Field homomorphism $\phi: F \to F$ with $F$ extension over $\mathbb{Q}$ is identity map

Question

While reviewing exercises for introduction to algebra final I came across the following question: Consider a field extension $F$ over $\mathbb{Q}.$ Now suppose we have a field homomorphism $\phi: F\to F $ Prove that for all $x\in \mathbb{Q}$ one has $\phi(x)=x.$ There was also a follow up question where we considered $F=\mathbb{Q}(\sqrt[4]{2021}).$ We needed to determine all the field isomorphisms $\phi: F \to F $, and prove that they were isomorphisms.

Now I don't know how to tackle this problem. I know a field homomorphism is by definition injective, so in the first part $\phi: F\to F $ is an isomorphism. I don't really know what to do with that information.

Answer 1

The proof of $\forall x\in\mathbb Q\colon\phi(x)=x$ can be done "by hand" in the following manner:

• First note that $\phi(0)=0$ and $\phi(1)=1$ by definition of field homomorphism.
• From this we can deduce $\phi(n)=n$ for every $n\in\mathbb N$ and actually every $n\in\mathbb Z$ since for field homomorphisms we know $\forall x\in\mathbb Q\colon\phi(-x)=-\phi(x)$.
• Now let $\frac{p}{q}\in\mathbb Q$ with $p,q\in\mathbb Z$ and $q\neq0$. Then we have $q\cdot\phi\left(\frac pq\right)=\phi(q)\cdot\phi\left(\frac pq\right)=\phi\left(q\cdot\frac pq\right)=\phi(p)=p$ so we end up with $\phi\left(\frac pq\right)=\frac pq$ and are done.

This has actually shown that the automorphism group of $\mathbb Q$ is the trivial group and it's a fun exercise to prove the same for $\mathbb R$ (via checking that any such automorphism is order-preserving and following one of the routes mentioned in the comments).

Answer 2

For the second part of your exercise, we can do the following.

Suppose $\phi: F \to F$ is a field morphism, where $F = \Bbb{Q}(\sqrt[4]{2021})$. By the first question, since $F$ is a field extension of $\Bbb{Q}$, $\phi$ is the identity on $\Bbb{Q}$.

Now consider $2021 = \phi(2021) = \phi(\sqrt[4]{2021}^4) = \phi(\sqrt[4]{2021})^4$. Hence we only have two options for $\phi(\sqrt[4]{2021})$. It is either $\sqrt[4]{2021}$ or $-\sqrt[4]{2021}$ (note that the other two fourth roots are impossible since the image $\phi$ is, in particular, real).

Because the field extension is characterized by $\sqrt[4]{2021}$, its image fixes the entire morphism (every element in $F$ is written as some rational expression in $\sqrt[4]{2021}$). Hence there are exactly two morphisms; \begin{align} &\phi_1: F \to F: \sqrt[4]{2021} \mapsto \sqrt[4]{2021} \\ &\phi_2: F \to F: \sqrt[4]{2021} \mapsto -\sqrt[4]{2021} \end{align} Note that both of these are isomorphisms (in fact, $\phi_1 = \operatorname{Id}$).