Why any homomorphism between fields over $\mathbb Q$ fixes $\mathbb Q$ pointwise?

I am studying section 14.1 in Dummit and Foote "Abstract Algebra" third edition and I am trying to prove that:

$\mathbb Q(\sqrt{2})$ and $\mathbb Q(\sqrt{3})$ are not isomorphic.

But I want to use the fact that any homomorphism between fields over $\mathbb Q$ fixes $\mathbb Q$ pointwise and I do not know the proof of this fact, Could someone tell me the proof of it or at least tell me where it is in section 14.1 in the book?

Thanks in advance!

Solution 1:

The proof is not difficult, and you should definitely be able to figure out yourself. I will show you the basic idea. Let $K$ and $L$ be fields containing $\mathbb Q$, and $\phi: K \rightarrow L$ a homomorphism of fields. We want to show that $\phi(x) = x$ for all $x \in \mathbb Q$.

Since $\phi$ is ring homomorphism, we know that $\phi(1) = 1$. Therefore, $\phi(2) = \phi(1+1) = \phi(1) + \phi(1) = 1+1=2$.

Similarly, $\phi(3) = \phi(1+1+1)= \phi(1) + \phi(1) + \phi(1) = 1+1+1 = 3$.

By the same reasoning, $\phi(n) = n$ for any natural number $n$.

It is not difficult to argue from here that $\phi(n) = n$ for any integer $n$. Since every rational number $x$ is equal to $\frac{n}{m}$ for a pair of integers $n$ and $m$, it is not difficult to argue furthermore that $\phi(x) = x$ for all rational numbers $x$.