Is it true that $\mathbb{Q}(\sqrt{2}) \cap \mathbb{Q}(i) = \mathbb{Q}$?

Question

Is it true that $\mathbb{Q}(\sqrt{2}) \cap \mathbb{Q}(i) = \mathbb{Q}$?

I know that \begin{align*} \mathbb{Q}(\sqrt{2}) &= \{a+b\sqrt{2} \mid a,b \in \mathbb{Q}\}, \\ \mathbb{Q}(i) &= \{a+bi \mid a,b \in \mathbb{Q}\} \end{align*}

I tend to believe it is, since every element in $\mathbb{Q}(\sqrt{2})$ belongs to $\mathbb{Q}(i)$ iff $b=0$. Also, an element in $\mathbb{Q}(i)$ belongs to $\mathbb{Q}(\sqrt{2})$ iff $b = 0$.

Is that enough for a formal proof?

Answer 1

Well, if you've actually proved the biconditional statements you mentioned, then you're done.

Alternatively, show that $$\Bbb Q\subseteq\Bbb Q(i)\cap\Bbb Q\bigl(\sqrt2\bigr),$$ which I leave to you. Then, suppose $z\in\Bbb Q(i)\cap\Bbb Q\bigl(\sqrt2\bigr).$ Since $z\in\Bbb Q\bigl(\sqrt2\bigr),$ then $z\in\Bbb R.$ From there, we can use the fact that $z\in\Bbb Q(i)$ to readily show that $z\in\Bbb Q,$ completing the proof.

Answer 2

yes it is true $Q\sqrt2$ and $Q(i)$ are extension of degree $2$ of $Q$ so the degree of $Q(i)\cap Q(\sqrt2)$ is either 1 or $2$, if it is 2 it implies that $Q(i)=Q(\sqrt2)$ thus $\sqrt2=a+bi, a,b\in Q$, by writing $(\sqrt2-a)^2=-b^2$, you obtain $a=0$ unless $\sqrt2\in Q$ a fact which is not true. If $a=0$, $\sqrt2=bi$ thus the $2=-b^2$ this not true also.