Consider a field $\mathbb Q (\sqrt5, \sqrt7,\sqrt{35})$ as an extension over $\mathbb Q$. What is the degree of the extension?
I am confused by the notation here. Does $\mathbb Q (\sqrt5, \sqrt7,\sqrt{35})$ mean a vector space with $(\sqrt5, \sqrt7,\sqrt{35})$ as a basis? If so, what does that signify?
I know to find the degree of $[\mathbb Q (\alpha):\mathbb Q]$ you find the degree of the minimal polynomial of $\alpha$ over $\mathbb Q$, but what does this mean when I have three terms inside the brackets? Do I find all the minimal polynomials?
Also, in terms of finding it's degree, does $\sqrt {35}=\sqrt {3*5}$ mean anything?
It means the smallest field containing $\mathbb{Q}$ and the three elements $\sqrt5, \sqrt7,\sqrt{35}$.
By general results one knows that this coincides with the smallest ring with these properties. As $\sqrt5 \sqrt7=\sqrt{35}$, $ \sqrt7\sqrt{35}= 7 \sqrt5$ and $ \sqrt5\sqrt{35}= 7 \sqrt7$ while the square of each of the elements is rational, you see that each element in that ring is of the form $q_1 + q_2 \sqrt5 + q_3 \sqrt7 +q_4 \sqrt{35}$ with rationals $q_i$. In other words a basis is given by $1,\sqrt5, \sqrt{7} ,\sqrt{35}$ and the degree is $4$.
Also note that $\mathbb Q (\sqrt5, \sqrt7) = \mathbb Q (\sqrt5, \sqrt7,\sqrt{35})$, as once you have $\sqrt5, \sqrt7$ in a field you must have their product too. The same is true omitting any one of the three roots.