Lecture. The Fundamental Theorem of Galois Theory (cont.) [l14]

  • May 18, 2026

Theorem 1. bijection between sub-extensions and subgroups [2302]

    Let $K / F$ be a finite Galois extension. There is a bijection $$ \begin{aligned} \{ \text{sub-extensions } F \subseteq E \subseteq K \} &\longleftrightarrow \{ \text{subgroups } H \leq G \} \\ E &\longmapsto \mathrm{Gal}(K / E) \\ K^{H} &\longleftarrow\!\shortmid H \end{aligned} $$ with the two maps inverse to each other.

    Proof. sketch (TODO) [2302A]

      Theorem 2. more statements of the Fundamental Theorem [2303]

        Let $K / E$ be a finite Galois extension.

        1. (Inclusion reversing) Suppose that $E_{1}, E_{2}$ are subfields of $K$ containing $F$ and $H_{1}, H_{2}$ are subgroups of $G = \mathrm{Gal}(K / F)$. Then $$\begin{aligned}E_{1} \subseteq E_{2} &\iff \mathrm{Gal}(K / E_{1}) \geq \mathrm{Gal}(K / E_{2}), \\ H_{1} \leq H_{2} &\iff K^{H_{1}} \supseteq K^{H_{2}}\end{aligned}$$

        2. For all $H \leq G$, $K / K^{H}$ is Galois with $\mathrm{Gal}(K / K^{H}) = H$.

          For all $F \subseteq E \subseteq K$, $K / E$ is Galois with $K^{\mathrm{Gal}(K / E)} = E$.

        3. Given $H \leq G = \mathrm{Gal}(K / F)$, $K^{H} / F$ is Galois if and only if $H \trianglelefteq G$. If this is the case, then $\mathrm{Gal}(K^{H} / F) \cong G / H$.

          Given $F \subseteq E \subseteq K$, $E / F$ is Galois if and only if $\mathrm{Gal}(K / E) \trianglelefteq G = \mathrm{Gal}(K / F)$. If this is the case, then $\mathrm{Gal}(E / F) \cong G / \mathrm{Gal}(K / E)$.

        Proof. sketch (TODO) [2303A]

          Remark 3. some remarks [2304]

            The above theorem is not true of $K / F$ is Galois but not finite. In that case, we need to equip $\mathrm{Gal}(K / F)$ with a profinite topology, and the bijection is between subextensions and closed subgroups of $\mathrm{Gal}(K / F)$. See this note.

            Example 4. some examples [2305]

              Example 4.1. $\mathbb{Q}\left( \sqrt[3]{ 2 }, \sqrt{ -3 } \right) / \mathbb{Q}$ [2305A]

