# Finding quadratic extensions in $\mathbb{Q}\left(\sqrt{i+2}\right)$

Let $\alpha=\sqrt{i+2}$ and let $F=\mathbb{Q}\left(\alpha\right)$. Note that $\left[F:\mathbb{Q}\right]=4$ since the minimal polynomial of $\alpha$ over $\mathbb{Q}$ is $x^{4}-4x^{2}+5$. Show that $F$ contains 3 distinct quadratic extensions of $\mathbb{Q}$.

I can find only one. We have $i=\left(\sqrt{i+2}\right)^{2}-2\in F$ and hence $\mathbb{Q}\left(i\right)\subseteq\mathbb{Q}\left(\sqrt{i+2}\right)$. I am not sure though about the other two. Any help would be appreciated.

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What are the other roots of $x^4 - 4x^2 + 5$? – Qiaochu Yuan Jul 19 '12 at 16:41
They are $\pm\sqrt{2\pm i}$. How does this help though? – Galois Jul 19 '12 at 16:55
Okay. Can you show that they (edit: some of them!) lie in $F$? Can you construct other quadratic extensions from them? – Qiaochu Yuan Jul 19 '12 at 16:58
Oh.. and the third quadratic extension is $\mathbb{Q}(\sqrt{-5})$. – Galois Jul 19 '12 at 17:08
Two of the other roots do not lie in $F$: your quartic extension is not normal. In general, you should expect that a quadratic extension of a quadratic extension will not be normal. – Lubin Jul 19 '12 at 17:09

First task: verify that your extension is not normal. As you’ve seen, the complete set of roots of your irreducible equation is $\{\pm\sqrt{2+i},\pm\sqrt{2-i}\}$. I’ll call these $\pm\xi$ and $\pm\eta$. To see that $\eta$ is not in your field $F= {\mathbb{Q}}(i,\xi)$, you need only check that the ratio between $\xi^2$ and $\eta^2$ as elements of ${\mathbb{Q}}(i)$ is not a square. This ratio is $(2+i)^2/5$, not a square since $5$ isn’t.
Next step is easy, to describe carefully what the normal closure (over ${\mathbb{Q}}$) is, and it’s simply ${\mathbb{Q}}(i,\xi,\eta)$, which you’ve already observed contains $\sqrt5$.
Now for the Galois group. Let's call $a$ the automorphism that exchanges $\xi$ and $\eta$, and $b$ the one that leaves $\xi$ fixed, sends $\eta$ to $-\eta$. They’re both involutions (square to the identity), and $ab$, by which I mean $b$ first, then $a$, sends $\xi\mapsto\eta\mapsto-\xi\mapsto-\eta\mapsto\xi$, so is of period four. You’ve got the dihedral group of order eight. This has five subgroups of order two, only one of them being normal, and three subgroups of order four, all of them normal, being of index two. I won’t go through the details of enumerating the elements of each subgroup, rather I include a (crude) field diagram, in which you see that the only quartic field that has two quadratic subfields is the normal one, ${\mathbb{Q}}(i,\sqrt5)$. Of course you have to verify all this yourself.