Hoffman and Kunze Exercise From exercise 8 in sec 1.2 page 5. 
Prove that
Each field of characteristic zero contains a copy of the rational number field.
I saw an answer that starts from definition. It goes:
since in a field of characteristic zero, 1 != 0, 1 != 1+1 != 1+1+1 ... Hence, the natural numbers are in this field. However, 1+1=2, 1+1+1=3 ... are not proved in this proof. The axioms of field does not say that 1+1=2, 1+1+1=3 etc.
So, is there a better proof?

To put this exercise in a more "formal" way, you should try to prove the following:

If a field $F$ has characteristic zero, then there exists an injective ring homomorphism $\varphi:\mathbb{Q}\rightarrow F$.

By a field homomorphism, I mean a function $\varphi$ which preserves addition and multiplications, obviously. The copy of $\mathbb{Q}$ in $F$ will be $\varphi(\mathbb{Q})$.
You can define $\varphi$ the following way: First define $\widetilde{\varphi}$ inductively by


*

*$\widetilde{\varphi}(0)=0$.

*$\widetilde{\varphi}(n+1)=\widetilde{\varphi}(n)+1$ for $n\in\mathbb{Z}^+$,


and for $n\in\mathbb{Z}^-\setminus\left\{0\right\}$ define $\widetilde{\varphi}(n)=-\widetilde{\varphi}(-n)$ (since $-n\in\mathbb{Z}^+)$.
You can show by induction that $\widetilde{\varphi}:\mathbb{Z}\rightarrow F$ is an injective ring homomorphism.
Now, if $q=\dfrac{n}{m}\in\mathbb{Q}$, with $n\in\mathbb{Z}$ and $m\in\mathbb{Z}^+\setminus\left\{0\right\}$, then, since $\widetilde{\varphi}$ as above is injective, we have $\widetilde{\varphi}(m)\neq 0$ in $F$. Define
$$\varphi(q)=\dfrac{\widetilde{\varphi}(n)}{\widetilde{\varphi}(m)}.$$
You can show that $\varphi$ is well-defined, that is, $\varphi(q)$ does not depend of the choice of $n$ and $m$, and that it is a ring homomorphism. Then, $\varphi$ is the function you seek.
