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 $\mathbb{R}$.

Solution 1:

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 $f(i)$?

Solution 2:

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 $\mathbb{Q}$.