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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 $\mathbb{R}$?2012-01-23
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    In which category?2012-01-23
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    As rings -- just edited.2012-01-23
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    Hopefully not $\mathsf{Set}$.2012-01-23
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    you guessed it right... What is $(\sqrt{-1})^2+1$ in $C$? Where would it be mapped by a ring isomorphism?2012-01-23
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    @Dylan: Or $\mathbb{Q}-\mathbf{VectorSpace}$...2012-01-23
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    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
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    @Aaron: Note everyone requires ring homomorphisms to be unital.2012-01-23
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    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
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    @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
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    Ahh, of course, thanks.2012-01-23
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    @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

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