Question

Prove that a finite dimensional extension field K of F is normal if and only if it has this property whenever L is an extension field of K and $\sigma :K\to L$ an injective homomorphism such that $\sigma \left(c\right)=c$ for every then $\sigma \left(K\right)\subseteq K$.

Short answer

$\sigma \left(K\right)\subseteq K$

Step-by-step solution

Definition of third isomorphism theorem 

Third isomorphism theorem state that a ring R and and two ideal righl, J of that ring are such that ideal ring are sunset of each other then ratio of both ideal ring I/J is an ideal of $R/J$ and$\frac{R/J}{I/J}\cong R/I$

Showing that  σK⊆K

Let $\alpha$be a root of f in k . Since k follow closure property in $F^n$of F then there exist some root $\beta \in F^n$.

For all f which is not containg in K and the isomerism$F\left(\alpha \right)=F\left(\beta \right)$.

Over the field F . Both field are isomotphisc to $F\left[X\right]/\left[f\right]$. Now since $K/F$ is algebraic. Thus,

$\sigma :K\to F^n$

Such that

$\sigma \left(a\right)\notin K$ and $\sigma \left(K\right)\subseteq K$

Since $K/F$ is algebraic and $\sigma :K\to K$ is identity on F then sigma is surjective. Hence $\sigma$ is surjective so $\sigma \left(K\right)\subseteq K$.