- Thread starter
- #1

- Mar 10, 2012

- 834

**QUESTION:**Is is 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?