Lecture. Composite extensions [l15]

  • May 20, 2026

Composite extensions

Definition 1. composite field [2401]

    Let $K_{1}, K_{2}$ be subfields of $K$. The composite field of $K_{1}$ and $K_{2}$, denoted $K_{1} K_{2}$, is the smallest subfield of $K$ containing both $K_{1}$ and $K_{2}$. Similarly, if $K_{1}, \dots, K_{n}$ are subfields of $K$, the composite field of $K_{1}, \dots, K_{n}$, denoted $K_{1}K_{2}\cdots K_{n}$ is the smallest subfield of $K$ containing both $K_{1}, \dots, K_{n}$.

    It follows from the definition that $K_{1} K_{2} = K_{1}(K_{2}) = K_{2}(K_{1})$.

    Example 2. some examples [2403]

      1. $K_{1} = \mathbb{Q}(\sqrt{ 2 }), K_{2} = \mathbb{Q}(\sqrt{ 3 })$. Then $K_{1} K_{2} = \mathbb{Q}(\sqrt{ 2 }, \sqrt{ 3 })$.
      2. $K_{1} = \mathbb{Q}(\sqrt{ 2 }), K_{2} = \mathbb{Q}(\sqrt[3]{ 2 })$. Then $K_{1} K_{2} = \mathbb{Q}(\sqrt[6]{ 2 })$. We know that $K_{1}K_{2} = \mathbb{Q}(\sqrt{ 2 }, \sqrt[3]{ 2 })$, so we need to check that $\mathbb{Q}(\sqrt{ 2 }, \sqrt[3]{ 2 }) = \mathbb{Q}(\sqrt[6]{ 2 })$. $\sqrt[3]{ 2 } = \left( \sqrt[6]{ 2 } \right)^{2}$ and $\sqrt{ 2 } = \left( \sqrt[6]{ 2 } \right)^{3} \implies \mathbb{Q}(\sqrt{ 2 }, \sqrt[3]{ 2 }) \subseteq \mathbb{Q}\left( \sqrt[6]{ 2 } \right)$. $\sqrt[6]{ 2 } = \sqrt{ 2 }\left( \sqrt[3]{ 2 } \right)^{-1} \implies \mathbb{Q}(\sqrt[6]{ 2 }) \subseteq \mathbb{Q}(\sqrt{ 2 }, \sqrt[3]{ 2 })$.

      Proposition 3. $\mathrm{Gal}(KE / E) \cong \mathrm{Gal}(K / K \cap E)$ [2404]

        Let $K / F$ and $E / F$ be finite field extensions. Suppose that $K / F$ is Galois. Then $KE / E$ is Galois and $\mathrm{Gal}(KE / E) \cong \mathrm{Gal}(K / K \cap E)$.

        Proof. sketch [2404A]

          $KE / E$ is Galois: Use Proposition 17012 and (3) of Proposition 2205 and the premise that $K / F$ is Galois.

          $\mathrm{Gal}(KE / E) \cong \mathrm{Gal}(K / K \cap E)$: The map $$ \mathrm{Gal}(KE / E) \to \mathrm{Gal}(K / K \cap E), \qquad \sigma \mapsto \sigma |_{K} $$ is a group homomorphism. The kernel is $\{ \operatorname{id}_{KE} \}$. Let $H$ be its image. Then $$ \begin{aligned} \alpha \in K^{H} &\iff \alpha \in K \text{ and } \sigma(\alpha) = \alpha \text{ for all } \sigma \in \mathrm{Gal}(KE / E) \\ &\iff \alpha \in K \cap \left( KE \right)^{\mathrm{Gal}(KE / E)}. \end{aligned} $$ $KE / K$ is Galois $\implies (KE)^{\mathrm{Gal}(KE / E)} = E$. Hence, $$ \alpha \in K^{H} \iff \alpha \in K \cap E. $$ It follows that $K^{H} = K \cap E$. Then again by the Fundamental Theorem of Galois Theory, we have $$ H = \mathrm{Gal}(K / K^{H}) = \mathrm{Gal}(K / K \cap E). $$

          Proposition 4. the degree of $K_1K_2/F$ [2405]

            Let $K_{1}$ and $K_{2}$ be two finite extensions of a field $F$ contained in $K$.

            1. $[K_{1}K_{2} : F] \leq \dfrac{[K_{1} : F][K_{2} : F]}{[K_{1} \cap K_{2} : F]}$.
            2. If $[K_{1} : F]$ and $[K_{2} : F]$ are coprime, then $K_{1} \cap K_{2} = F$ and $[K_{1}K_{2} : F] = [K_{1} : F][K_{2} : F]$.
            3. If $K_{1} / F$ is Galois, then the equality in (1) holds.

