$F$ is purely inseparable over $K$ if and only if the only $K$-monomorphism $F \to C$ is the inclusion map.

Question

Let $C$ be an algebraically closed field, and $F,K$ are fields with $K\subseteq F\subseteq C$. How to show that $F$ is purely inseparable over $K$ if and only if the only $K$-monomorphism $F \to C$ is the inclusion map?

Answer 1

This statement is not true . Consider F be a finite extension of degree $n(>0)$ over $K=\mathbb{Q}$, which is the field of rational numbers. Then any $\mathbb{Q}$ algebra from $K$ to the algebraic closure of $\mathbb{Q}$ is always injective. But F is never be a inseperableble extension of $\mathbb{Q}$. Hence the statement is false

If you consider the map $F\hookrightarrow C$ is just a vector space homomorphism, then above example shows that the converse is not true.

Answer 2

Couple of hints.

In one direction, if $\Phi\colon F\to C$ is a homomorphism of $K$-algebras, then for any element $\alpha\in F$ you have $p_\alpha(\Phi(\alpha))=0$, where $p_\alpha$ is the minimal polynomial of $\alpha$ over $F$.

In the other direction, if $p$ is an irreducible polynomial over $K$ with distinct roots $\alpha_1,\alpha_2\in C$, then $K[\alpha_1]\cong K[\alpha_2]$ and the isomorphism extends to an automorphism of $C$.

Answer 3

Repeating again $K=\mathbb{Q}\subset F=\mathbb{Q}[\sqrt{3}]\subset C=\bar{\mathbb{Q}}$ is a seperable extension and any nonzero K algebra homomorphism $F\hookrightarrow C$ is injective.