Suppose $a,b\in E$ and $\sigma\in \textrm{Gal}(E/F)$ with $\sigma(a)=b$. Show that $a$ and $b$ have the same minimal polynomial in $F[x]$ Let $E/F$ be a Galois extension. Suppose $a,b\in E$ and $\sigma\in \textrm{Gal}(E/F)$ with $\sigma(a)=b$. I need to show that $a$ and $b$ have the same minimal polynomial in $F[x]$.
Here's what I have so far:
Let $m(x)$ be the minimal polynomial for $b$. Then $m(b)=m(\sigma(a))=0$. If $\sigma$ is the identity function, then $a=b \Rightarrow a$ and $b$ have the same minimal polynomial. If $\sigma$ is not the identity function, then $\sigma(a)\neq b$ since $m$ must have distinct roots in $E/F$. If $\sigma(a)=a$, then $a,b$ have the same minimal polynomial. If $\sigma(a)\neq a$, then...
At this point, I am a little stuck. Any hints would be greatly appreciated!
 A: The main step you are missing is the equality $m\left(\sigma\left(a\right)\right) = \sigma\left(m\left(a\right)\right)$. This is a particular case of the more general equality
\begin{align}
p\left(\sigma\left(a\right)\right) = \sigma\left(p\left(a\right)\right)
\qquad \text{ for every polynomial } p \in F\left[x\right] .
\label{darij1.eq.1}
\tag{1}
\end{align}
To prove \eqref{darij1.eq.1}, you can fix a polynomial $p \in F\left[x\right]$. Write $p$ in the form $p = p_n x^n + p_{n-1} x^{n-1} + \cdots + p_0 x^0$ for some $n \geq 0$ and some $p_0, p_1, \ldots, p_n \in F$. Thus, $p\left(a\right) = p_n a^n + p_{n-1} a^{n-1} + \cdots + p_0 a^0$ but also
\begin{align}
& p\left(\sigma\left(a\right)\right) \\
& = p_n \left(\sigma\left(a\right)\right)^n + p_{n-1} \left(\sigma\left(a\right)\right)^{n-1} + \cdots + p_0 \left(\sigma\left(a\right)\right)^0 \\
& \qquad \left(\text{since } p = p_n x^n + p_{n-1} x^{n-1} + \cdots + p_0 x^0 \right) \\
& = \sigma\left(p_n\right) \left(\sigma\left(a\right)\right)^n + \sigma\left(p_{n-1}\right) \left(\sigma\left(a\right)\right)^{n-1} + \cdots + \sigma\left(p_0\right) \left(\sigma\left(a\right)\right)^0 \\
& \qquad \left(\text{since each $i$ satisfies } p_i = \sigma\left(p_i\right) \text{ (because } p_i \in F \text{)} \right) \\
& = \sigma\left( p_n a^n + p_{n-1} a^{n-1} + \cdots + p_0 a^0 \right) \\
& \qquad \left(\text{since } \sigma \text{ is a ring homomorphism}\right) \\
& = \sigma\left(p\left(a\right)\right)
\end{align}
(since $p_n a^n + p_{n-1} a^{n-1} + \cdots + p_0 a^0 = p\left(a\right)$). Hence, \eqref{darij1.eq.1} is proven.
Once you have $m\left(\sigma\left(a\right)\right) = \sigma\left(m\left(a\right)\right)$, you can use this to show that the minimal polynomial of $a$ divides the minimal polynomial of $b$ (why?). But the same argument, with $a$, $b$ and $\sigma$ replaced by $b$, $a$ and $\sigma^{-1}$, yields that the minimal polynomial of $b$ divides the minimal polynomial of $a$. Combine these two facts to obtain your claim.
