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There is a field $\mathbb K$. I've got an injective homomorphism $\varphi: \mathbb R \rightarrow \mathbb K$. Also I got $i \in \mathbb K$ with $i\cdot i = -1$.

I have to show, that there are exactly 2 field-isomorphisms $\varphi: \mathbb C \rightarrow \mathbb K$ with $\varphi(a,0) = \varphi(a)$ for all $a \in \mathbb R$. These two isomorphisms are $\varphi_1(a, b) = \varphi(a) + i\varphi(b)$ and $\varphi_2(a, b) = \varphi(a) - i\varphi(b)$.

I showed, that $\varphi_1$ and $\varphi_2$ are field-isomorphisms, but I fail at proving, that there are no other isomorphisms.

I proofed (by using the isomorphism-rules), that $\varphi(a, b) = \varphi(a) + \varphi(0,1)\varphi(b)$. But I don't see, why $\varphi(0,1)$ has to be $i$ or $-i$. The only thing I know is that $\varphi(0,1)\varphi(0,1) = \varphi(-1)$.

Can someone give me a hint?

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...but you must have $\,\phi(-1)=-\phi(1)=-1 \,$ , don't you? –  DonAntonio Mar 11 '13 at 23:05
2  
P.S. you can't prove $\varphi$ is a field isomorphism, unless you have some additional hypotheses. e.g. if $\mathbb{K} = \mathbb{C}(x)$.... –  Hurkyl Mar 11 '13 at 23:08
    
I guess $\phi|_{\mathbb{R}} = \varphi $ and $\mathbb{K} = \mathbb{R}(\mathbb{i})$ are presumed –  user32847 Mar 11 '13 at 23:13

2 Answers 2

up vote 2 down vote accepted

Hint 1: You know some values of $\varphi(x)$

Hint 2: You can solve for $\varphi(0,1)$.


Alternatively,

Hint 3: Homomorphisms from $\mathbb{R}[x] / \langle x^2 + 1\rangle$ to any ring $S$ are....

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Thanks, i've got it. –  user38034 Mar 11 '13 at 23:21

Hint 1: $\varphi$ is already determined by the image of $i$.

Hint 2: For the image of $i$, there are only two possibilities.

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