Injective homomorphism between two fields of characteristic $0$ for $x\in\mathbb{Q}$

Let $F$ and $K$ be two fields of characteristic $0$, and $\phi: F\to K$ be an injective homomorphism. I want to prove that $\phi(x)=x$ for all $x\in\Bbb{Q}$.

I know that each field of characteristic $0$ contains a copy of $\Bbb{Q}$. But I'm afraid I'm not quite sure what I should read on in order to be fully ready to prove this result. Does this have something to do with algebraic field extensions?

This is how I would begin my proof... Let $x\in \mathbb{Q}$, then $x\in F$. Since $\phi$ is injective, $!\exists y\in K$ such that $\phi(x)=y$. Let $x=\frac{a}{b}$ and suppose that $\phi(x)\in K\backslash \mathbb{Q}$.

What would be the property of these two fields and/or the homomorphism that I should pay my attention to in order to understand how to prove this result? Thanks you for your suggestions.

• Because $\phi$ is a homomorphism, it must send $1$ to $1$ and $0$ to $0$. Now use that $\phi$ preserves the additive structure to see what must happen to $\Bbb Z$. After this, multiplicitivity will tell you where $\Bbb Q\subseteq F$ must map. – Stahl Mar 23 '17 at 7:36
• The set of fixed points is a field and $\mathbb Q$ is contained in any subfield. – MooS Mar 23 '17 at 7:38
• Hint: you just need that $\phi$ is an injective group homomorphism with respect to addition such that $\phi(1)=1$. – egreg Mar 23 '17 at 8:08
• The beginning of your proof doesn't make any sense. You say: "Let $x\in \mathbb{Q}$, then $x\in F$". It's not that $F$ literally contains $\mathbb{Q}$, it's that the subfield generated by $1$ is isomorphic to $\mathbb{Q}$. Understanding this properly will make the exercise above easy. – Mathematician 42 Mar 23 '17 at 8:11
• @Mathematician42 I politely disagree. Viewing $\mathbb Q$ as subfied of any other field of characteristic $0$ does not only make this exercise easier (the exercise precisely does this just by its formulation), it also makes your whole life easier. And since there is a unique monomorphism $\mathbb Q \to F$, this is not an issue at all. – MooS Mar 23 '17 at 8:17

If $1_K \in K$ and $1_F \in F$ are the multiplicative neutral elements of $K$ and $F$, notice that $\Bbb Q$ naturally embeds in $K$ by the map $\Bbb Q \ni \frac p q \mapsto (p \cdot 1_K) \cdot (q \cdot 1_K)^{-1} \in K$. This is where we use the fact that the characteristic is $0$: this guarantees that $q \cdot 1_K \ne 0_K$, so we may invert it. Similar considerations hold for $F$.

Let now $\phi : F \to K$ be an injective field homomorphism. In general, $\phi (1_F) = \phi (1_F ^2) = \phi (1_F)^2$, whence it follows that either $\phi (1_F) = 0_K$ or that $\phi (1_F) = 1_K$. Injectivity eliminates the first possibility, whence it follows that $\phi (1_F) = 1_K$. We have, for every $m \in \Bbb N \setminus \{0\}$ that

$$\phi (m \cdot 1_F) = \phi ( \underbrace {1_F + \dots + 1_F} _{m \text{ times}}) = \underbrace {\phi (1_F) + \dots + \phi (1_F)} _{m \text{ times}} = \underbrace{ 1_K + \dots + 1_K} _{m \text{ times}} = m \cdot 1_K$$

It is obvious that $\phi (0 \cdot 1_F) = \phi (0_F) = 0_K = 0 \cdot 1_K$, so we deduce that $\phi (m \cdot 1_F) = m \cdot 1_K$ for all $m \in \Bbb N$.

Mow, if $m \in \Bbb N$ then $0_K = \phi (0_F) = \phi(m \cdot 1_F - m \cdot 1_F) = \phi (m \cdot 1_F) + \phi (- m \cdot 1_F)$, whence we deduce that $\phi (-m \cdot 1_F) = - \phi (m \cdot 1_F) = - m \cdot 1_K$, which finally allows us to say that $\phi (m \cdot 1_F) = m \cdot 1_K$ for all $m \in \Bbb Z$.

For $p \in \Bbb Z$ and $q \in \Bbb Z \setminus \{0\}$ we may now write

$$\phi \big( (p \cdot 1_F) (q \cdot 1_F)^{-1} \big) = \phi \big( (p \cdot 1_F) \big) \cdot \phi \big( (q \cdot 1_F)^{-1} \big) = (p \cdot 1_K) \cdot (q \cdot 1_K) ^{-1} ,$$

which shows that the copy of $\Bbb Q$ inside $F$ is taken isomorphically onto the copy of $\Bbb Q$ inside $K$.