Is my proof correct? (Isomorphism)

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November 25th 2009, 10:17 AM
sfspitfire23

Is my proof correct? (Isomorphism)

We must show that $\mathbb{R}$ and $\mathbb{C}$ are NOT isomorphic. Consider the mapping $\alpha : \mathbb{C} \rightarrow \mathbb{R}$ . Then we have $\alpha(i)=a$ which can be written as $\alpha(i)^2=a^2$ . So, $\alpha(i^2)=a^2$ which is the same as $\alpha(-1)=a^2$ . We can split this apart to get $\alpha(-1)+\alpha(0)=a^2$ . This shows that $-1=a^2$ . But nothing in $\mathbb{R}$ squares to $-1$ . Therefore they are not isomophic.
November 25th 2009, 11:07 AM
Jose27

$\alpha(-1)=-1$ since $\alpha$ is an homomorphism, I don't know why you split this in the last term. But, yeah your proof is fine.
November 25th 2009, 11:08 AM
tonio

Quote:

Originally Posted by sfspitfire23

We must show that $\mathbb{R}$ and $\mathbb{C}$ are NOT isomorphic. Consider the mapping $\alpha : \mathbb{C} \rightarrow \mathbb{R}$ . Then we have $\alpha(i)=a$ which can be written as $\alpha(i)^2=a^2$ . So, $\alpha(i^2)=a^2$ which is the same as $\alpha(-1)=a^2$ . We can split this apart to get $\alpha(-1)+\alpha(0)=a^2$ . This shows that $-1=a^2$ . But nothing in $\mathbb{R}$ squares to $-1$ . Therefore they are not isomophic.

Almost: why $\alpha(-1)=-1$ ?

Tonio
November 25th 2009, 11:41 AM
sfspitfire23

Quote:

Originally Posted by tonio

Almost: why $\alpha(-1)=-1$ ?

Tonio

Because it's a homomorphism? I think this is where I slip up. I'm not sure how to say that it goes to -1.
November 25th 2009, 01:36 PM
tonio

Quote:

Originally Posted by sfspitfire23

Because it's a homomorphism? I think this is where I slip up. I'm not sure how to say that it goes to -1.

I supose you were trying to show $\mathbb{C}\ncong \mathbb{R}$ as rings, and then $\alpha(1)=1\,,\,\alpha(0)=0\,\,\Longrightarrow\,\, 0=\alpha(0)=\alpha(1+(-1))=\alpha(1)+\alpha(-1)=1+\alpha(-1)$ and we're done.

Tonio