Find the degree $[K:F]$ of the following field extensions: (a) $K=\mathbb{Q}(\sqrt{7})$, $F=\mathbb{Q}$ (b) $K=\mathbb{C}(\sqrt{7})$, $F=\mathbb{C}$ (c) $K=\mathbb{Q}(\sqrt{5},\sqrt{7},\sqrt{35})$, $F=\mathbb{Q}$ (d) $K=\mathbb{R}(s)$, $F=\mathbb{R}$, where $s^3-4s^2+6s+6=0$
Firstly, I should mention that I am very new to field theory; this is for an algebraic number theory book that I'm trying to follow, but the only algebra I know is group theory.
I know the degree is the dimension of $K$ as an $F$-vector space, so I need to find a basis for each of the above $K$'s. I'm pretty sure (a) has degree 2, as does (c), because $\sqrt{35}=\sqrt{7}\sqrt{5}$, so we might as well simply look at $\mathbb{Q}(\sqrt{35})$.
I'm not sure what to do about (d), and (b) puzzles me because the complex numbers already contain $\sqrt{7}$. Does this mean it has degree 1?
I'm sorry my question can't be more specific.
You are correct about (a), its degree is 2.
For (b), your suspicion is also correct, its degree is 1 since $\sqrt{7}$ already belongs to $\mathbb{C}$ ($\mathbb{C}$ is algebraically closed so it has no finite extensions).
Your reasoning for (c) isn't quite right. Yes, $\sqrt{5} \cdot \sqrt{7} = \sqrt{35}$ but $\mathbb{Q}(\sqrt{35})$ is strictly smaller than $K$. Consider instead $L=\mathbb{Q}(\sqrt{5})$. $L$ is of degree 2 over $F=\mathbb{Q}$. However, $\sqrt{7}$ doesn't belong to $L$. $K=L(\sqrt{7})$ has degree 2 over $L$. Thus $[K:F] = [K:L]\cdot [L:F] = 2 \cdot 2 =4$.
As for (d), I'm confused as well. Maybe your question is missing something, but if $s^3-4s^2+6s+6=0$, then $s$ can be either real of one of two (conjugate) complex roots. If $s$ is real then $K=\mathbb{R}$ and the extension has degree 1. If $s$ is complex, then $K=\mathbb{C}$ and the extension has degree 2.