Any field that contains the integers contains the rationals as a subfield (Analysis proof)

Question

I'm an intro analysis student trying to prove that any field $f$ that contains the integers contains the rationals as a subfield. I'm asked to proceed by proving that there exists a subset $F'$ of $F$ and a function $Φ: F'→\mathbb{Q}$ such that $\mathbb{Z} ⊂ \mathbb{Q}$ and $Φ$ satisfies the three following properties:

1. $Φ $ is bijective

2. $Φ(a + b) = Φ(a) + Φ(b)$, for $a$, $b$ ∈ $F$

3. $Φ(a.b) = Φ(a).Φ(b)$, for $a$, $b ∈ F$

I have gone through a few answers for similar questions but I am not sure how to proceed. We have not covered isomorphisms in the course yet so I am unsure how this is related to it (though other posts like this one and this one indicate that there's a connection). Any help would be greatly appreciated.

Answer 1

Hint: Suppose $\mathbb{Z} \subset F$. Given $n \in \mathbb{Z}$ and $m \in \mathbb{Z} \setminus\{0\}$, the fact that $F$ is a field means that $nm^{-1} \in F$. How should we associate $nm^{-1}$ with a rational number?

You'll need to prove that this association is actually an isomorphism, which really just comes down to doing a few calculations.

Answer 2

If $F$ is a field and $\mathbb Z \subset F$ then for any integers $m,n\ne 0$ then $m\cdot \frac 1n = \frac mn \in F$. But $\{\frac mn|m,n\in \mathbb Z; n\ne 0\} = \mathbb Q$. So $\mathbb Q \subset F$.

Or to look from a slightly different angle:

If $q\in \mathbb Q$ then there exist integers $m,n$ so that $q = \frac mn$ and as $m,n\in \mathbb Z\subset F$ and $F$ is a field then $\frac 1n\in F$ and $m*\frac 1n = \frac mn=q\in F$. So $\mathbb Q \subset F$