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.

  • 1
    $\begingroup$ 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. $\endgroup$ – Mark Sep 10 at 16:21
  • $\begingroup$ I shouldn't have missed that point. Thank you for pointing it out. $\endgroup$ – Subhasis Biswas Sep 10 at 16:21
  • $\begingroup$ @Mark, Is it correct now? $\endgroup$ – Subhasis Biswas Sep 10 at 16:29
  • $\begingroup$ Yes, it is fine. $\endgroup$ – Mark Sep 10 at 16:37
  • 3
    $\begingroup$ Duplicate of A field with characteristic $0$ contains $\mathbb Q$ $\endgroup$ – 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 $\Bbb Z$.

  • $\begingroup$ I want the proof to be very basic here. I don't have much depth in field theory. This answer will be of help when I proceed further :) $\endgroup$ – Subhasis Biswas Sep 10 at 16:28

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