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

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
As rings -- just edited. –  Emir 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

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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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 $\mathbb{Q}$.
Isomorphic as vector spaces over $\mathbb{Q}$, yes... but not over $\mathbb{R}$ (just to clarify). –  Skolem Jan 23 '12 at 21:30