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?