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?

  • 1
    $\begingroup$ In general it is risky to ask questions about intersections, because it is heavily dependent on the meaning. For some rings, list $\mathbb Q[\sqrt[3]{2}]$, there ar emultiple embeddings into $\mathbb C$, so whether the rings intersects another field extension of $\mathbb Q$ is ambiguous. For quadratic extensions such as these, though, the image of the embeddings are unique (even if the embedding functions are not.) $\endgroup$ Feb 8 '16 at 1:08
  • $\begingroup$ Where did this question come from? Homework! $\endgroup$ Feb 8 '16 at 1:08
  • $\begingroup$ you know $Q \subset Q(√2)$ and $Q\subset Q(i)$ then $Q\subset Q(√2)\cap Q(i) $ then to see $=$ you have to show "give $a,b\in Q$ $(b\noteq 0)$ that $a+b√2 \notin Q(i)$ and so... The answer it is true $\endgroup$
    – Elll
    Feb 8 '16 at 1:09
  • $\begingroup$ @MhenniBenghorbal I found this question in an old exam while studying for my exam (Linear Algebra). $\endgroup$
    – Kobi T
    Feb 8 '16 at 1:13

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.


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.


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