Prove that $|\text{Aut}(F(\alpha)/F)| = |\text{Aut}(F(\beta)/F)|$ and the groups $\text{Aut}(F(\alpha)/F)$ and $\text{Aut}(F(\beta)/F)$ are isomorphic Let $F$ be a subfield of $\mathbb{C}$. Suppose that $f(x) \in F[x]$ is irreducible over $F$, $\alpha, \beta \in \mathbb{C}$ and $f(\alpha) = f(\beta) = 0$.
(a) Prove that $|\text{Aut}(F(\alpha)/F)| = |\text{Aut}(F(\beta)/F)|$.
(b) Prove that the groups $\text{Aut}(F(\alpha)/F)$ and $\text{Aut}(F(\beta)/F)$ are isomorphic.
Which $\text{Aut}(S/T)$ is denoted as the set of automorphism of the extension $S$ of $T$.
Which $|\text{Aut}(S/T)|$ is denoted as the order of the group $\text{Aut}(S/T)$.
$\textbf{My Attempt:}$
(a) Since, $F$ be a subfield of $\mathbb{C}$.
Then, $\mathbb{C}$ is an extension field of $F$ which $\mathbb{C}/F$.
Since, $f(x) \in F[x]$, $\alpha, \beta \in \mathbb{C}$ and $f(\alpha) = f(\beta) = 0$.
Then, this means that both $\alpha$ and $\beta$ are algebric over $F$.
Let $m_{\alpha, F}(x)$ denotes as the minimal polynomial of $\alpha$ over $F$.
Let $m_{\beta, F}(x)$ denotes as the minimal polynomial of $\beta$ over $F$.
Since, $f(x) \in F[x]$ is irreducible over $F$, and $f(\alpha) = f(\beta) = 0$.
Then, this means $f(x) = m_{\alpha, F}(x) = m_{\beta, F}(x)$. Which means that $|F(\alpha)| = |F(\beta)|$
Since, notice that $|\text{Aut}(F(\alpha)/F)|$ is equal to the number of roots of $m_{\alpha, F}(x)$ in the field $F(\alpha)$.
and $|\text{Aut}(F(\beta)/F)|$ is equal to the number of roots of $m_{\beta, F}(x)$ in the field $F(\beta)$.
Then, because of $|F(\alpha)| = |F(\beta)|$.
So the number of roots of $m_{\alpha, F}(x)$ in the field $F(\alpha)$ and the number of roots of $m_{\beta, F}(x)$ in the field $F(\beta)$.
Hence, $|\text{Aut}(F(\alpha)/F)| = |\text{Aut}(F(\beta)/F)|$.
(b) Since, from part (a), we proved that $f(x) = m_{\alpha, F}(x) = m_{\beta, F}(x)$.
Then, this means $\beta$ is a root of $m_{\alpha, F}(x)$ and also $\alpha$ is a root of $m_{\beta, F}(x)$.
Since, in part (a), we also proved that $\alpha, \beta$ are algebric over $F$.
So, this means $F(\alpha)$ is isomorphic to $F(\beta)$.
Which means the set of automorphisms of $F(\alpha)$ denote as $\text{Aut}(F(\alpha))$ is isomorphic to
$~\hspace{10mm}$ the set of automorphisms of $F(\beta)$ denote as $\text{Aut}(F(\beta))$.
Since, notice that $F$ is a subfield of $F(\alpha)$ and $F$ is a subfield of $F(\beta)$.
Then, $\text{Aut}(F(\alpha)/F)$ is a subgroup of $\text{Aut}(F(\alpha))$ and $\text{Aut}(F(\beta)/F)$ is a subgroup of $\text{Aut}(F(\beta))$.
So, $\text{Aut}(F(\alpha)/F)$ is isomorpic to $\text{Aut}(F(\alpha)/F)$.
$\textbf{Are the prove of part (a) and (b) coreect ? Is there anything I can change to make it better ?}$
 A: The key is to understand that both the fields $F(\alpha), F(\beta)$ are isomorphic to the field $F[x] /(f(x)) $ and hence they are isomorphic to each other. The isomorphism $\phi$ between $F(\alpha), F(\beta) $ is given by $\phi(\alpha) =\beta $.
Let $\sigma\in\textrm{Aut} (F(\alpha) /F) $. Then $\tau:F(\beta) \to F(\beta) $ given by $$\tau(x) =\phi(\sigma(\phi^{-1}(x)))$$ for all $x\in F(\beta) $ is an automorpshism of $F(\beta) $ which leaves $F$ fixed and hence $\tau\in\textrm {Aut} (F(\beta) /F) $.
You should now check that $\psi:\textrm{Aut} (F(\alpha) /F) \to\textrm{Aut} (F(\beta) /F) $ given by $$\psi(\sigma) =\phi\circ \sigma\circ\phi^{-1}$$ is a bijection. This proves part a) of the question.
The same $\psi$ is also a group isomorphism and that handles part b).
