$K$-monomorphism that is not $K$-automorphism?

Question

I am confused by the terminology where $K$ precedes terms such as $K$-monomorphism and $K$-automorphism in Galois theory. I am trying to come up with a simple example about $K$-monomorphism that is not $K$-automorphism to understand the purpose of preceding the words such as the monomorphism with the field $K$. How is the example different from example without the preceding $K$ in the terms such as here?

Answer 1

Consider subfields of $\mathbb{C}$. The polynomial $X^3 - 2$ is irreducible over $\mathbb{Q}$. Hence $F = \mathbb{Q}[X]/(X^3-2)$ is field. The roots of $X^3-2$ are $\sqrt[3]{2}, \omega \sqrt[3]{2},$ and $\omega^2\sqrt[3]{4}$, where $\omega$ is a primitive $3$rd root of unity. The ring homomorphisms $\mathbb{Q}[X] \rightarrow \mathbb{Q}(\sqrt[3]{2}), X \mapsto \sqrt[3]{2}$ and $\mathbb{Q}[X] \rightarrow \mathbb{Q}(\omega \sqrt[3]{2})$ are both surjective with kernel $(X^3-2)$, hence there are isomorphisms $\mathbb{Q}[X]/(X^3-2) \cong \mathbb{Q}(\sqrt[3]{2}), \mathbb{Q}(\omega \sqrt[3]{2})$. By composing, we get an isomorphism of fields $$\mathbb{Q}(\sqrt[3]{2}) \cong \mathbb{Q}(\omega \sqrt[3]{2})$$ given by $\sqrt[3]{2} \mapsto \omega \sqrt[3]{2}$ which is actually a $\mathbb{Q}$-isomorphism. But this is not a $\mathbb{Q}$-automorphism, since that would say that the fields are literally equal, not just isomorphic.