- $K_{1} = \mathbb{Q}(\sqrt{ 2 }), K_{2} = \mathbb{Q}(\sqrt{ 3 })$. Then $K_{1} K_{2} = \mathbb{Q}(\sqrt{ 2 }, \sqrt{ 3 })$.
- $K_{1} = \mathbb{Q}(\sqrt{ 2 }), K_{2} = \mathbb{Q}(\sqrt[3]{ 2 })$. 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 $\sqrt{ 2 } = \left( \sqrt[6]{ 2 } \right)^{3} \implies \mathbb{Q}(\sqrt{ 2 }, \sqrt[3]{ 2 }) \subseteq \mathbb{Q}\left( \sqrt[6]{ 2 } \right)$. $\sqrt[6]{ 2 } = \sqrt{ 2 }\left( \sqrt[3]{ 2 } \right)^{-1} \implies \mathbb{Q}(\sqrt[6]{ 2 }) \subseteq \mathbb{Q}(\sqrt{ 2 }, \sqrt[3]{ 2 })$.