A: For (a):
Since, $F$ is a subfield of $\mathbb{C}$.
Then, $\mathbb{C}$ is an extension field of $F$ which $\mathbb{C}/F$.
Since, $f(x) \in F[x]$, $\alpha,\beta \in \mathbb{C}$ and $f(\alpha)=f(\beta)=0$.
Then, both $\alpha$ and $\beta$ are algebraic over $F$.
Let $m_{\alpha,F}(x)$ denotes as the minimal polyn of $\alpha$ over $F$.
Let $m_{\beta,F}(x)$ denotes as the minimal polyn of $\beta$ over $F$.
Since, $f(x)\in F[x]$ is irreducible over $F$ and $f(\alpha)=f(\beta)=0$.
Then, $f(x)=m_{\alpha,F}=m_{\beta,F}$ $\implies$ $|F(\alpha)| = |F(\beta)|$.
Since, $F$ is a subfield of $\mathbb{C}$. Also, both $\alpha$, $\beta$ are algebraic over $F$.
Then, $|Aut(F(\alpha)/F)|$ and $|Aut(F(\beta)/F)|$ equal to the number of roots of $m_{\alpha,F}(x)$ and $m_{\alpha,F}(x)$ respectively.
Since, the minimal polyn are unique in $F[x]$.
Also, $F(\alpha)$ and $F(\beta)$ are finite extension of $F$.
Then, number of root of $m_{\alpha,F}(x)$ in $F(\alpha)$ $=$ number of roots of $m_{\beta,F}(x)$ in $F(\beta)$.
Therefore, we proved that $|Aut(F(\alpha)/F)|=|Aut(F(\beta)/F)|$.
For (b):
Define $\phi : Aut(F(\alpha)/F) \rightarrow Aut(F(\beta)/F)$.
First, prove there is a bijection from $Aut(F(\alpha)/F)$ to $Aut(F(\beta)/F)$.
Notice if $\sigma \in F(\alpha)/F$, then every $s\in F(\alpha)$, we have $s=\sum_{i=0}^{n} c_i \alpha^i$, where $c_i$ are constants in $F$.
Then, $\sigma(s) = \sum_{i=0}^n c_i \sigma(\alpha)^i$.
So, $\sigma$ is determined by $\sigma(\alpha)$.
Assume $\sigma(\alpha)=f_1(\alpha)$ for some $f_1(x)\in F[x]$.
Then, $\phi(\sigma) \in F(\beta)/F$, such that $\sigma(\beta) = f_1(\beta)$.
Since, if there exists some $\sigma_1,\sigma_2 \in F(\alpha)/F$ and $\sigma_1=\sigma_2$, then $f_1(\alpha) = f_2(\alpha)$, where for some $f_1(x), f_2(x)\in F[x]$.
Then, $f_1(\alpha)-f_2(\alpha)=(f_1-f_2)(\alpha)=0$.
Then, there is a injection and surjection from $\sigma$ to $\phi(\sigma)$.
Hence, there is a bijection from $Aut(F(\alpha)/F)$ to $Aut(F(\beta)/F)$.
Next, the groups $Aut(F(\alpha)/F)$ and $Aut(F(\beta)/F)$ are homomorphic.
Since, $Aut(F(\alpha)/F)$ and $Aut(F(\beta)/F)$ are groups.
So, $\phi(\sigma_1 \circ \sigma_2) = \phi\sigma_1 \circ \phi\sigma_2$, for all $\sigma_1$ and $\sigma_2 \in Aut(F(\alpha)/F)$.
Assume $\sigma_1(\alpha)=f_1(\alpha)$ and $\sigma_2(\alpha)=f_2(\alpha)$, for some $f_1(x), f_2(x) \in F[x]$.
Then, $\sigma_1 \circ \sigma_2(\alpha)=f_1 \circ f_2(\alpha)$.
Then, $\phi(\sigma_1\circ\sigma_2(\alpha))=f_1\circ f_2(\beta)$.
Since, $\phi\sigma_1(\beta)=\phi(\sigma_1(\beta))=f_1(\beta)$ and $\phi\sigma_2(\beta)=\phi(\sigma_2(\beta))=f_2(\beta)$.
Then, $\phi\sigma_1 \circ \phi\sigma_2(\beta)=\phi\sigma_1(f_2(\beta))=\phi\sigma_1\circ f_2(\beta)= f_1 \circ f_2 (\beta)$.
So, $\phi(\sigma_1 \circ \sigma_2)=\phi\sigma_1\circ\phi\sigma_2$, for all $\sigma_1$ and $\sigma_2 \in Aut(F(\alpha)/F)$.
Since, there is a bijection from $Aut(F(\alpha)/F)$ to $Aut(F(\beta)/F)$,
and the groups $Aut(F(\alpha)/F)$ and $Aut(F(\beta)/F)$ are homomorphic.
Therefore, $Aut(F(\alpha)/F)$ and $Aut(F(\beta)/F)$ are isomorphic.
