# Is my proof correct? (Isomorphism)

• Nov 25th 2009, 11:17 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.
• Nov 25th 2009, 12:07 PM
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.
• Nov 25th 2009, 12:08 PM
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
• Nov 25th 2009, 12:41 PM
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.
• Nov 25th 2009, 02:36 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