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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?

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  • $\begingroup$ There's a lattice diagram: imgur.com/XuyO8hx so it seems there is some significance to the relationship between $\sqrt2$ , $\sqrt3$, and $\sqrt6$ but that seems outside the scope of the question. $\endgroup$ – Burnsba Oct 4 '18 at 0:24
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Adjoin $\sqrt{2}$ : $\mathbb{Q}(\sqrt{6})(\sqrt{2})=\mathbb{Q}(\sqrt{2}, \sqrt{3})$. You can also adjoin $\sqrt{3}$.

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