Question

Let's have some fun: Part III
Abstract Algebra

Answer 1

Let k be a positive integer:
Define
\[\huge k\mathbb{Z}={\left\{ n \ \left|\right| \ \frac{n}{k} \in \mathbb{Z} \right\} }\]
Or in other words, all multiples of k.

Answer 2

Show that
\[\huge 3 \mathbb{Z} \ \ncong \ 2 \mathbb{Z}\]
as rings...

(Show that 3Z and 2Z are not isomorphic as rings)

Answer 3

consider a mapping f from 3Z to 2Z and check whether it is is one-one,onto and homomorphism.....if it doesnt satisfy any one of the conditions.that means it is not isomorphic

Answer 4

Of course... but the task was to show that no such isomorphism exists :D

Answer 5

It'll literally take an eternity to check all the mappings :D

Answer 6

i dont think so...

Answer 7

Well, your solution, then? :)

Answer 8

why dont you start the question with a contradiction supposing that the mapping is isomorphism...

Answer 9

Mhmmm... and then...?

Answer 10

and then proving it wrong...

Answer 11

How?

Answer 12

function is one one that is sure..

Answer 13

check for onto and homomorphism....

Answer 14

for ring homomorphism...
f(x+y)=f(x)+f(y)
and f(xy)=f(x)f(y) for all x,y belong to Z

Answer 15

You're still not being very specific as to how you're going to go about proving this, though... ^^

Answer 16

I think I'll go with contradiction :)
What if there IS an isomorphism from 3Z to 2Z ?

Answer 17

Continue, though

Answer 18

Well, if they're isomorphic as rings, they have to be isomorphic as groups, right? :)

Answer 19

I... should think so?

Answer 20

Well, they're both cyclic, and we know with cyclic groups, the image of a generator under an isomorphism is a generator of the other group. 3Z is generated by 3, so its image under the isomorphism must be 2, or -2. Let's go with 2 :)

Answer 21

Let F be that isomorphism.

F(3) = 2
That means
F(3+3) = F(6) = F(3) + F(3) = 2 + 2 = 4
F(6) = 4
F(3 x 6) = F(18) = F(3) x F(6) = 2 x 4 = 8
F(3 + 3 + 3 + 3 + 3 + 3) = F(18) = 6 x F(3) = 6 x 2 = 12.
F(18) = 12
But it has been shown earlier that
F(18) = 8

This function is therefore not a very good one (LOL) and this isomorphism is bogus :D

Answer 22

That was messy, but I guess it'll do.