At the end of https://youtu.be/Buv4Y74_z7I he asks a question, "What do you adjoin on ${Q}(\sqrt6)$ to get $\mathbb{Q}(\sqrt2,\sqrt3)$?"
So $\mathbb{Q}(\sqrt6)$ has elements like $a + b\sqrt6$ with $a,b \in \mathbb{Q}$.
And $\mathbb{Q}(\sqrt2,\sqrt3)$ has elements like $a + b\sqrt2 + c\sqrt3 + d\sqrt6$ with $a,b,c,d \in \mathbb{Q}$.
That's all I know.
I guess the variables $a,b,c,d$ are not the same in both equations, but it seems it doesn't really matter as long as they're rational?
Well, "Q adjoin x" is going to be of the form $a + bx$ with $a,b \in \mathbb{Q}$. And then "(Q adjoin x) adjoin y" is going to look like $(a + bx) + (c + dx)y$ with $a,b,c,d \in \mathbb{Q}$.
So I just need to solve for $x$ in the following:
$$
\mathbb{Q}(\sqrt2,\sqrt3) = \mathbb{Q}(\sqrt6)(x)
$$
$$
a + b\sqrt2 + c\sqrt3 + d\sqrt6 = (e + f\sqrt6)+(e+f\sqrt6)x
$$
Then I guess that means the answer to the question, "What do you adjoin on ${Q}(\sqrt6)$ to get $\mathbb{Q}(\sqrt2,\sqrt3)$?" is either $\frac{1}{\sqrt2}$ or $\frac{1}{\sqrt3}$.
Is that right?
Best Answer
Adjoin $\sqrt{2}$ : $\mathbb{Q}(\sqrt{6})(\sqrt{2})=\mathbb{Q}(\sqrt{2}, \sqrt{3})$. You can also adjoin $\sqrt{3}$.