# 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.

• Start instead by a ring homomorphism $\mathbb Z\to F$. If it has a kernel, then... – Thomas Andrews Sep 26 '16 at 22:24

## 2 Answers

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 $r\in\mathbb{Q}$.

• You said that the set on the left is itself a subfield of $F$ so it must be the intersection of all subfields. Why is that? – user342914 Sep 26 '16 at 22:27
• @user342914 It is a subfield of $F$, and any subfield of $F$ must contain it. – arkeet Sep 26 '16 at 22:29
• Thank you. Why must $A$ contain $e,2e,...,(p-1)e$? – user342914 Sep 26 '16 at 22:40
• Every subfield $A$ must contain $e$ by definition. Then $e+e=2e$ is also in $A$ according to subfield axioms. So is $2e+e=3e$ and so on. If char(F)=p this stops at $(p-1)e$ since $pe=0$. – Olivier Moschetta Sep 26 '16 at 23:13
• @OlivierMoschetta Why does the fact that every subfield contains $ne$ imply that it must contain $re$ for $r\in \mathbb{Q}$? – user342914 Sep 27 '16 at 10:15

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.