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A polynomial \(f(x)\in F[x]\) is called separable if it has no repeated zeros in \(\bar F\).
Every irreducible polynomial of a finite field or a field with characteristic zero is separable.
In particular, if \(F\) is an infinite field with \(\text{char}F \not = 0\), \(f(x)\in F[x]\) may have repeated zeros.
Every finite extension \(E\) over a field \(F\) of characteristic zero is a simple extension.
primitive element 這個詞有好幾個不同意思，見 wiki。
Proof
Let \(E\) and \(F\) be two fields. If \(\rho: F\to E\) is an injective homomorphism of \(F\) into \(E\), then \(\rho\) is called an embedding of \(F\) into \(E\).
\(F\)-embedding: \(\rho(x) = x\) for all \(x\in F\).
\(F\)-embedding 代表 \(F\) 被固定住！
Consider the case when \(F=\Bbb Q\) and \(\alpha=\sqrt[3]{2}\). Note that other zeros of \(\text{Irr}(\sqrt[3]{2}, \Bbb Q)(x)\) are \(\sqrt[3]{2}\omega\) and \(\sqrt[3]{2}\omega^2\), which are non-real. This means the only one zero of \(\text{Irr}(\sqrt[3]{2}, \Bbb Q)(x)\) lies in \(\Bbb Q(\sqrt[3]{2})\). From this theorem, we can conclude that
\[\text{Aut}\Big(\Bbb Q(\sqrt[3]{2})/\Bbb Q \Big) = \{e\}.\]There is nothing useful in the automorphism group. Instead, consider the following field isomorphism:
\[\Bbb Q(\sqrt[3]{2}) \cong \Bbb Q[x]/\langle x^3-2\rangle \cong \Bbb Q(\sqrt[3]{2}\omega)\subset \bar{\Bbb Q}.\]We now have
\[\begin{align*} \rho: \Bbb Q(\sqrt[3]{2}) &\to \bar{\Bbb Q}, \\ a_0+a_1\sqrt[3]{2}+a_2\sqrt[3]{4} &\mapsto a_0+a_1\sqrt[3]{2}\omega + a_2\sqrt[3]{4}\omega^2, \end{align*}\]which is an injective ring homomorphism that fixes \(\Bbb Q\). Thus \(\rho\) is a \(\Bbb Q\)-embedding of \(\Bbb Q(\sqrt[3]{2})\)!
好像沒有說得很清楚，再想想。(2023/5/9 更新)
Let \(E/F\) be a finite extension. Define \(\text{Emb}(E/F)\) be the set of \(F\)-embeddings of \(E\) into \(\bar F\). In particular, \(\text{Aut}(E/F)\subseteq \text{Emb}(E/F)\).
\(\text{Emb}(E/F)\) 只是個集合！
Let \(E/F\) be a field extension, and let \(\alpha\) be a zero of \(f(x)\) in \(E\). Then, for all \(\rho\in \text{Emb}(E/F)\), \(\rho(\alpha)\) is a zero of \(f(x)\) in \(\bar F\).
Simplified Proof
Let \(f(x) = \sum_i a_ix^i\) and \(\alpha\) be one of its zeros. Since \(\rho\) is a ring homomorphism, we have
\[\begin{align*} f(\rho(\alpha)) &= \sum_i a_i\rho(\alpha)^i \\ &= \sum_i a_i \rho(\alpha^i) \\ &= \rho\Big(\sum_i a_i \alpha^i\Big) = 0. \end{align*}\]This prove that \(\rho(\alpha)\) is a zero of \(f(x)\). ◼
Let \(E/F\) be a finite field extension, and let \(f(x)=\text{Irr}(\alpha, F)(x)\) in \(E\). Let \(\beta\) be a zero of \(f(x)\) in \(\bar F\). Then there exists some \(\rho\in\text{Emb}(E/F)\) such that \(\rho(\alpha)=\beta\).
Proof
Consider
\[\begin{gather} \rho: E=&F(\alpha)& \cong &F[x]/\langle f(x)\rangle& \cong &F(\beta)& \\ &g(\alpha)& \ &\longrightarrow& \ &g(\beta)& \end{gather}\]where \(\rho\) is a composition of two field isomorphisms, which is also a field isomorphism. ◼
\(g(x) \in F[x]\)!
