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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.

Jenny
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3 Answers3

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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 $a\in\mathbb{R}$.

egreg
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  • 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? – Jenny Dec 06 '13 at 18:39
  • @Jenny Yes, correct. – egreg Dec 06 '13 at 18:45
  • How do we know that f(1)=1 though? – Jenny Dec 06 '13 at 18:55
  • @Jenny Set $x=f(1)$; then $x=f(1)=f(1^2)=f(1)^2=x^2$. So either $x=0$ or $x=1$; but $f(1)=0$ implies $f(a)=0$ for all $a$ (why?). Impossible, because $f$ is bijective. – egreg Dec 06 '13 at 18:57
  • I do not understand why f(1) cannot be 0. Could you help me understand this? – Jenny Dec 06 '13 at 19:04
  • @Jenny If $f(1)=0$, then $f(a)=f(a\cdot1)=f(a)f(1)=f(a)\cdot0=0$ for all $a$. – egreg Dec 06 '13 at 19:06
  • One last question, how can we do a mini proof of why f(-1)=-1? I tried and figured out how to do it using f(0)=0 but we have not shown that either. – Jenny Dec 06 '13 at 19:50
  • @Jenny That's true for any ring homomorphism that maps $1$ into $1$. And it's easy. – egreg Dec 06 '13 at 22:10
  • I do not understand why f(1) = 1 implies f(-1) = -1. – Geoffrey Critzer May 22 '15 at 16:14
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    @GeoffreyCritzer Because $f(0)=0$ and $f(0)=f(1+(-1))=f(1)+f(-1)=1+f(-1)$. – egreg May 22 '15 at 16:15
  • @egreg OK thanks, and do I understand correctly that : f(0) = 0 because for any isomorphism the identity elements must be paired together. – Geoffrey Critzer May 22 '15 at 16:46
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    @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. – egreg May 22 '15 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?)

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

Martin Brandenburg
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