I'm going over the proof of Theorem 25.1 in Abstract Algebra, and I'm trying to understand a particular detail concerning separable polynomials. Here's the relevant portion of the theorem statement:
Let $E$ be a finite extension of a field $F$, and suppose that $E$ is the splitting field over $F$ for some separable $f(X) \in F[X]$.
Now to the relevant snippet of the proof: Let $f(X) \in F[X]$ be as stated above. To ease notation let $K = \Phi(\Gamma(E/F))$. The proof states that $f(X)$ is separable over $K$. Why is this? Here is the textbook's definition of a separable polynomial:
If $f(X) \in F[X]$ we say $f(X)$ is separable over $F$ if every irreducible factor of $f(X)$ in $F[X]$ has distinct roots.
It's clear that $F \subseteq K$, so $f(X) \in F[X] \implies f(X) \in K[X]$. So to show that $f(X)$ is separable over $K$, we must show that if $g(X)$ is a factor of $f(X)$ that is irreducible in $K[X]$, then $g(X)$ has distinct roots. Now if such a factor $g(X)$ happens to be in $F[X]$, then we immediately know it has distinct roots (because $f(X)$ is separable of $F$). But what about when $g(X) \in K[X] \setminus F[X]$? I suspect we need to somehow use the fact that $E$ is the splitting field for $f(X)$ over $F$ (or that $K = \Phi(\Gamma(E/F))$, but I'm not sure how to proceed with this...any help would be appreciated. Apologies if this is a really dumb question.
Edit 5/11/21: Solution Credit to @user363464 for this; I present here the solution in my own words.
First, I give the precise statement of Euclid's Lemma for polynomials.
Euclid's Lemma: Let $F$ be a field, let $g(X) \in F[X]$ be irreducible over $F$, and let $f(X) = f_1(X) f_2(X) \cdots f_n(X)$ for some $f_1(X) f_2(X),\ldots,f_n(X) \in F[X]$. If $g(X)$ divides $f(X)$ in $F[X]$ (i.e., if $f(X) = g(X) q(X)$ for some $q(X) \in F[X]$), then $g(X)$ divides $f_i(X)$ in $F[X]$ for some $i \in \{1,2,\ldots,n\}$.
Let $g(X)$ be a factor of $f(X)$ in $K[X]$ that is irreducible over $K$. By the Factorization Theorem for polynomial rings, $f(X) = f_1(X) \cdots f_n(X)$ for some polynomials $f_1(X),\ldots,f_n(X) \in F[X]$ which are irreducible over $F$. Then by Euclid's Lemma (applied to $f(X), g(X) \in K[X]$), $$g(X) | f(X) = f_1(X) \cdots f_n(X) \implies g(X) | f_i(X) \quad \text{for some } i \in \{1,2,\ldots,n\}.$$
Therefore, $f_i(X) = g(X) q(X)$ for some $q(X) \in K[X]$. Then since $f(X)$ is separable over $F$ and $f_i(X)$ is a factor of $f(X)$ in $F[X]$ that is irreducible over $F$, we know that $f_i(X)$ has distinct roots. (This is what it means for $f(X)$ to be separable over $F$.) Now all the roots of $g(X)$ are roots of $f_i(X)$ (because $f_i(X) = g(X) q(X)$), so the roots of $g(X)$ are distinct. Hence, $f(X)$ is separable over $K$. QED.
Let $g$ be an irreducible factor of $f$ over $K$. By euclid's lemma for polynomials, $g$ divides an irreducible factor $h$ of $f$ over $F$. Now $h$ has distinct roots since $f$ is separable over $F$. So does $g$, given that $g$ divides $h$.