# How to adjoin field from $\mathbb{Q}(\sqrt6)$ to $\mathbb{Q}(\sqrt2,\sqrt3)$

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?

• 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. – Burnsba Oct 4 '18 at 0:24

Adjoin $$\sqrt{2}$$ : $$\mathbb{Q}(\sqrt{6})(\sqrt{2})=\mathbb{Q}(\sqrt{2}, \sqrt{3})$$. You can also adjoin $$\sqrt{3}$$.