I solved the following exercise but have trouble making sense of the result:
If $\widetilde{\mathbb C}$ is another field of complex numbers and $\varphi : \mathbb C \to \widetilde{\mathbb C}$ is a ring homomorphism fixing the real numebrs then it follows that $\varphi$ is bijective and therefore the field of complex numbers is essentially uniquely determined.
I had no trouble showing that $\varphi$ is bijective: it follows directly from the fact that it is a ring homomorphism and that for $x \in \mathbb R$ we have $\varphi (x) = x$.
But I have trouble understanding what it means that then the complex numbers are ''essentially uniquely determined''.
Please could someone explain to me what this means?
It seems to me that even though any homomorphism fixing the reals turns out to be bijective we could still find a field of complex numbers such that there is no $\varphi : \mathbb C \to \widetilde{\mathbb C}$ that preserves $\mathbb R$ or maybe even no homomorphism $\varphi$ at all. Then wouldn't this mean that the complex numbers are not unique?
Edit
In the (complex analysis) book there is a theorem which states that there exists a field $\mathbb C$ that
(1) Contains $\mathbb R$ as a subfield
(2) has a solution to $x^2 + 1$
(3) if $i$ denotes one of these solutions then the map $\mathbb R \times \mathbb R \to \mathbb C$, $(x,y) \mapsto x + iy$ is bijective.
Edit 2
I believe there should be an answer without using ''algebraicity'' because it is not mentioned in the book which is a book about complex analysis and ''algebraicity'' is an algebraic property. I believe it should be possible to explain this using only basic complex analysis since it's an exercise in the first chapter of the book.