The extension KE/F is finite and separable

[mathmari]

Hey!

Let $C$ be an algebraic closure of $F$ and let $f\in F[x]$ be separable.
Let $K\leq C$ be the splitting field of $f$ over $F$ and let $E\leq C$ be a finite and separable extension of $F$.
I want to show that the extension $KE/F$ is finite and separable. We have that $KE$ is the smallest field that contains $K$ and $E$.

Since $K$ is the splitting field of a polynomial over $F$, we have that the extension $F\leq K$ is normal, right? (Wondering)

We have the following two chains of extensions:
$$F\leq K\leq KE\leq C \\ F\leq E \leq KE\leq C$$ right? (Wondering)

Since $C$ is an algebraic closure of $F$, we have that the extension $C/F$ is algebraic, and so finite.
We have that $[C:F]=[C:KE][KE:F]$, and so the extension $KE/F$ is also finite.

How can we show that the extension $KE/F$ is also separable? (Wondering)

 

[jvicens]

Hi there,

Great question! To show that the extension $KE/F$ is separable, we can use the fact that $f$ is separable over $F$. Since $f\in F[x]$ is separable, we know that all of its roots in $C$ are distinct. This means that the minimal polynomial of any element $\alpha \in KE$ over $F$ is separable. Therefore, the extension $KE/F$ is separable.

Also, you are correct in saying that the extensions $F\leq K$ and $F\leq E$ are normal, as they are both splitting fields of polynomials over $F$. And your chains of extensions are also correct.

I hope this helps! Let me know if you have any other questions.