            Proposition. sketch [2405A]

              1. What we want to show is $$\begin{aligned}&[K_{1}K_{2} : F] \leq \frac{[K_{1} : F][K_{2} : F]}{[K_{1} \cap K_{2} : F]} \\\iff &[K_{1}K_{2} : K_{1} \cap K_{2}][K_{1} \cap K_{2} : F] \leq [K_{1} : F][K_{2} : K_{1} \cap K_{2}] \\\iff & \frac{[K_{1}K_{2} : K_{1} \cap K_{2}]}{[K_{2} : K_{1} \cap K_{2}]} \leq \frac{[K_{1} : F]}{[K_{1} \cap K_{2} : F]} \\\iff & [K_{1} K_{2} : K_{2}] \leq [K_{1} : K_{1} \cap K_{2}].\end{aligned}$$ Let $E = K_{1} \cap K_{2}$. Suppose $[K_{1} : E] = n$ and $K_{1} = E\alpha_{1} + \cdots + E\alpha_{n}$ where $\alpha_{1}, \dots, \alpha_{n} \in K_{1}$. Then $K_{1} = E(\alpha_{1}, \dots, \alpha_{n}) \implies K_{1}K_{2} = K_{2}(\alpha_{1}, \dots, \alpha_{n})$. Now, $[K_{1} : E] < \infty$ implies that $\alpha_{1}, \dots, \alpha_{n}$ are algebraic over $E$, hence algebraic over $K_{2}$. Then we have $K_{1}K_{2} = K_{2}[\alpha_{1}, \dots, \alpha_{n}]$.

                To show that $[K_{1}K_{2} : K_{2}] < n$, it suffices to show that $K_{1}K_{2} \subseteq K_{2}\alpha_{1} + \cdots + K_{2}\alpha_{n}$. Now every element of $K_{1}K_{2}$ is of the form of linear combinations of $K_{2}\alpha_{1}^{r_{1}}\cdots\alpha_{n}^{r_{n}}$.Since each $\alpha_{i} \in K_{1}$, their multiple is also in $K_{1}$, i.e. in $E\alpha_{1} + \dots + E\alpha_{n} \subseteq K_{2}\alpha_{1} + \cdots + K_{2}\alpha_{n}$. So is the combinations, i.e. the element of $K_{1}K_{2}$. Hence we have shown $K_{1}K_{2} \subseteq K_{2}\alpha_{1} + \cdots + K_{2}\alpha_{n}$.

              2. Omitted

              3. $K_{1} / F$ Galois $\implies K_{1} / K_{1} \cap K_{2}$ Galois. Then use Proposition 2404.

              Proposition 5. $K_{1} / F$ and $K_{2} / F$ Galois $\implies K_{1} \cap K_{2} / F$ and $K_{1}K_{2} / F$ Galois [2406]

                Let $K_{1}, K_{2}$ be finite Galois extensions of $F$.

                1. $K_{1} \cap K_{2}$ and $K_{1}K_{2}$ are Galois over $F$.
                2. $\mathrm{Gal}(K_{1}K_{2} / F)$ is isomorphic to$$H = \{ (\sigma, \tau) \mid \sigma |_{K_{1} \cap K_{2}} = \tau_{K_{1} \cap K_{2}} \} \leq \mathrm{Gal}(K_{1} / F) \times \mathrm{Gal}(K_{2} / F).$$In particular, if $K_{1} \cap K_{2} = F$, then $\mathrm{Gal}(K_{1}K_{2} / F) \cong \mathrm{Gal}(K_{1} / F) \times \mathrm{Gal}(K_{2} / F)$.

                It is true that $K_{1} / F, K_{2} / F$ Galois, $K_{1} \cap K_{2} = F \implies \mathrm{Gal}(K_{1}K_{2} / F) \cong \mathrm{Gal}(K_{1} / F) \times \mathrm{Gal}(K_{2} / F)$, without assuming the extensions are finite. Probably can be proved by taking direct limits. Another proof is using $\mathrm{Gal}(K_{1} / F) \cong \mathrm{Gal}(K_{1}K_{2} / K_{2})$ and $\mathrm{Gal}(K_{2} / F) \cong \mathrm{Gal}(K_{1}K_{2} / K_{1})$. To prove that $\mathrm{Gal}(K_{1}K_{2} / K_{2}) \to \mathrm{Gal}(K_{1} / F)$ is surjective, letting $H$ be the image, we have $K_{1}^{H} \subseteq (K_{1}K_{2})^{\mathrm{Gal}(K_{1}K_{2} / K)} = K_{2}$, so $K_{1}^{H} = F$. $\mathrm{Gal}(K_{1}K_{2} / K_{2})$ is compact, so $H$ is compact, closed in $\mathrm{Gal}(K_{1} / F)$, so $K_{1}^{H} = F \implies H_{1} = \mathrm{Gal}(K_{1} / F)$.