# 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!

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 $$a+b\theta+c\theta^2+d\sqrt{2}+e\theta\sqrt{2}+f\theta^2\sqrt{2}.$$
You're right. You need $$\mathbb Q(\theta,\sqrt2)=K(\sqrt 2)$$.
The elements of this field are of the form $$a+b\sqrt 2+c\theta+d\theta^2+e\theta\sqrt 2+f\theta^2\sqrt 2$$, with $$a,b,c,d,e,f \in \mathbb Q$$.
This is simple because $$\theta$$ has degree $$3$$ and $$\sqrt 2$$ has degree $$2$$, which are coprime.