# Show that $\mathbb{Q}(\sqrt{n},\sqrt{-n})\subseteq\mathbb{Q}(\sqrt{n}+\sqrt{-n})$ by evaluating $\alpha^3+2n\alpha$ and $\alpha^3-2n\alpha$

Let $n$ be a positive integer that is not a perfect square and let $\alpha=\sqrt{n}+\sqrt{-n}$

Previous parts of the question have yielded that $\sqrt{n}\notin\mathbb{Q}(\sqrt{-n}), \sqrt{-n}\notin\mathbb{Q}(\sqrt{n})$ and $[\mathbb{Q}(\sqrt{n},\sqrt{-n}):\mathbb{Q}]=4$.

I have evaluated the expressions as asked and got $\alpha^3+2n\alpha=4n\sqrt{-n}$ and $\alpha^3-2n\alpha=-4n\sqrt{n}$. How does evaluating these tell me that $\mathbb{Q}(\sqrt{n},\sqrt{-n})\subseteq\mathbb{Q}(\sqrt{n}+\sqrt{-n})$? I'm not sure what the missing step is.

• Hint: $1/4n\in\mathbb Q$. – Wojowu Mar 21 '17 at 19:47
• Hmm, my thoughts are that $\alpha^3+2n\alpha$ is the minimal polynomial for $\mathbb{Q}(\sqrt{-n})$ and $\alpha^3-2n\alpha$ is the minimal for $\mathbb{Q}(\sqrt{n})$. Maybe I should multiply the two evaluated expressions together to get something in the form $(\sqrt{n}+\sqrt{-n})$ and so it belongs in $\mathbb{Q}(\sqrt{n}+\sqrt{-n})$? – Mike A Mar 21 '17 at 20:14

You have generated $\sqrt{n}$ and $\sqrt{-n}$ as polynomials in $\sqrt{n} + \sqrt{-n}$ with rational coefficients.
$\mathbb{Q}(x)$ is defined as the field generated by polynomials in $x$ with rational coefficients.
Since $\sqrt{n}$ and $\sqrt{-n}$ can be generated by polynomials in $\sqrt{n} + \sqrt{-n}$, any number generated by polynomials in $\sqrt{n}$ and $\sqrt{-n}$ can be generated in $\mathbb{Q}(\sqrt{n} + \sqrt{-n})$ by substituting in their respective polynomials in terms of $\sqrt{n} + \sqrt{-n}$.