If $\min(\alpha,F)$ has only one root in $E$, must $\min(p(\alpha),F)$ have only one root in $E$

Question

Let $F<E$ 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 $0$ ?

Thank you a lot.

(Note: $\min(\alpha,F)$ denotes the minimal monic polynomial of $\alpha$ over $F$)

Bounty is for the answer to both questions :)

Answer 1

The answer to both your questions is NO. Here is a counterexample in characteristic zero. Consider $Q=X^3-X-1$. Its discriminant is $-23$ , let $w$ be a (necesarily complex) square root of this discriminant and $E={\mathbb Q}(w)$, $P=Q(X^3)$, finally let $\alpha$ be a root of $P$.

Then, putting $g=\alpha+w$ we have $(g-w)^9-(g-w)^3-1=0$ or $$g^9 - 828g^7 + 66654g^5 - 1022029g^3 + 2518638g - 1\\ =w(9g^8 - 1932g^6 + 66654g^4 - 438015g^2 + 279864)$$, whence $$(g^9 - 828g^7 + 66654g^5 - 1022029g^3 + 2518638g - 1)^2+23(9g^8 - 1932g^6 + 66654g^4 - 438015g^2 + 279864)^2=0$$. Since this polynomial is irreducible over $\mathbb Q$ (to check this, type P<g> := PolynomialRing(IntegerRing());Factorization((g^9 - 828*g^7 + 66654*g^5 - 1022029*g^3 + 2518638*g - 1)^2+23*(9*g^8 - 1932*g^6 + 66654*g^4 - 438015*g^2 + 279864)^2); in the Magma calculator online ) it follows that $P$ is irreducible over $E$. The Magma calculator also tells us that the decomposition field (let us call it $\mathbb D$) of $P$ over $\mathbb Q$ has degree $108$ over $\mathbb Q$ (to check this, type P<x> := PolynomialRing(IntegerRing());GaloisGroup(x^9-x^3-1);). This fact will be useful later on.

Suppose that $\beta\neq\alpha$ is a root of $P$ in ${\mathbb L}=E(\alpha)$. There are two cases. If $\alpha$ and $\beta$ correspond to the same root of $Q$ (i.e. $\alpha^3=\beta^3$), then $\frac{\beta}{\alpha}$ is a primitive third root of unity, so $E(\alpha)$ contains $j=e^{\frac{2\pi i}{3}}$. Since $2$ is coprime to $9$, it would follow that $j\in E$ which is easily seen to be false. Otherwise $\alpha^3$ and $\beta^3$ are distinct roots of $Q$ ; the third root is $\gamma^3$ where $\gamma=-\frac{1}{\alpha\beta}$. The roots of $P$ are then the numbers of the form $uv$ with $u\in\lbrace 1,j,j^2 \rbrace$ and $v\in\lbrace \alpha,\beta,\gamma\rbrace$. So it suffices to add $j$ to $\mathbb L$ to get all the roots of $P$, and hence ${\mathbb D}={\mathbb L}[j]$. But this is impossible as $[{\mathbb L}[j]:{\mathbb Q}] \leq 9\times 2\times 2=36$ while $[{\mathbb D}:{\mathbb Q}]=108$.

We have thus shown that $E(\alpha)$ cannot ontain a root of $P$ other than $\alpha$. On the other hand, the minimal polynomial of $a=\alpha^3$ over $E$ is $Q$ and $E(\alpha^3)$ contains all the roots of $Q$, by the choice of $w$ (explicitly, the two other roots of $Q$ are $-\frac{a}{2}\pm pw$ where $p=\frac{6a^2-9a-4}{46}$).