algebraic number theory - Field Extensions with $sqrt{2}$

If I already have a field $\mathbb{K} = \mathbb{Q}(\theta)$ where $\theta$ is the root of an irreducible cubic polynomial, then what field extension is necessary to access elements of the form $\theta ( \theta + \sqrt{2})$. My guess is that we need $\mathbb{K}(\sqrt{2})$ but I am unsure how to see or manipulate this field. Any help would be appreciated!

Answer

If $K=\Bbb{Q}(\theta)$ and $L\supset K$ is a field extension with $\alpha:=\theta(\theta+\sqrt{2})\in L$, then because $\theta\in L$ also

$$\frac{\alpha}{\theta}-\theta=\sqrt{2}\in L,$$
and conversely if $\sqrt{2}\in L$ then $\theta(\theta+\sqrt{2})\in L$. So $L$ contains $K(\sqrt{2})$, and $K(\sqrt{2})$ is the smallest extension of $K$ containing $\theta(\theta+\sqrt{2})$.

Because $\theta$ is a zero of an irreducible cubic, the extension $K/\Bbb{Q}$ has degree $3$. In particular it has no quadratic subfield, so $\sqrt{2}\notin K$. This means that for every $z\in K(\sqrt{2})$ there exist unique $x,y\in K$ such that $z=x+y\sqrt{2}$.

For every element $x\in K$ there exist unique $a,b,c\in\Bbb{Q}$ such that $x=a+b\theta+c\theta^2$ because $\theta$ is a zero of an irreducible cubic. Hence $K(\sqrt{2})$ consists of expressions of the form
$$a+b\theta+c\theta^2+d\sqrt{2}+e\theta\sqrt{2}+f\theta^2\sqrt{2}.$$