$\Bbb Q$ is lowest field of characteristic zero

Teorem: field of characteristic zero contains a subfield that is isomorphic to $\Bbb Q$; so $\Bbb Q$ is lowest field of characteristic zero.

proof: I did following: let char(K)=0. and $$\beta :\Bbb Q \to \Bbb K$$ $$\frac{a}{b}\to (a1_k)(b1_k)^{-1}$$ so it becomes bijective homeomorphism. then $\Bbb Q \equiv\beta(\Bbb Q) \le \Bbb K$

now, assume $F \subseteq \Bbb Q$ and char (F)=0. we know there is $F' (subfield)\le F$. how can I continue?

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You have proven the theorem, right? What are you asking now?? –  awllower Jan 13 '13 at 18:46
no I havent. I want to show the second part. –  emmett Jan 13 '13 at 18:48
But your first part defines an isomorphism from Q to F, hence F is isomorphic with Q. Is there something wrong? –  awllower Jan 13 '13 at 18:52
I should point out that homeomorphism is an isomorphism if topological spaces. You mean to say a bijection homomorphism. –  Asaf Karagila Jan 13 '13 at 19:35

1 Answer

First let me point out that using $\Bbb C$ in the context of field theory is probably not a good idea if you're not talking about the complex numbers.

Secondly note that there aren't many ways of defining this embedding really. If $K$ is a field of characteristics zero, and we want $f\colon\Bbb Q\to K$ to be an embedding then $0_\Bbb Q$ has to be $0_K$, and similarly for $1_\Bbb Q$. Now by induction we have that for every natural number $f(n)$ is decided, that is $f(1+1)=f(1)+f(1)$, and so on.

From this we deduce that $f$ is completely determined on $\mathbb Z$ (applying $0_K=f(0_\Bbb Q)=f(n+(-n))=f(n)-f(n)$). And from that we can deduce that $f$ is completely determined on every rational number (the same logic applied to multiplication).

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