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Minimal polynomial trouble

caffeinemachine

Well-known member
MHB Math Scholar
Mar 10, 2012
834
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 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?
 

caffeinemachine

Well-known member
MHB Math Scholar
Mar 10, 2012
834
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 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?
Here's an answer. abstract algebra - Minimal Polynomial Trouble - Mathematics Stack Exchange