                In (2) of Example 1505, we have seen that $\mathbb{Q}\left( \sqrt[3]{ 2 }, \sqrt{ -3 } \right)$ is the splitting field of $x^{3} - 2$ over $\mathbb{Q}$, so $\mathbb{Q}\left( \sqrt[3]{ 2 }, \sqrt{ -3 } \right)/ \mathbb{Q}$ is Galois. By (5) of Example 2106, we know that $\mathrm{Gal}\left( \mathbb{Q}\left( \sqrt[3]{ 2 }, \sqrt{ -3 } \right) / \mathbb{Q} \right) = \left< \sigma_{1}, \sigma_{2} \right> \cong S_{3}$ with $$\begin{aligned} \sigma_1 : \sqrt[3]{2} &\longmapsto \sqrt[3]{2}\frac{-1 + \sqrt{-3}}{2}, &\qquad \sigma_2 : \sqrt[3]{2} &\longmapsto \sqrt[3]{2}, \\ \sqrt{-3} &\longmapsto \sqrt{-3}, &\qquad \sqrt{-3} &\longmapsto -\sqrt{-3}. \end{aligned}$$ The nontrivial proper subgroups of $\mathrm{Gal}\left( \mathbb{Q}\left( \sqrt[3]{ 2 }, \sqrt{ -3 } \right) / \mathbb{Q} \right)$ are $\left< \sigma_{1} \right>, \left< \sigma_{2} \right>, \left< \sigma_{1}\sigma_{2} \right>, \left< \sigma_{1}^{2}\sigma_{2} \right>$, and $\left< \sigma_{1} \right>$ is normal of order $3$, $\left< \sigma_{2} \right>, \left< \sigma_{1} \sigma_{2} \right>, \left< \sigma_{1}^{2} \sigma_{2} \right>$ are not normal and of order $2$. $$\begin{aligned} \mathbb{Q}(\sqrt[3]{2}, \sqrt{-3})^{\langle \sigma_1 \rangle} = \mathbb{Q}(\sqrt{-3}), &\qquad \mathbb{Q}(\sqrt[3]{2}, \sqrt{-3})^{\langle \sigma_2 \rangle} = \mathbb{Q}(\sqrt[3]{2}), \\ \mathbb{Q}(\sqrt[3]{2}, \sqrt{-3})^{\langle \sigma_1\sigma_2 \rangle} = \mathbb{Q}\left(\sqrt[3]{2}\frac{-1-\sqrt{-3}}{2}\right), &\qquad \mathbb{Q}(\sqrt[3]{2}, \sqrt{-3})^{\langle \sigma_1\sigma_2^2 \rangle} = \left(\sqrt[3]{2}\frac{-1+\sqrt{-3}}{2}\right), \end{aligned}$$ and these are all the proper nontrivial sub-extensions of $\mathbb{Q}\left( \sqrt[3]{ 2 }, \sqrt{ -3 } \right) / \mathbb{Q}$.

                Example 4.2. $\mathbb{Q}\left( \sqrt{ 2 }, \sqrt{ 3 } \right)$ [2305B]

                  (D&F p.576) Since every subgroup of the Klein-4 group is normal, all the subfields of $\mathbb{Q}(\sqrt{ 2 }, \sqrt{ 3 })$ are Galois extensions of $\mathbb{Q}$.

                  Example 4.3. splitting field of $x^{8} - 2 \in \mathbb{Q}[x]$ [2305C]

                    (D&F Example p.577-581)

                    Example 4.4. $\mathbb{F}_{p^{n}}$ [2305D]

                      Consider $\mathbb{F}_{p^{n}}$, the unique subfield of $\overline{\mathbb{F}}_{p}$ of cardinality $p^{n}$.

                      By Example 22016, $$ \mathrm{Gal}\left( \mathbb{F}_{p^{n}} / \mathbb{F}_{p} \right) = \left< \operatorname{Fr} \right> \cong \mathbb{Z} / n\mathbb{Z}. $$ The subgroups of $\mathrm{Gal}\left( \mathbb{F}_{p^{n}} / \mathbb{F}_{p} \right)$ are $\left< \operatorname{Fr}^{d} \right> \cong d\mathbb{Z} / n\mathbb{Z}, d \mid n, d \geq 1$. (They are all normal subgroups since $\mathbb{Z} / n\mathbb{Z}$ is abelian.) $$ \mathbb{F}_{p^{n}}^{\left< \operatorname{Fr}^{d} \right> } = \{ \alpha \in \mathbb{F}_{p^{n}} \mid \alpha^{p^{d}} = \alpha \} = \mathbb{F}_{p^{d}}. $$ By the Fundamental Theorem, we have $$ \mathrm{Gal}\left( \mathbb{F}_{p^{d}} / \mathbb{F}_{p} \right) = \mathrm{Gal}\left( \mathbb{F}_{p^{n}}^{\left< \operatorname{Fr}^{d} \right> } / \mathbb{F}_{p} \right) \cong \mathrm{Gal}\left( \mathbb{F}_{p^{n}} / \mathbb{F}_{p} \right) / \left< \operatorname{Fr}^{d} \right> = \left< \operatorname{Fr} \right> / \left< \operatorname{Fr}^{d} \right> \cong \mathbb{Z} / d\mathbb{Z}, $$ compatible with what we have proved in Example 22016: $\mathrm{Gal}\left( \mathbb{F}_{p^{d}} / \mathbb{F}_{p} \right) \cong \mathbb{Z} / d\mathbb{Z}$.