# Minimal Polynomial Trouble

Let $L$ be an extension of a field $F$. Let $\alpha_1, \alpha_2\in L$ be such that both of them are algebraic over $F$ and have the same minimal polynomial $m$ over $F$. We know that there is an isomorphism $\phi:F(\alpha_1)\to F(\alpha_2)$ defined as $\phi(\alpha_1)=\alpha_2$ and $\phi(x)=x$ for all $x\in F$. We can extend $\phi$ to an isomorphism $\psi:F(\alpha_1)[x]\to F(\alpha_2)[x]$ defined as $\psi(\sum_{j=0}^k a_jx^j)=\sum_{j=0}^k \phi(a_j)x^j$. Let $l_1$ be the minimal polynomial of $\alpha_2$ over $F(\alpha_1)$ and define $l_2(x)=\psi(l_1(x))$.

QUESTION. Is it necessarily true that $\alpha_1$ has $l_2$ as its minimal polynomial over $F(\alpha_2)$?

I have some evidence which says that $l_2$ indeed is the minimal polynomial of $\alpha_1$ over $F(\alpha_2)$.

1. Say $l_2^*$ is the minimal polynomial of $\alpha_1$ over $F(\alpha_2)$. Then $[F(\alpha_1,\alpha_2):F(\alpha_2)]=\deg l_2^*$. Which gives $[F(\alpha_1,\alpha_2):F]=\deg l_2^*\cdot\deg m$. Similarly $[F(\alpha_1,\alpha_2):F]=\deg l_1\cdot\deg m$ and we conclude that $\deg l_1=\deg l_2^*$.

2. Since $l_1$ is the minimal polynomial of $\alpha_2$ over $F(\alpha_1)$, we must have that $l_1(x)|m(x)$ in $F(\alpha_1)[x]$. Thus there is an element $q_1(x)\in F(\alpha_1)[x]$ such that $l_1(x)q_1(x)=m(x)$. Operating $\psi$ on both the sides we get $l_2(x)q_2(x)=m(x)$, where $q_2(x)=\psi(q_1(x))$. Now since $m(\alpha_1)=0$, we have $l_2(\alpha_1)q_2(\alpha_1)=0$. If from here we can show that $l_2(\alpha_1)=0$ then we'd be done. But I can't rule out the possibility that $q_2(\alpha_1)=0$.

Can any one help?

No, I think maybe you were taking examples that were too easy (or maybe too hard!). Let’s look at a specific situation, where $L$ is normal and abelian over $K$. Take indeed a cubic cyclic extension $L$ of $\mathbb Q$. Let $\zeta\in L\setminus\mathbb Q$, so that $L=\mathbb Q(\zeta)$. Then $\zeta$ has two conjugates, say $\zeta'$ and $\zeta''$, both in $L$. Now call $\phi$ the automorphism of $L$ that sends $\zeta$ to $\zeta'$; it’ll send $\zeta'$ to $\zeta''$, of course, since it’s not an involution. And what’s the minimal polynomial for $\zeta'$ over $L=\mathbb Q(\zeta)$? It’s $X-\zeta'$, since $\zeta'$ is already in $L$. And when you operate on $X-\zeta'$ by $\phi$, you get $X-\zeta''$, the minimal polynomial for $\zeta''$, not for $\zeta$.
In case you haven’t worked with cubic cyclic fields over $\mathbb Q$, the easiest one is the cubic subfield of $\mathbb Q(\omega_7)$, where $\omega_7$ is a primitive seventh root of unity. It’s not hard to find an element in this field and its minimal polynomial.
EDIT: To help @caffeinemachine with the example I had in mind, let me work it out more fully. We start with $\omega=\omega_7$, primitive seventh root of unity, whose minimal polynomial over $\mathbb Q$ is $(X^7-1)/(X-1)$ We therefore have the fundamental “zero-relation” $\omega^3+\omega^2+\omega+1+\omega^{-1}+\omega^{-2}+\omega^{-3}=0$. Now let $\zeta=\omega+\omega^{-1}$, and expand $\zeta^3$ using Binomial Theorem, and then subtract the fundamental zero from the right-hand side of what you just got to find that $$\zeta^3=-\omega^2+2\omega-1+2\omega^{-2}-\omega^{-2}\,.$$ Next write $$\zeta^3+\zeta^2=2\omega+1+2\omega^{-2}\,,$$ so that the minimal polynomial for $\zeta$ is $F(X)=X^3+X^2-2X-1$. By calculating $F(X+2)$ and seeing that the result is $7$-Eisenstein, you verify that $F$ itself is irreducible. Now, the Galois group of $\mathbb Q(\omega)$ over $\mathbb Q$ is cyclic, generated by (for instance) $\omega\mapsto\omega^3$. I won’t go through the details, but you see that this substitution sends $\zeta$ to $-\zeta^2-\zeta+1$, and we can call this quantity $\zeta'$, and the automorphism of $\mathbb Q(\zeta)$ that induced it we can call $\phi$. You can check that $\phi(\zeta')=\zeta^2-2$, which is what I had in mind for $\zeta''$.