If $\min(\alpha,F)$ has only one root in $E$, must $\min(p(\alpha),F)$ have only one root in $E$ Let $F\subseteq E$ be an algebraic field extension. Let $\alpha\in E$ be such that $\min(\alpha,F)$ has only one root in $E$ (which will be $\alpha$). Is it true that for any $p(x)\in F[x]$ we must have:
"$\min(p(\alpha),F)$ has only one root in $E$"
Another question: Does the above conjecture at least hold in characteristic $0$?
(Note: $\min(\alpha,F)$ denotes the minimal monic polynomial of $\alpha$ over $F$)
 A: This is not true in general, even in characteristic $0$:
Example. Let $\alpha = \sqrt[3]{1+\sqrt{8}} \in \mathbb R$, and let $L = \mathbb Q(\alpha)$. The minimal polynomial of $\alpha$ over $\mathbb Q$ is
$$(x^3-1)^2 - 8 = x^6 - 2x^3 - 7.$$
If $\beta = \sqrt[3]{1-\sqrt{8}}$ and $\zeta_3 = e^{2\pi i/3}$, then the roots of $f$ in $\mathbb C$ are
\begin{align*}
\alpha, & & \zeta_3 \alpha, & & \zeta_3^2 \alpha,\\
\beta, & & \zeta_3 \beta, & & \zeta_3^2 \beta.
\end{align*}
Since $L \subseteq \mathbb R$, the only other root that could possibly be in $L$ is $\beta$ (the other ones aren't even in $\mathbb R$). But
$$(1+\sqrt{8})(1-\sqrt{8}) = -7,$$
showing that $(1-\sqrt{8})$ and $(1+\sqrt{8})$ are the primes of $K = \mathbb Q(\sqrt{2})$ above $7$. For distinct principal primes $(p)$ and $(q)$ (to be safe let's say not lying above $(2)$ or $(3)$), the extensions $K(\sqrt[3]{p})$ and $K(\sqrt[3]{q})$ are linearly disjoint (look at ramification behaviour).
In particular, $\sqrt[3]{1-\sqrt{8}} \not\in K(\sqrt[3]{1+\sqrt{8}})$, so $\alpha$ is the only root of $f$ in $L$. On the other hand, $\sqrt{8} = \alpha^3 - 1$ is a polynomial in $\alpha$, and its minimal polynomial $x^2 - 8$ has two roots in $L$. $\square$
