A basic question on factorization Is the following true? If not, can anyone add some reasonable assumptions to make it true?
Statement. Let $E/F$ be a field extension and let $\alpha \in E \setminus F$ be algebraic over $F$. If $f(x) \in F[x]$ is irreducible, then $f(x/\alpha) \in F[\alpha][x]$ is irreducible.
Attempt to prove the statement. Suppose for contradiction that $f(x/\alpha) = p(x)q(x)$ where $p(x), q(x) \in (F[\alpha])[x]$ and $\deg(p(x)), \deg(q(x)) > 0$. Over a splitting field $K$ that contains $\alpha$, we can write $f(x) = a(x - r_{1}) \cdots (x - r_{n})$ for some $a \in F^{*}$, so we have 
$f(x/\alpha) = (a/\alpha^{n})(x - \alpha r_{1}) \cdots (x - \alpha r_{n})$.
Without loss of generality, we can write $p(x) = \gamma(x - \alpha r_{1}) \cdots (x - \alpha r_{j})$ and $q(x) = \delta(x - \alpha r_{j+1}) \cdots (x - \alpha r_{n})$ for some $\gamma, \delta \in F[\alpha]$ such that $\gamma\delta = a / \alpha^{n}$. Since $p(x), q(x) \in (F[\alpha])[x]$, we have 
$p_{0}(x) := \delta p(x) = a(x/\alpha - r_{1}) \cdots (x/\alpha - r_{j}) \in (F[\alpha])[x];$
$q_{0}(x) := (\gamma/a) q(x) = (x/\alpha - r_{j+1}) \cdots (x/\alpha - r_{n}) \in (F[\alpha])[x].$
We have $f(x) = p_{0}(\alpha x)q_{0}(\alpha x)$, and I was wishing that $p_{0}(\alpha x), q_{0}(\alpha x)$ were polynomials over $F$, but there are no reasons to believe so.
Motivation. The motivation was to show $\Phi_{p}(x/\alpha) \in \mathbb{Q}[\alpha][x]$ is irreducible where $\alpha := 2^{1/p}$ and $\Phi_{p}(x) \in \mathbb{Q}[x]$ is the $p$th cyclotomic polynomial where $p$ is a prime number. I did attempt this in a different way but I am not sure how to conclude whether this is true.
A connection to the statement. Write $\zeta := e^{2\pi i/p}$, and we have $\mathbb{Q}[\zeta, \alpha]$ as a splitting field of $x^{p} - 2 \in \mathbb{Q}[x]$ of degree $(p-1)p$, which shows that $[\mathbb{Q}[\alpha\zeta, \alpha] : \mathbb{Q}[\alpha]] = p-1$, thus $\alpha \zeta$ must have the minimal polynomial $g(x) \in \mathbb{Q}[\alpha][x]$ whose degree is $p-1$. 
If the previous statement is true, there is no choice but $g(x) = \alpha^{p}\Phi_{p}(x/\alpha)$, but like I said, I am not sure.
 A: I am a little regretful that I have not thought about this much before I asked this question. After talking to Gerry Myerson, I realized that the statement that I stated was very far from being true. Instead, we have the following.
Lemma. A polynomial $f \in F[x]$ is irreducible if and only if it is a minimal polynomial of some $\alpha$ in some extension $E/F$.
Proof. Suppose $f$ is irreducible. Then take $E = F[x]/(f)$. The converse is trivial. Q.E.D.
This seems like a very naive lemma, but I think the following variation looked very nontrivial to me before.
Corollary. A polynomial $f \in F[x]$ is irreducible if and only if there exists an extension $E/F$ and $\alpha \in E$ such that $f(\alpha) = 0$ and $\deg(f) = [E:F]$.
Proof. If we have an extension $E/F$ and $\alpha \in E$ such that $f(\alpha) = 0$, then the minimal polynomial of $\alpha$ must divide $f$ since it is the generator of the kernel of evaluation $(F[x], x) \rightarrow (F[\alpha], \alpha)$. The rest follows from Lemma. Q.E.D.
My previous question was as follows. 
Question. Let $\alpha \in E$ is algebraic over $F$ where $E/F$ is an extension. How can we tell $f(x/\alpha)$ is irreducible over $F[\alpha]$ given $f(x)$ is irreducible over $F$?
Answer. By Corollary, the polynomial $f(x/\alpha)$ is irreducible over $F[\alpha]$ if and only if we have an extension $L/F[\alpha]$ and $\beta \in L$ such that $f(\beta/\alpha) = 0$ and $\deg(f) = [L:F[\alpha]]$.
