Thread: Ring Isomorphism

I need help with the following proof:

Show that if :Z Z is a ring isomorphism, then must be the identity mapping. Is there an additive group isomorphism :Z Z other than the identity mapping?

Thanks

and how do you Z that represents the set of integers in Latex.

Originally Posted by hercules

I need help with the following proof:

Show that if :Z Z is a ring isomorphism, then must be the identity mapping. Is there an additive group isomorphism :Z Z other than the identity mapping?

Thanks

and how do you Z that represents the set of integers in Latex.

first \mathbb{Z} will give the integers symbol.

for the 2nd part consider group

then is an isomorphism


and



For the first part i am not sure. I will think about it.

Good luck.

Originally Posted by TheEmptySet

first \mathbb{Z} will give the integers symbol.

for the 2nd part consider group

then is an isomorphism


and



For the first part i am not sure. I will think about it.

Good luck.

Thanks Emptyset.

Can someone please help me with the first part. My major problem.

Originally Posted by hercules

I need help with the following proof:

Show that if :Z Z is a ring isomorphism, then must be the identity mapping. Is there an additive group isomorphism :Z Z other than the identity mapping?

Thanks

and how do you Z that represents the set of integers in Latex.

I have been thinking about this and this is what i came up with...

proof by contradiction

let an isomorphsim that is not the identity mapping.

Then there exists such that



I will assume a is positive(it can be done if a is negative)

we can write a in a few different ways



since Phi is an isomorphism we know that



so know we get




This is a contradiction.

Originally Posted by TheEmptySet

I have been thinking about this and this is what i came up with...

proof by contradiction

let an isomorphsim that is not the identity mapping.

Then there exists such that



I will assume a is positive(it can be done if a is negative)

we can write a in a few different ways



since Phi is an isomorphism we know that



so know we get




This is a contradiction.

I was thinking the same thing. I don't know if that makes you more or less confident in your solution...

If is a ring isomorphisms then certainly it is a group automorphism of . Group isomorphism on cyclic groups are completely determined on the generator. Since we must have (for that is the definition of a commutative unitary ring homomorphism) it is completely determined, i.e. unique, and in fact the identity mapping.