Consider $\displaystyle \mathbb{Q}(\sqrt{2}) $. Is the dimension of $\displaystyle \mathbb{Q}(\sqrt{2}) $ equaled to $\displaystyle 3 $? Because $\displaystyle \{1, \sqrt{2} \} $ is a basis for $\displaystyle \mathbb{Q}(\sqrt{2}) $ over $\displaystyle \mathbb{Q} $.

Also why does $\displaystyle [\mathbb{Q}(\sqrt{2}+\sqrt{3}): \mathbb{Q}] = 4 $. Is it because $\displaystyle \text{irr}(\sqrt{2}+\sqrt{3}, \mathbb{Q}) = x^4-10x^{2}+1 $? And the degree of this polynomial is 4?