3
$\begingroup$

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 $F$)

$\endgroup$
3
$\begingroup$

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 $L$. $\square$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.