\(\vert \text{Emb}(E/F)\vert \le [E:F]\).
\(\vert \text{Emb}(E/F)\vert\): number of distinct zeros of \(\text{Irr}(\alpha, F)(x)\) in \(\bar F\).
Let \(\beta\) be another zero of \(\text{Irr}(\alpha, F)\) in \(E = F(\alpha)\), and let \(\beta = \rho(\alpha)\). Then,
\[\beta\in E \implies \rho \in \text{Aut}(E/F) \tag{1}\] \[\beta\not\in E \implies \rho \in \text{Emb}(E/F)\backslash\text{Aut}(E/F)\tag{2}\]Let \(E/F\) be a field extension. If
\[\begin{align*} \text{Aut}(E/F) &= \text{Emb}(E/F) \tag{1} \\ \vert\text{Emb}(E/F)\vert &= [E:F] \tag{2} \end{align*}\]are satisfied, then \(E/F\) is called a Galois extension. Moreover, \(\text{Aut}(E/F)\) is called the Galois group of \(E/F\), denoted by \(\text{Gal}(E/F)\).
第一式代表所有 zero 都在 \(E\)（splitting）；第二式代表沒有重根（separable）！
Let \(E/F\) be a finite Galois extension with Galois group \(G\). Suppose \(E = F(\alpha)\). Then \(G\) acts transitively on the set of zeros of the irreducible polynomial \(\text{Irr}(\alpha, F)(x)\) of \(\alpha\). Moreover,
\[\prod_{g\in G}(x-g(\alpha)) = \text{Irr}(\alpha, F)(x).\]
\(G\) 其實就是 \(\text{Aut}(E/F)\) 呦！
\(G\) acts on the set of zeros as permutations!
Proof
For all \(k\in E\), \(k\) can be written as
\[k = a_0 + a_1\alpha + \cdots + a_{n-1}\alpha^{n-1}\]with \(a_i \in F\). Then for some \(g\in G\),
\[g(k) = a_0 + a_1g(\alpha) + \cdots + a_{n-1}g(\alpha)^{n-1},\]which means that \(g\) is uniquely determined by \(g(\alpha)\), i.e., there is a one-to-one correspondence between \(g\) and \(g(\alpha)\) for all \(g\in G\). Moreover, since \(\vert G\vert = [E:F] = n\) and \(g(\alpha)\) is also a zero of \(\text{Irr}(\alpha, F)(x)\), we have proved the theorem! ◼
Let \(E/F\) be a Galois extension with Galois group \(G\). Let \([E:F] = n\) and \([F(\alpha):F] = m\), for some \(\alpha \in E\). Then
\[\prod_{g\in G}(x-g(\alpha)) = f(x)^{n/m}.\]
Proof
Let \(H = \text{Emb}(F(\alpha)/F)\). We have \([G:H] = [E:F(\alpha)] = n/m\), and can write \(G\) as the union of \(H\) cosets:
\[G = H \sqcup h_1H \sqcup \cdots \sqcup h_{n/m-1}H.\]Moreover, by this theorem, we can write \(f(x)\) as
\[\prod_{h\in H}(x-h(x)) = f(x).\]Equivalently, for any coset of \(H\),
\[\prod_{h' \in h_iH}(x-h'(x)) = \prod_{h\in H}(x-h_ih(x)) = \prod_{h\in H}(x-h(x)) = f(x),\]since left multiplication only permutes \(h\)s! Thus,
\[\prod_{g\in G}(x-g(\alpha)) = \Big(\prod_{h\in H}(x-h(x))\Big)^{n/m} = f(x)^{n/m}. \tag*{$\blacksquare$}\]任何二次擴張都是 Galois！有兩種觀點可以說明：根與係數和 normal subgroup。
假設在 \(F\) 中加入 \(\alpha\)，因為 \(\text{Irr}(\alpha, F)(x) = (x-\alpha)(x-\alpha')\) 的係數都在 \(F\)，利用根與係數的關係，可以發現 \(\alpha'\) 由 \(\alpha\) 生成。
再來，因為是二次擴張，所以 \([\text{Aut}(F(\alpha)/F): G_{F(\alpha)}]=2\)，\(G_{F(\alpha)}\) 是 normal subgroup。根據 the fundamental theorem of Galois theory，\(F(\alpha)/F\) 就是 Galois！