relationship between monomorphism and automorphism

Question

If I have a monomorphism $\phi : L \to M$ where $L,M$ are fields with $L \subset M$ and I know that $\phi(L) \subset L$ does that necessairly mean that $\phi : L \to L$ is an automorphism? I can understand it if $\phi(L) = L$, but we don't have that

Answer 1

Since $\phi(L) \subset L$ this question is the same as saying given a monomorphism $\phi : L \to L$ is $\phi$ necessarily an automorphism? As shown in the post linked in the comment, this is not true. However, let us suppose that $L$ is an algebraic extension of a field $F$, and $\phi$ is an $F$-monomorphism, i.e. $\phi(a) = a$ for all $a \in F$. Then $\phi$ is indeed an automorphism.

We need only show that $\phi$ is surjective. Let $\alpha \in L$. Since $L/F$ is algebraic, $\alpha$ is algebraic over $F$. Let $m(x)$ be the minimal polynomial of $\alpha$ over $F$ and let $S = \{\beta \in L : m(\beta) = 0\}$. You can check that since $\phi$ is an $F$-monomorphism, $\phi(\alpha) \in S$. Since $\phi$ is injective, we must have $\phi(S) = S$, i.e. $\phi$ permutes the roots of $m(x)$ in $L$. Can you see now why $\phi$ must be surjective?

Answer 2

The question linked by Mathmo123 in the comments answers this one, because we can take $L=M=K$ and $F$ to be the base field (either $\mathbb{Q}$ or $\mathbb{Z}/p$).

A standard counter-example, given by knsam in the comments, is the map $\mathbb{Q}(t)\to\mathbb{Q}(t)$ sending $t$ to $t^2$.

If your book claims that your statement is true, it is most likely considering the case where $L$ is a finite extension of some base field, like $\mathbb{Q}$. In that case, $L$ and $\phi(L)$ must have the same degree over $\mathbb{Q}$, and so if $\phi(L)\subset L$, then they are equal.