Basis for $\mathbb{Q}[\sqrt{8}]$ over $\mathbb{Q}[\sqrt{2}]$

Provided that $x^2-8$ is the minimal polynomial for $\mathbb Q[\sqrt8]$ and $x^2-2$ is minimal for $\mathbb Q[\sqrt 2]$ we should have a basis with four elements. Thus far I know $1$ and $\sqrt 2$ should be in the basis, but I am having trouble figuring out the other elements. It seems like they would just be multiples of $\sqrt 2$.

• Q[sqrt8] is a subfield of Q[sqrt2] though. Apr 14 '14 at 8:07
• Not just a subfield... Apr 14 '14 at 8:07

Since $\sqrt{8} = 2\sqrt{2}$, then $\mathbb{Q}[\sqrt{8}]$ is a degree $1$ (trivial) field extension over $\mathbb{Q}[\sqrt{2}]$. In other words, $\mathbb{Q}[\sqrt{8}] = \mathbb{Q}[\sqrt{2}]$. Therefore, you only need $1$ basis vector, which can be any nonzero element of $\mathbb{Q}[\sqrt{2}]$ that you'd like.