I just started reading Linear Algebra by Hoffman and Kunze, and I came across the following line:
The interested reader should verify that any subfield of $\mathbb C$
must contain every rational number.
Of course, I am interested, so I tried to come up with a proof. I just need verification on the correctness of my proof, and any feedback on how I can improve it. So here it goes.
Proof
A subfield is simply a subset of a field. A field $\mathbb F$ must obey the following axioms:
1. Addition is commutative, i.e. $x + y = y + x$ for all $x, y$ in $ \mathbb F$.
2. Addition is associative, i.e. $x + (y + z) = (x + y) + z$ for all $x, y, z$ in $\mathbb F$.
3. There exists a unique additive identity $0$ such that $x + 0 = x$ for all $x$ in $\mathbb F$.
4. There exists a unique additive inverse $-x$ such that $x + (-x) = 0$ for all $x$ in $\mathbb F$.
5. Multiplication is commutative, i.e. $x \cdot y = y \cdot x$ for all $x, y$ in $\mathbb F$.
6. Multiplication is associative, i.e. $x \cdot (y \cdot z) = (x \cdot y) \cdot z$ for all $x, y, z$ in $\mathbb F$.
7. There exists a unique multiplicative identity $1$ such that $x \cdot 1 = x$ for all $x$ in $\mathbb F$.
8. For every nonzero $x$ in $\mathbb F$, there exists a unique multiplicative inverse $x^{-1}$ such that $x \cdot x^{-1} = 1$.
9. Multiplication is distributive over addition, i.e. $x \cdot (y + z) = x \cdot y + x \cdot z$ for all $x, y, z$ in $\mathbb F$.
10. Closure under addition, i.e. for all $x, y$ in $\mathbb F$, $x + y$ must also be in $\mathbb F$.
11. Closure under multiplication, i.e. for all $x, y$ in $\mathbb F$, $x \cdot y$ must also be in $\mathbb F$.
From the above it follows that any subfield $\mathbb F$ of $\mathbb C$ must contain the elements $0$ and $1$ (as the additive and multiplicative identity respectively). This satisfies axioms $3$ and $7$. By our axiomatic definition of fields, any subfield of $\mathbb C$ must also satisfy axioms $1, 2, 5, 6, 9$. To satisfy axiom $10$, we see that $\mathbb F$ must contain $\mathbb Z^+$ by induction (I do not know how to prove this). To satisfy axiom $4$, it follows that $\mathbb F$ must contain $\mathbb Z^-$ as well. Hence $\mathbb F$ contains $\mathbb Z$. To satisfy axiom $8$, we need to include all scalars of the form $\frac 1x$, where $x \in \mathbb Z \setminus\{0\}$. To accomodate axiom $11$, we must include all scalars of the form $\frac 1x \cdot y$, where $x \in \mathbb Z \setminus \{0\}, y \in \mathbb Z$. Hence $\mathbb F$ must contain $\mathbb Q$. Now every axiom is satisfied and we see that any subfield of $\mathbb C$ must at least contain $\mathbb Q$.
Best Answer
What you have done so far is ok.
To prove that $\Bbb Z^+$ is in $\Bbb F$, just note that $1\in\Bbb F$ and that $n+1$ is in $\Bbb F$ whenever $n$ is in $\Bbb F$ (axioms 7 and 10).