To prove that $\mathbb{Q}$ is the smallest subfield of $\mathbb{C}$

Assumption: There exsits $$F$$ which is a subfield of $$\mathbb{C}$$ such that $$F\subsetneq \mathbb{Q}$$.

Claim: $$\mathbb{Z}\subset F$$.

Proof: Let $$m \in \mathbb{Z^+ }$$. We know, that $$1 \in F$$. Taking $$\displaystyle\underbrace{1+1+1+...+1}_{m \text{ times}}=m.1=m\in F$$.Again, $$F$$ being a field, for any $$m \in F \implies -m \in F$$. And $$0\in F$$ is trivial.

Hence, $$m\in \mathbb{Z} \implies m \in F \implies$$ $$\mathbb{Z}\subset F$$.

Now, by assumption, $$\exists \ w\in \mathbb{Q}$$ such that $$w\ \notin F$$. Now, $$w=p/q=pq^{-1}$$ for some $$p, q \in \mathbb{Z}$$, with $$q \neq 0$$.

As per our proven claim, $$p, q \in F$$. Again, $$F$$ being a field, $$w=pq^{-1}\in F$$. A contradiction.

Hence, $$\mathbb{Q} \subseteq F$$, for any subfield $$F$$ of $$\mathbb{C}$$.

Is this correct? Kindly verify.

• The idea is correct, though when you checked that $\mathbb{Z}\subseteq F$ you assumed that $m$ is positive. If you want a very formal proof then you have to write that since all elements of $F$ must have an additive inverse we conclude that the negative integers are in $F$ as well. – Mark Sep 10 at 16:21
• I shouldn't have missed that point. Thank you for pointing it out. – Subhasis Biswas Sep 10 at 16:21
• @Mark, Is it correct now? – Subhasis Biswas Sep 10 at 16:29
• Yes, it is fine. – Mark Sep 10 at 16:37
• – Bill Dubuque Sep 10 at 16:44

Firstly, the mapping $${\Bbb Z}\rightarrow{\Bbb C}:m\mapsto m\cdot 1$$, where $$1$$ is the unit element in $${\Bbb C}$$, is a ring monomorphism and so the image $$\{m\cdot 1\mid m\in{\Bbb Z}\}$$ can be identified with $$\Bbb Z$$. This probably better reflects your first part.
Secondly, you can prove straightforwardly that that $$\Bbb C$$ contains a copy of $$\Bbb Q$$ by starting with the copy of $$\Bbb Z$$.