Suppose $E$ and $F$ are fields, and $φ : F → E$ is a ring homomorphism such that $φ(1) = 1$. I've shown that $φ$ is injective. But if I have $\Bbb{Q} \subset E$ and $\Bbb{Q} \subset F$, how do I show that for every $r \in \Bbb{Q}$, $φ(r) = r$?
Any hints would be very helpful! Thanks!
The simplest thing to observe is that $\varphi(\frac{p}{q})=\varphi(pq^{-1})=\varphi(p)(\varphi(q))^{-1}=\frac{p}{q}$ and hence conclude. This is enough. But for a bit more rigorous argument see below.
First observe that since a field is a simple ring, any homomorphism $\varphi$ is either trivial or injective , i.e $\ker(f)=\{0\}$ or $F$.
So $\varphi(1)=1\implies \varphi$ is injective.
So since both the fields contain $\Bbb{Q}$ they are of characteristic $0$.
Now both $E$ and $F$ contain $\Bbb{Z}$.
Consider $\varphi\vert_{\Bbb{Z}}:\Bbb{Z}\to E$ . That is the restriction of $\varphi$ to $\Bbb{Z}\subset F$
So $\varphi(n)\vert_{\Bbb{Z}}=n\cdot 1=n\in E$
Thus $\varphi\vert_{\Bbb{Z}}$ is an injective ring homomorphism.
Now as $\Bbb{Q}$ is the field of fractions of of $\Bbb{Z}$ . $\varphi\vert_{\Bbb{Z}}$ can be extended uniquely to a homomorphism from the fraction field of $\Bbb{Z}$ i.e of $\Bbb{Q}$. See here
i.e.
$\bar{\varphi}:\Bbb{Q}\to E$ such that $\bar{\varphi}(\frac{p}{q})=\frac{\varphi(p)}{\varphi(q)}=\frac{p}{q}$
This is again an injective ring homomorphism and hence a bijection from $\mathbb{Q}(\subset F)\to \mathbb{Q}(\subset E)$ and it is precisely the identity map when restricted to $\Bbb{Q}$.
Now by uniqueness you have $\varphi\vert_{\Bbb{Q}}=\bar\varphi$