I'm trying to show the intersection of two finite Galois extension is still Galois while finding my own argument not plausible, so I'm just wondering if there are any errors in my argument.
Below is my argument:
Say $F$ is a field and $E_1,E_2$ are finite Galois extensions of $F$. Then, since $E_1$ is a finite Galois extension, by Artin's primitive element theorem, we have $E_1=F(\alpha)$ for some element $\alpha$. Similarly, $E_2=F(\beta)$ for some element $\beta$.
Then if $\alpha\in E_2$, then we have $F(\alpha)\subseteq E_2$ and so $E_1\cap E_2=F(\alpha)\cap E_2=F(\alpha)=E_1$ and thus the intersection is indeed Galois. Similarly if $\beta\in E_1$ then we are done.
Hence, we may assume $\alpha\notin E_2$ and $\beta\notin E_1$. However, in this case, we must have $E_1\cap E_2=F(\alpha)\cap F(\beta)$ where $F(\alpha)$ does not contain $\beta$ and $F(\beta)$ does not contain $\alpha$. Viz $E_1\cap E_2=F$, which is a Galois extension of $F$ trivially.
End of my argument.
The part I'm uncertain with is that is it necessary if $\alpha$ not in $F(\beta)$ and $\beta$ not in $F(\alpha)$ then their intersection must be $F$. Also, this solution seems too naive to be true, so I'm not sure.
Thanks in advance.
Your proof is indeed wrong, it is not necessarily true that if $\alpha\notin F(\beta)$ and $\beta\notin F(\alpha)$ then $F(\alpha)\cap F(\beta)=F$. The following is a counterexample: let $F=\mathbb Q$, $\alpha=i+\sqrt{2}$ and $\beta=\sqrt{2-\sqrt{2}}$. Then one can check that $F(\alpha)$ is a Galois extension of $\mathbb Q$ with Galois group $C_2\times C_2$ while $F(\beta)$ is a Galois extension of $\mathbb Q$ with Galois group $C_4$. Hence, $\alpha\notin F(\beta)$ and $\beta\notin F(\alpha)$. However, $F(\alpha)\cap F(\beta)=\mathbb Q(\sqrt{2})$.
To prove your claim you only need to check that $E_1\cap E_2$ is a normal and separable extension of $F$. Separability is obvious: take any element $\alpha$ in the intersection and look at its minimal polynomial. Since $\alpha\in E_1$ and $E_1/F$ is Galois by hypothesis, all the roots must be distinct. On the other hand, consider an irreducible $f\in F[x]$ and let $\alpha\in E_1\cap E_2$ be one of its roots. Since $\alpha\in E_1$ and $E_1/F$ is Galois, then also all other roots belong to $E_1$. But the same is true for $E_2$, and hence $f$ splits into linear factors in $(E_1\cap E_2)[x]$.