Subfields of $\mathbb{Q}$ How to prove that $\mathbb{Q}$ doesn't have any proper subfields? I have no idea how to prove it.
 A: Every subfield $F$ has to include $0,1$ and has to be closed under addition,so  you get $\mathbb{Z}\subseteq F$. Since for any $x\in F\setminus\lbrace 0\rbrace$ you have $\frac{1}{x}\in F$ and therefore $\lbrace \frac{1}{n}\vert n\in\mathbb{N}\rbrace\subseteq F$; furthermore $F$ has to be closed under multiplication, so  $m\cdot\frac{1}{n}=\frac{m}{n}\in F$ for all $m,n\in \mathbb{N}$. Since $F$ must contain the additive inverse of every positive fraction $F=\mathbb{Q}$.
A: HINT: Recall that $\Bbb Q$ is the field of fractions of $\Bbb Z$ (every element of $\Bbb Q$ can be written as a ratio of elements from $\Bbb Z$). 
Show that if $F$ is a field of characteristics zero there is exactly one monomorphism (of rings!) from $\Bbb Z$ into $F$, and show how this monomorphism extends to the field of fractions of $\Bbb Z$. In other words, if $F$ is a field of characteristics zero, there is an injective homomorphism of rings from $\Bbb Q$ into $F$.
Use this to conclude that if $F\subseteq\Bbb Q$ is a field, then $F=\Bbb Q$.
