Prime subfield is either isomorphic to $\mathbb{Q}$ or $F_p$ I'm trying to prove the following statement:
Let $F$ be a field. The intersection of all subfields of $F$ is a subfield which is isomorphic to $\mathbb{Q}$ if $\operatorname{char}(F)=0$, and isomorphic to $F_p$ if $\operatorname{char}(F)=p$.
I assume I need to set up a injective ring homomorphism $\varphi\colon \mathbb{Q}\to F.$ I don't really know where to go apart from that though. 
 A: The field either has a positive characteristic or characteristic $0$. In the former case, you have $p$ is the smallest number for which $p\cdot 1=0$, so there are at least $p$ elements in it, and the $\Bbb Z$ action on the field descends to a $\Bbb Z/p$ action, hence it is an $\Bbb F_p$ module, i.e. it is a field extension of $\Bbb F_p$. In the case there is no smallest such number, $\Bbb Z\subseteq F$ and therefore $\Bbb Q\subseteq F$ because $F$ is closed under the field operations.
A: Assume char(F)$=p$ to start with and let $e$ be the multiplicative unit in $F$.  Let $A$ be a subfield of $F$. According to subfield axioms, $A$ contains $0$, $e$, $2e$, ..., $(p-1)e$ which are distinct elements, hence 
$$\{0,e,...,(p-1)e\}\subset A$$
for every subfield $A$. The set on the left is itself a subfield of $F$ so it must be the intersection of all subfields. It should be clear that it is isomorphic to $\mathbb{F}_p$.
The case where the characteristic is infinite is similar except you can show that every subfield must contain $ne$ where $n\in\mathbb{N}$ and hence must contain $re$ where $r\in\mathbb{Q}$.
A: Define a map $f:\Bbb{Z}\to{F}$  by $f(n)=n.e$ , where $e$ is the unit element of $F$.
$f$ is homomorphism:
$$f(m+n)=(m+n).e=m.e+n.e=f(m)+f(n)$$
and
$$f(mn)=(mn).e=(m.e)(n.e)=f(m)f(n)$$.
By Fundamental theorem of homomorphism,
$$f(\Bbb{Z})\cong \frac{\Bbb{Z}}{\operatorname{Ker}{f}}\subseteq F$$.
Where $\operatorname{Ker}{f}$ is the ideal of $\Bbb{Z}$, in fact principal ideal of $\Bbb{Z}$.Then $\operatorname{Ker}f=<q> $, for some $q\in \Bbb{Z}$.
Now consider the needed cases and you are done!
