previous post, not this. :)