Isomorphism correction?

**sfspitfire23** · November 29th 2009, 03:01 PM

I'm having trouble showing how $\alpha(1/4)=a^2\Rightarrow 1/4=a^2$ . How would I do this?

We must show that $\mathbb{Q}$ and $\mathbb{Z}$ are not isomorphic. Consider the mapping $\alpha : \mathbb{Q} \rightarrow \mathbb{Z}$ . Then we have $\alpha(1/2)=a$ . Then we have $\alpha(1/2)^2=a^2$ which can be written as $\alpha((1/2)^2)=a^2$ . So, $\alpha(1/4)=a^2$ . We can pull this apart to get $\alpha(1/8)+\alpha(1/8)=a^2$ . to yield $1/4=a^2$ . But there is nothing in $\mathbb{Z}$ that squares to $1/4$ . Therefore, they are not isomorphic.

**tonio** · November 29th 2009, 03:51 PM

Originally Posted by sfspitfire23

I'm having trouble showing how $\alpha(1/4)=a^2\Rightarrow 1/4=a^2$ . How would I do this?

We must show that $\mathbb{Q}$ and $\mathbb{Z}$ are not isomorphic. Consider the mapping $\alpha : \mathbb{Q} \rightarrow \mathbb{Z}$ . Then we have $\alpha(1/2)=a$ . Then we have $\alpha(1/2)^2=a^2$ which can be written as $\alpha((1/2)^2)=a^2$ . So, $\alpha(1/4)=a^2$ . We can pull this apart to get $\alpha(1/8)+\alpha(1/8)=a^2$ . to yield $1/4=a^2$ . But there is nothing in $\mathbb{Z}$ that squares to $1/4$ . Therefore, they are not isomorphic.

You write your questions in a very sloppy way...this can't be good for you: I suppose you mean to prove $\mathbb{Q}\,,\,\mathbb{Z}$ aren't isomorphic...as rings, right?
Well, I can't understand why you think, or want it to be true, that if $\alpha(1\slash 8)+\alpha(1/8)=a^2$ , then $1\slash 4=a^2$ ...

Following your idea, I'd simply say $\alpha(1/2)=a\Longrightarrow a^2=\alpha(1/2)^2=\alpha(1/4)\Longrightarrow a=\alpha(1/2)=\alpha(2\cdot 1/4)=2\alpha(1/4)=2a^2$ , and get from here your contradiction.

Tonio

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