# Ring Isomorphism

• Sep 8th 2011, 04:42 AM
Bernhard
Ring Isomorphism
Can anyone help with the following problem:

Prove that the rings $\displaystyle 2\mathbb{Z}$ and $\displaystyle 3\mathbb{Z}$ are not isomorphic?

Be grateful for help

Peter
• Sep 8th 2011, 04:48 AM
HallsofIvy
Re: Ring Isomorphism
How many members does $\displaystyle 2\mathbb{Z}$? What about $\displaystyle 3\mathbb{Z}$?
• Sep 8th 2011, 04:53 AM
Bernhard
Re: Ring Isomorphism
2Z = { ... -4.-2, 0, 2, 4, 6, ....} so it is infinite

3Z = { ... -6. -3, 0, 3, 6, .... } so it is infinite

... or have I got something wrong?

Peter
• Sep 8th 2011, 05:51 AM
zoek
Re: Ring Isomorphism
Quote:

Originally Posted by Bernhard
Can anyone help with the following problem:

Prove that the rings $\displaystyle 2\mathbb{Z}$ and $\displaystyle 3\mathbb{Z}$ are not isomorphic?

Be grateful for help

Peter

Let $\displaystyle f:2\mathbb{Z} \to 3\mathbb{Z}$ be a ring isomorphism. Then $\displaystyle f(2) \in 3\mathbb{Z}$. Thus $\displaystyle f(2) = 3 x$, $\displaystyle x \in \mathbb{Z}$.

Therefore $\displaystyle f(4) = f(2+2) = f(2)+f(2) = 6x$ and $\displaystyle f(4)=f(2\cdot2)=f(2) f(2)=9x^2$.

So, $\displaystyle 6x=9x^2 \Rightarrow x=0$ or $\displaystyle x=2/3$.

$\displaystyle x=0$ means $\displaystyle f(2)=0$, so $\displaystyle f(2\lambda ) = f(2)f(\lambda )=0$, $\displaystyle \forall \lambda \in \mathbb{Z}$ which means that $\displaystyle f$ is not an isomorphism,

and $\displaystyle x=2/3$ with $\displaystyle x\in \mathbb{Z}$ is Absurd.

So, the rings $\displaystyle 2\mathbb{Z}$ and $\displaystyle 3\mathbb{Z}$ are not isomorphic.
• Sep 8th 2011, 07:20 AM
HallsofIvy
Re: Ring Isomorphism
Sorry, I was thinking "$\displaystyle \mathbb{Z}$ modulo 2 and 3". zoek has it right.
• Sep 8th 2011, 03:55 PM
Bernhard
Re: Ring Isomorphism
No problems. Thanks anyway

And thanks for help with previous posts as well!

Peter