7
$\begingroup$

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 $\mathbb{R}$.

  • 2
    Not isomorphic as *what*? Additive groups? Rings? Vector spaces over $\mathbb{Q}$? Vector spaces over $\mathbb{R}$?2012-01-23
  • 0
    In which category?2012-01-23
  • 0
    As rings -- just edited.2012-01-23
  • 0
    Hopefully not $\mathsf{Set}$.2012-01-23
  • 2
    you guessed it right... What is $(\sqrt{-1})^2+1$ in $C$? Where would it be mapped by a ring isomorphism?2012-01-23
  • 1
    @Dylan: Or $\mathbb{Q}-\mathbf{VectorSpace}$...2012-01-23
  • 0
    Your intuition is right. Any map of rings sends $1$ to $1$, and hence $n$ to $n$ for any integer $n$. Given $i\in \mathbb C$, we must have $\phi(i)^2=\phi(i^2)=\phi(-1)=-1$, but since $\mathbb R$ has no square root of $-1$, there exists no homomorphism (let alone isomorphism) from $\mathbb C$ to $\mathbb R$2012-01-23
  • 0
    @Aaron: Note everyone requires ring homomorphisms to be unital.2012-01-23
  • 0
    How about as abelian groups? Are they isomorphic in that category? (My intuition would be no, but I can't find an easy argument)2012-01-23
  • 1
    @Tobias: Since they are isomorphic as $\mathbb{Q}$-vector spaces (both have dimension $|\mathbb{R}|$), they are necessarily isomorphic as abelian groups.2012-01-23
  • 0
    Ahh, of course, thanks.2012-01-23
  • 0
    @Tobias: You may also be interested in the "structure theorem for divisible abelian groups"; any divisible abelian group must be a direct sum of copies of $\mathbb{Q}$ and copies of Prufer $p$-groups (for possibly many different primes). So you can also get it out of that.2012-01-23

2 Answers 2

8

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 $f(i)$?

7

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 $\mathbb{Q}$.

  • 0
    Isomorphic as vector spaces over $\mathbb{Q}$, yes... but not over $\mathbb{R}$ (just to clarify).2012-01-23
  • 0
    @Skolem: Thanks, I said it once, should have said it twice.2012-01-23