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I want to prove this using contradiction, supposing that there is a ring isomorphism between the two and then finding a contradiction.

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Is there an element $a\in\mathbb{R}$ such that $a^2=-1$?

Added from comments.

Suppose $f\colon\mathbb{C}\to\mathbb{R}$ is an isomorphism. It's usually required that field homomorphisms map $1$ into $1$, but it's not necessary to make this assumption here. Indeed, set $x=f(1)$. Then $$ x=f(1)=f(1^2)=f(1)f(1)=x^2 $$ so $x-x^2=0$, which means $x=0$ or $x=1$ because we're in a field. If $f(1)=0$, then, for all $z\in\mathbb{C}$, we have $$ f(z)=f(z\cdot1)=f(z)f(1)=f(z)\cdot0=0 $$ which is impossible, because $f$ is bijective (injectivity is sufficient).

Thus $f(1)=1$, hence $f(-1)=-1$. Now $a=f(i)$ would have the property that $a^2=f(i^2)=f(-1)=-1$ and there's no such element $a\in\mathbb{R}$.

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  • $\begingroup$ No there is not, so we know f(1)=f(1^2)=f(1)^2=1^2=1, this implies that f(-1)=-1 because 0=f(0)=f(1-1)=f(1)+f(-1) and since f(1)=1, f(-1) must equal -1. So if we consider f(i)^2=f(i^2)=f(-1)=-1, this implies that -1 is the square of some number in the real numbers which is impossible. Is this correct? $\endgroup$
    – Jenny
    Commented Dec 6, 2013 at 18:39
  • $\begingroup$ @Jenny Yes, correct. $\endgroup$
    – egreg
    Commented Dec 6, 2013 at 18:45
  • $\begingroup$ How do we know that f(1)=1 though? $\endgroup$
    – Jenny
    Commented Dec 6, 2013 at 18:55
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    $\begingroup$ @GeoffreyCritzer Because $f(0)=0$ and $f(0)=f(1+(-1))=f(1)+f(-1)=1+f(-1)$. $\endgroup$
    – egreg
    Commented May 22, 2015 at 16:15
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    $\begingroup$ @GeoffreyCritzer $f(0)=f(0+0)=f(0)+f(0)$, so an additive group homomorphism must send $0$ to $0$. Since $f(1)=f(1^2)=(f(1))^2$, we get that $f(1)=0$ or $f(1)=1$ (because $\mathbb{C}$ is a field). Thus, even if we don't assume that $f(1)=1$, we get it, because $f(1)=0$ implies $f(x)=0$ for all $x$, so certainly $f$ wouldn't be an isomorphism. $\endgroup$
    – egreg
    Commented May 22, 2015 at 17:14
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Hint: If the field of complex numbers were isomorphic to the field of real numbers, there would be no reason to define the notion of complex numbers when we already have the real numbers. But there is such a reason. (What is it?)

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$\mathbb{C}$ has a nontrivial ring automorphism, $\mathbb{R}$ not.

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