I am new to Galois theory and trying to show that the number of automorphisms of a field extension $F/M$ with $[F:M] = n$ (fixing $M$ of course) is $≤ n$. As I understand it, there are exactly $n$ if (and only if?) the extension is Galois, which I hope follows from a proof of the above.
I understand that the minimal polynomial of any algebraic element $v$ in a $F$ has a degree $≤ n$, since there are $n+1$ elements
$$1, v, v^2... v^n$$
which are necessarily linearly dependent and thus can used to formulate a degree n polynomial with root $v$. I also understand that an automorphism $\phi$ must permute each element with the other roots of it's minimal polynomial, since
$$p(v) = a_0 + a_1v+\cdots+ a_nv^n = 0 $$ implies $$\phi(p(x)) = a_0 + a_1\phi(v)+\cdots+ a_n\phi(v)^n = 0$$
However, I don't think the result follows from this; I know it is possible for multiple different automorphisms to send a given element to the same destination (thinking of examples like $\mathbb{Q}[\sqrt2, i]$ which I believe has four automorphisms, two of which send $\sqrt2$ to itself and two of which send $i$ to itself). Clearly there is something I'm missing; I would really appreciate a pointer here. Thanks :)
How are we sure that $F = M(v_1,...,v_n)$ ? where $M(v_1,...v_n)$ is the field obtained by adding $n$ elements $v_1,...,v_n$ to $M$ . Since $[F: M] = [F:M(v_1)] \times [M(v_1):M]$, since $[F:M]$ is finite, we have that $n$ is finite i.e., on adding each extra element $v_i \in F \setminus M(v_1,...,v_{i-1})$, $[F:M(v_1,...,v_i)]$ strictly reduces and in the end becomes $1$ for a finite $n$.
Let the field $F/M$ be written as $F = M(v)$ i.e., $F$ is obtained by adding an element $v$ to $M$ and hence $$F = \{\sum_{i=0}^{n-1} m_i v^i : m_i \in M\}.$$
Now let $\phi$ be an automorphism on $F/M$. $\phi$ fixes all elements of $M$ i.e., $\phi(m) = m$ by definition. So now for $f \in F$, $$\phi(f) =\phi(\sum_{i=0}^{n-1} m_i v^i) = \sum_{i=0}^{n-1} \phi(m_i v^i) = \sum_{i=0}^{n-1} m_i \phi(v^i) = \sum_{i=0}^{n-1} m_i \phi(v)^i .$$
Since every element $f \in F$ can be represented as $f = \sum_{i=0}^{n-1} m_i v^i$, we have that $\phi$ is completely determined by its image of $v$ since $\phi(f) = \sum_{i=0}^{n-1} m_i \phi(v)^i$ for any $f \in F$. As you already know $\phi(v)$ can have atmost $[F:M]$ possibilities since $\phi(v)$ must satisfy minimal polynomial of $v$.
Now what happens if we write $F = M(v) = M(w)$. The answer is it will take care for $F = M(w)$ on its own as long as you stick to one representation $F = M(v)$.
So what happens when $F = M(v_1,v_2)$. Then write it as $F = M(v_1)(v_2)$ i.e., generate the field by adding $v_1$ first and then on top of the resulting field add $v_2$. Now write $[F = M(v_1)(v_2):M(v_1)] = n_2$ and $[M(v_1):M] = n_1$ then $$F = \{\sum_{i=0}^{n_2-1} r_i v_2^i : r_i = \sum_{j=0}^{n_1-1} m_j v_1^j, m_j \in M\} .$$ Let minimal polynomial of $v_2$ over $M(v_1)$ be $m_{v_2}(x) = \sum_{j=0}^{n_2} \ell_j x^i$ with $\ell_j \in M(v_1)$. Now for $f \in F$, $\phi(f) = \sum_{i=0}^{n_2-1} \phi(r_i) \phi(v_2)^i$. Now $\phi(r_i)$ is completely determined by the image $\phi(v_1)$ and for each such image $\phi(v_2)$ must satisfy a minimal polynomial of degree $n_2$ i.e., $m_{v_2} = \sum_{j=0}^{n_2} \phi(\ell_j) \phi(v_2)^i = 0$ which has atmost $n_2$ possible images $\phi(v_2)$ for a fixed $\phi(v_1)$ or fixed $\phi(\ell_j)$. We already know that there are atmost $n_1$ possible images for $\phi(v_1)$. So there are a total of atmost $n_1n_2$ possible images for $(v_1,v_2)$. We know that $\phi$ is completely determined by its image on $v_1$ and $v_2$.
Note here that for each possible image $\phi(v_1)$ there are $n_2$ possible images for $\phi(v_2)$ so several automorphisms maps to same image $\phi(v_1)$ (atmost $n_2$ automorphisms can have same image $\phi(v_1)$) but no two automorphism will have same image simultaneously both on $v_1$ and $v_2$.
Extend the above to $F=M(v_1,...,v_n)$.