I have some short examples but don't exactly know how they work.
ex1. dim $\mathbb{C}$ over $\mathbb{R}$ is $2$
ex2. dim $\mathbb{R}(\sqrt2)$ over $\mathbb{R}$ is $1$
ex3. Basis of $\mathbb{Q}(\sqrt[3]{2}) $ over $\mathbb{Q}$ is $1, 2^{1/3}, 2^{2/3}$.
ex4. Basis of $F(\sqrt3)$ over $F$ is $1, 3^{1/2}$
ex5. Basis of $\mathbb{Q}(\sqrt[3]2)(\sqrt3)$ over $\mathbb{Q}$ is $1, 2^{1/3}, 2^{2/3}, \sqrt3, \sqrt3 *2^{1/3}, \sqrt3*2^{2/3}$. So it's $6$ dimensional vector space.
This is my first time studying "Extending Fields" but the solutions are too short for me to understand.
EX1. I think $i$ is the solution to $x^2+1$. Is that why dimension is $2$?
EX2. If I were to do the same thing as in EX1, $x^2-1$ has $\sqrt2$ as a root. Then shouldn't the dimension be $2$?
EX3. The polynomial that has $\sqrt[3]2$ as a solution is $x^3 - 2$. And then what do I do?
EX4. Same as EX3. Found the polynomial $x^2 -3$. Can't figure out the next step.
EX5. No idea for this one..