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Check here for some basic information.
Let \(E\) be a field extension over a field \(F\). The group of automorphisms on \(E/F\) is defined as
\[\begin{align*} \text{Aut}(E/F) &= \{\rho: E\to E\mid \rho \text{ is a field isomorphism and } \rho(x)=x \text{, for all } x\in F \} \\ &= \{\rho: E\to E\mid \rho \text{ is a } F \text{-linear field isomorphism}\}. \end{align*}\]
How to see that the above two definitions are equivalent? Suppose \(\rho\) is a field ispmorphism and \(\rho(x)=x\), for all \(x\in F\). Then for any \(k\in F\) and \(x, y\in E\), we have
\[\rho(kx+y) = \rho(k)\rho(x)+\rho(y) = k\rho(x)+\rho(y),\]which indeed shows that \(\rho\) is \(F\)-linear.
\(\rho(z)=\bar z\) is an element if \(\text{Aut}(\Bbb C/\Bbb R)\).
Let \(f(x)\in F[x]\), and let \(\alpha\) be a root of \(f(x)\) in \(E\). Then \(f(\rho(\alpha)) = \rho(f(\alpha)) = \rho(0) = 0\). (Why? Exapnd everything and use the definition of automorphism!). Therefore, \(\rho(\alpha)\) is also a root of \(f(x)\)!. This means that \(\text{Aut}(E/F)\) can act on the set of roots of \(f(x)\) in \(E\).
Suppose \(E=F(\alpha)\) and \([E:F]=\alpha\). From here, we know that every element in \(E\) can be written as
\[a_0+a_1\alpha+\cdots+a_{n-1}\alpha^{n-1}, a_i \in F.\]Then
\[\rho(a_0+a_1\alpha+\cdots+a_{n-1}\alpha^{n-1}) = a_0+a_1\rho(\alpha)+\cdots+a_{n-1}\rho(\alpha)^{n-1},\]which means that \(\rho\) is uniquely determined by \(\rho(\alpha)\).
Let \(E=F(\alpha)\) be a simple algebraic field extension over \(F\). Suppose \(\text{Irr}(\alpha, F)(x)\) has \(m\) distinct roots in \(E\). Then
\[\vert \text{Aut}(E/F)\vert=m\le[E:F].\]In particular, \(\vert \text{Aut}(E/F)\vert\) is the number of distinct zeros of \(\text{Irr}(\alpha, F)\) in \(E\).
注意！以下討論的 \(F\) 都是 finite field！
Check out Finite Field and Characteristic.
Let \(F=\Bbb F_q\) and \(E = \Bbb F_{q^n}\). Consider the map \(\sigma:E\to E\) given by \(\sigma(x)=x^q\). We can show that \(\sigma\) is an element of \(\operatorname{Aut}(E/F)\) by showing
In fact, \(\sigma\) is called the Frobenius automorphism, also denoted by \(\text{Frob}_{E/F}\).
For a finite extension \(E\) over a finite field \(F\), the Frobenius automorphism \(\text{Frob}_{E/F}\) generates the automorphism group \(\text{Aut}(E/F)\), i.e., \(\text{Aut}(E/F) = \langle \text{Frob}_{E/F}\rangle\), which is of order \([E:F]\).
In particular, if \([E:F]=n\), then
\[\text{Aut}(E/F) = \langle \text{Frob}_{E/F}\rangle \cong \Bbb Z_n.\]
Finite field extension 的 automorphism group 才會被 \(\text{Frob}_{E/F}\) 生成！
Proof
Suppose the order of \(\sigma = \text{Frob}_{E/F}\) is \(m\). Recall from this theorem, the size of \(\text{Aut}(E/F)\) is bounded by \(n\), which means that \(m\le n\).
Let \(\alpha\) be a generator of \(E^\times\) of order \(q^n - 1\). Then
\[\begin{align*} \alpha = \sigma^m(\alpha) = \alpha^{q^m} &\implies a^{q^m-1}=1 \\ &\implies (q^n-1)\vert (q^m-1) \\ &\implies m\ge n. \end{align*}\]Hence, we conclude that \(m = n\).
Since \(n = m = \vert\langle \text{Frob}_{E/F}\rangle\vert \le \vert\text{Aut}(E/F)\vert\le n\), we have \(\langle \text{Frob}_{E/F}\rangle = \text{Aut}(E/F)\). ◼
Let \(d\) be a positive integer. We have \(q^d-1\vert q^n-1\) iff \(d\vert n\).
Proof Hint
Since \(q\) can be anything, we can even use it for polynomials: \(x^d-1\vert x^n-1\) iff \(d\vert n\). Use long division on polynomials!
Suppose \(E=F(\alpha)\). Then the automorphisms in \(\text{Aut}(E/F)\) act transitively on the roots of \(\text{Irr}(\alpha, F)(x)\).
Proof
Since each automorphism in \(\text{Aut}(E/F)\) is uniquely determined by its action on \(\alpha\), \(\sigma^i(\alpha)\) are distinct roots of \(\text{Irr}(\alpha, F)(x)\), for all \(0\le i\le n-1\). Hence, we have
\[\text{Irr}(\alpha, F)(x) = \prod_{i=0}^{n-1}\big(x - \sigma^i(\alpha) \big). \tag*{$\blacksquare$}\]記得 \(\langle \sigma \rangle = \langle \text{Frob}_{E/F}\rangle = \text{Aut}(E/F)\)！
From the definition of Frobenius automorphism (\(\sigma(x) = x^q\)), we can see that it is determined by the underlying field \(F\), with size \(q\). Thus, for an intermediate field \(K\) between \(E\) and \(F\), we have
\[\text{Frob}_{K/F}(x) = x^q = \text{Frob}_{E/F}(x)\]for all \(x\in K\). Thus \(\text{Frob}_{E/F}\Big\vert_K = \text{Frob}_{K/F}\).
