Ring Isomorphism

**Bernhard** · September 8th 2011, 04:42 AM

Can anyone help with the following problem:

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

Be grateful for help

Peter

**HallsofIvy** · September 8th 2011, 04:48 AM

How many members does $2\mathbb{Z}$ ? What about $3\mathbb{Z}$ ?

**Bernhard** · September 8th 2011, 04:53 AM

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

**zoek** · September 8th 2011, 05:51 AM

Originally Posted by Bernhard

Can anyone help with the following problem:

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

Be grateful for help

Peter

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

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

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

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

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

So, the rings $2\mathbb{Z}$ and $3\mathbb{Z}$ are not isomorphic.

**HallsofIvy** · September 8th 2011, 07:20 AM

Sorry, I was thinking " $\mathbb{Z}$ modulo 2 and 3". zoek has it right.

**Bernhard** · September 8th 2011, 03:55 PM

No problems. Thanks anyway

And thanks for help with previous posts as well!

Peter

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