$f$ is factored into many same degree irreducible polynomials. I met a problem when I study about Galois field and do this exercise. Hopefully, someone can help me.
Suppose that $L/K$ is normal extension and $f$ is an irreducible polynomial in $K[X]$. Prove that if $f$ is reducible in $L$ then $f$ is factored into many same degree irreducible polynomials. 
The first solution:
Because $f$ is an irreducible polynomial then $f$ can be written by $f_{1}f_{2}...f_{m}$ with $f_{i}$ is irreducible.
$\alpha_{i}$ and $\alpha_{j}$ are roots of $f_{i}$ and $f_{j}$. We need to show that $degf_{i}=degf_{j}$
It suffices to show that $[L(\alpha_{i}) : L]=[L(\alpha_{j}) : L]$


*

*$[L(\alpha_{i}) : L][L : K]=[L(\alpha_{i}) : K]=[L(\alpha_{i}) : K]= [L(\alpha_{i} : K(\alpha_{i})][K(\alpha_{i}) :K]$.

*$[L(\alpha_{j}) : L][L : K]=[L(\alpha_{j}) : K]=[L(\alpha_{j}) : K]= [L(\alpha_{j} : K(\alpha_{j})][K(\alpha_{j}) :K]$.


But $f$ is irreducible in $K[X]$ then $degf=[K(\alpha_{j}) :K]=[K(\alpha_{i}) :K]$.
Moreover, $\phi$ is homomorphism from $K(\alpha_{i})$ to $K(\alpha_{j})$, $\phi(a)=a$, $a\in K$, $\phi(\alpha_{i})=\alpha_{j}$. It is isomorphism then creates an isomorphism from $L(\alpha_{i})$ to $L(\alpha_{j})$.
WED.
I do not understand why "It is isomorphism then creates an isomorphism from $L(\alpha_{i})$ to $L(\alpha_{j}).$"
However, I saw another solution that is:
Call that $H$ is the seperate field of $L$ and $G=Gal(H/L)$. Take $\phi$ in $G$ we have $\phi(\alpha_{i})=\alpha_{j}$ then $\phi(f_{i})$ is irreducible and $\alpha_{j}$ is a root of it. Because $f_{i}$, $f_{j}$ and $\phi_(f_{i})$ are irreducible so degree of them is equal. 
I do not know why $\phi(f_{i})$ is irreducible and $\alpha_{j}$ is a root of it.
$\mathbb{Q} \subseteq \mathbb{Q}(\sqrt[3]{2} ,$i$)$.
$f(x)=x^{3}-2$ take $\sqrt[3]{2}$ being a root but $i$ is not when we effect $\phi$ on it . 
I really wanna see the explantion.
 A: I am also not really satisfied with the given solution, since it does not point out, that normality of $L/K$ is important.
This is my proof:
First, note that we can assume $L$ to be finite over $K$, if not, we can always replace it by the normal hull of $K(\text{coefficients of the } f_i)$. Also, note that your solution also implicitly assumes $L/K$ to be finite without saying it and without giving an argument why this is allowed...
$\alpha_i$ and $\alpha_j$ have the same minimal polynomial over $K$, hence there is an $K$-isomorphism $$K(\alpha_i) \to K(\alpha_j), \alpha_i \mapsto \alpha_j$$
Note that $L(\alpha_i)/K(\alpha_i)$ is algebraic. By a fundamental lemma about algebraic field extensions, we can extend this homomorphism to a $K$-homomorphism
$$\sigma: L(\alpha_i) \to \overline K$$
Here comes the crucial point: A priori we do not know that the image of this map is contained in $L(\alpha_j)$ (and it might be false if $L/K$ is not normal). But in the normal case, we have the following argument to show it:
We have $\sigma(K(\alpha_i)) = K(\alpha_j) \subset L(\alpha_j)$ and $\sigma(L) \subset L \subset L(\alpha_j)$, where the crucial inclusion $\sigma(L) \subset L$ holds since $L/K$ is normal!
Thus we have shown $\sigma(L(\alpha_i)) \subset L(\alpha_j)$, i.e. we have found a $K$-homomorphism $L(\alpha_i) \to L(\alpha_j)$. Any field homomorphism is injective, hence we obtain $[L(\alpha_i):K] \leq [L(\alpha_j):K]$. We can repeat the argument with $i,j$ swapped to obtain $[L(\alpha_j):K] \leq [L(\alpha_i):K]$, hence $[L(\alpha_i):K] = [L(\alpha_j):K]$ and in particular $[L(\alpha_i):L] = [L(\alpha_j):L]$. This completes the proof.
A: The elements you are adding ($\alpha_i,\alpha_j$) are roots of $f_i$ and $f_j$. Now there exists as you say $\phi:L(\alpha_i)\to L(\alpha_j)$ which is the identity on $K$ and which sends $\alpha_i\mapsto \alpha_j$ (this is because $\alpha_i$ and $\alpha_j$ have the same polynomial over $K$, so your counterexample does not apply here). Now $f_i(\alpha_i)=0$ so applying $\phi$ $f_i$ is sent to a polynomial which is zero in $\phi(\alpha_i)=\alpha_j$, so it is a multiple of $f_j$, the minimal polynomial of $\alpha_j$. It is irreducible because a factorization after appliying $\phi$ yields a factorization in $L(\alpha_i)$ just by applying $\phi^{-1}$. So you have $f_i=f_j$.
If you still have questions, just ask!
