# Ring Isomorphism

• Apr 26th 2008, 02:59 PM
hercules
Ring Isomorphism - help needed
I need help with the following proof:

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

Thanks

and how do you Z that represents the set of integers in Latex.
• Apr 26th 2008, 03:08 PM
TheEmptySet
Quote:

Originally Posted by hercules
I need help with the following proof:

Show that if $\theta$:Z $\rightarrow$ Z is a ring isomorphism, then $\theta$ must be the identity mapping. Is there an additive group isomorphism $\theta$:Z $\rightarrow$ 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 $\mathbb{Z},+$

then $\theta(x)=-x$ is an isomorphism

$\theta(a+b)=-(a+b)=(-a)+(-b)=\theta(a)+\theta(b)$
and

$\theta (0)=-0=0$

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

Good luck.
• Apr 26th 2008, 06:31 PM
hercules
Quote:

Originally Posted by TheEmptySet
first \mathbb{Z} will give the integers symbol.

for the 2nd part consider group $\mathbb{Z},+$

then $\theta(x)=-x$ is an isomorphism

$\theta(a+b)=-(a+b)=(-a)+(-b)=\theta(a)+\theta(b)$
and

$\theta (0)=-0=0$

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

Good luck.

Thanks Emptyset.

• Apr 26th 2008, 08:25 PM
TheEmptySet
Maybe this is it?
Quote:

Originally Posted by hercules
I need help with the following proof:

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

Thanks

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

let $\Phi (x)$ an isomorphsim that is not the identity mapping.

Then there exists $a \in \mathbb{Z}$ such that

$a \ne \Phi (a)$

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

we can write a in a few different ways

$a = 1 \cdot a=\underbrace{1+1+1+1+...+1}_{a Times}$

since Phi is an isomorphism we know that

$\Phi (0)=0,\Phi(1)=1$

so know we get

$\Phi(a)=\Phi(1+1+1+...+1)= \Phi(1)+\Phi(1)+ \Phi(1)+ \Phi(1)+ ...\Phi(1)=a \cdot \Phi(1) =a \cdot 1=a$
$\Phi(a)=\Phi(1\cdot a)=\Phi(1) \cdot \Phi(a)=1\cdot \Phi(a)=\Phi(a)$

• Apr 26th 2008, 08:27 PM
Jhevon
Quote:

Originally Posted by TheEmptySet

let $\Phi (x)$ an isomorphsim that is not the identity mapping.

Then there exists $a \in \mathbb{Z}$ such that

$a \ne \Phi (a)$

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

we can write a in a few different ways

$a = 1 \cdot a=\underbrace{1+1+1+1+...+1}_{a Times}$

since Phi is an isomorphism we know that

$\Phi (0)=0,\Phi(1)=1$

so know we get

$\Phi(a)=\Phi(1+1+1+...+1)= \Phi(1)+\Phi(1)+ \Phi(1)+ \Phi(1)+ ...\Phi(1)=a \cdot \Phi(1) =a \cdot 1=a$
$\Phi(a)=\Phi(1\cdot a)=\Phi(1) \cdot \Phi(a)=1\cdot \Phi(a)=\Phi(a)$

If $\theta: \mathbb{Z}\mapsto \mathbb{Z}$ is a ring isomorphisms then certainly it is a group automorphism of $\left< \mathbb{Z},+\right>$. Group isomorphism on cyclic groups are completely determined on the generator. Since we must have $\theta (1) = 1$ (for that is the definition of a commutative unitary ring homomorphism) it is completely determined, i.e. unique, and in fact the identity mapping.