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prove that $\mathbb{C}$ and $\mathbb{R}$ are not isomorphic as rings

My guess is that the proof for this has something to do with the fact that $\sqrt{-1}\in\mathbb{C}$ cannot be mapped to $\mathbb{R}$.

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Not isomorphic as what? Additive groups? Rings? Vector spaces over $\mathbb{Q}$? Vector spaces over $\mathbb{R}$? – Arturo Magidin Jan 23 '12 at 20:45
In which category? – Norbert Jan 23 '12 at 20:46
you guessed it right... What is $(\sqrt{-1})^2+1$ in $C$? Where would it be mapped by a ring isomorphism? – N. S. Jan 23 '12 at 20:48
@Dylan: Or $\mathbb{Q}-\mathbf{VectorSpace}$... – Arturo Magidin Jan 23 '12 at 20:48
@Tobias: Since they are isomorphic as $\mathbb{Q}$-vector spaces (both have dimension $|\mathbb{R}|$), they are necessarily isomorphic as abelian groups. – Arturo Magidin Jan 23 '12 at 20:58

If $f\colon\mathbb{C}\to\mathbb{R}$ is a ring homomorphism, then since $f(1)=f(1^2) = f(1)^2$, we must have either $f(1)=1$ or $f(1)=0$.

If $f(1)=0$, then $f(\mathbb{C})=\{0\}$.

If $f(1)=1$, then what is $f(-1)$? And what is $f(i)$?

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It depends on what kind of isomorphism you are looking for. Certainly they are not isomorphic as rings. The argument is not hard to write out, and starts from your observation.

Suppose to the contrary that $\phi$ is a ring isomorphism from $\mathbb{C}$ to $\mathbb{R}$. Note that $\phi(1)=1$, and therefore $\phi(-1)=-1$.

Let $\phi(i)=a$. Then $\phi(i \cdot i)=\phi(-1)=-1$. But also $\phi(i\cdot i)=a^2\ne -1$.

However, $\mathbb{R}$ and $\mathbb{C}$ are isomorphic as groups under addition. We can also view each of them as a vector space over the field $\mathbb{Q}$ of rational numbers. They are isomorphic as vector spaces over $\mathbb{Q}$.

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Isomorphic as vector spaces over $\mathbb{Q}$, yes... but not over $\mathbb{R}$ (just to clarify). – Rex Butler Jan 23 '12 at 21:30
@Skolem: Thanks, I said it once, should have said it twice. – André Nicolas Jan 23 '12 at 21:36

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