Suppose \(K=F(\beta)\) for some \(\beta\in E\) and \([K:F] = m\). Then, all the roots of \(\text{Irr}(\beta, F)(x)\) are \(\beta, \beta^q, \cdots, \beta^{q^{(m-1)}}\) (by this theorem), and we have \(\beta^{q^m} = \beta\). Therefore,
\[\prod_{i=0}^{n-1}\Big(x-\text{Frob}^i_{E/F}(\beta)\Big) = \prod\Big(x-\beta^{q^i}\Big) = \Big[\text{Irr}(\beta, F)(x)\Big]^{n/m},\]where each root occurs \(n/m\) times.
Here comes a theorem:
Let \(E\) be a finite extension of a finite field \(F\), and let \(f(x)\) be an irreducible polynomial in \(F[x]\) with a root in \(E\). Then, all the roots of \(f(x)\) in \(\bar F\) lie in \(E\), and \(\text{Aut}(E/F)\) acts transitively on these roots.
一個在 \(E\)，全都在 \(E\)。
因為乘法有封閉性！\((\sigma(x) = x^q)\)
設 \(F=\Bbb Z_2\)、\(E = F(\alpha)\)，且 \([E:F]=4\)。對於所有 \(\gamma \in E\)，在什麼情況底下等式會成立？
\[\text{Irr}(\gamma, F)(x) \stackrel{?}{=} \prod_{\sigma\in\text{Aut}(E/F)}\big(x - \sigma(\gamma) \big)\]從上述說明可以得知，\(\gamma\) 加入 \(F\) 後必須促成四次擴張；而因為 \(\vert E\vert = 2^4\)，所以 \(E\) 和 \(F\) 之間的 intermediate field 只有 \(\Bbb F_{2^2} = \Bbb F_4\)。
因為只有 \(2\vert 4\)！（除了 \(1\)、\(4\)）
也就是說，所有 \(\gamma \in E\backslash\Bbb F_4\) 都可以使該等式成立！
另外，令 \(\sigma = \text{Frob}_{E/F}\)。計算其實很方便：
\[\sigma(\alpha + 1) = (\alpha + 1)^4 = \alpha^4 + 1^4\]因為是 automorphism，直接把次方分配進去！
Let \(K = F(\alpha)\) for some \(\alpha \in E\), and \([K:F] = m\). Let \(H\) be a subgroup of \(G =\text{Aut}(E/F)\). Then, \(H = \langle\sigma^m\rangle\) for a unique integer \(m\vert n\), where \(\sigma = \text{Frob}_{E/F}\). Observe that
\[\sigma^m(x) = x^{q^m} = \text{Frob}_{E/K}(x),\]and thus
\[H = \text{Aut}(E/K).\]Now consider the subset \(E^H\) of \(E\) fixed by \(H\):
\[E^H := \{a\in E\mid h(a) = a, \forall h\in H \},\]and note that
\[\begin{align*} E^H &= \{a\in E\mid \sigma^{m}(a) = a \} \\ &= \{a\in E\mid a^{q^m}=a \} = \Bbb F_{q^m} = K. \end{align*}\]怪怪的？！\(E^H\) 的定義和最後那段不太一樣？
Let \(K\) be an intermediate field between \(E\) and \(F\), so \(K = \Bbb F_{q^m}\) for some \(m\vert n\). Consider the subgroup of \(G\) fixing \(K\):
\[G_K = \{g\in G\mid g(x) = x\forall x\in K\} = \langle \sigma^{m'}\rangle\]for some unique \(m'\vert n\). We now show that \(m = m'\), which implies that
\[G_k = \text{Aut}(E/K).\]Sine \(\sigma^m\) fixes \(K\), \(\sigma^m\in G_K\) and \(m' \le m\). Let \(\alpha\) be a generator of \(K^\times\) of order \(q^m-1\). Then we have
\[\alpha = \sigma^{m'(\alpha)} = \alpha^{q^{m'}} \implies \alpha^{q^{d'}-1} = 1,\]which implies that \(d \le d'\). Therefore \(d=d'\).
With the above arguments, we now come to the following theorem.
For a finite extension \(E\) over a finite field \(F\), let \(G = \text{Aut}(E/F)\) and \(\sigma = \text{Frob}_{E/F}\). Then, there is a bijection between intermediate fields \(K\), between \(F\) and \(E\), and subgroups \(H\) of \(G\), which is given by:
\[\begin{align*} K = \Bbb F_{q^d} &\xrightarrow[]{\rho} G_K = \langle\sigma^d\rangle \\ E^H = \Bbb F_{q^d} &\xleftarrow[]{\tau} H = \langle \sigma^d \rangle. \end{align*}\]Moreover, \(\rho\circ \tau\) and \(\tau \circ\rho\) are both the identity maps.
Let \(F\) be a finite field. Then any finite field extension \(E/F\) must be a Galois extension.
Proof
By the above two observations, \(E/F\) is Galois. ■
其實從這個定理就可以看出來了！