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Let $N$ be a Galois extension of $\mathbb{Q}$ such that $G=\text{Gal}(N/\mathbb{Q})\cong \mathbb{Z}_4$. How can we decide whether or not $\mathbb{Q}(\sqrt{a})$ can be embedded in $N$, given that $a>0$ and $a$ is not a square in $\mathbb{Q}$?

I can say that $\mathbb{Z}_4$ has $\mathbb{Z}_2$ as a subgroup, so there is a corresponding subfield with extension degree $2$, and if $\sigma$ generates $\mathbb{Z}_4$, then $\sigma^2$ generates the subgroup $\mathbb{Z}_2$, so $\sigma^2$ should fix $\sqrt{a}$. I wasn't sure how to proceed from here, so I decided to just try to construct such an extension.

Let $b\in\mathbb{Q}$ be such that $b-\sqrt{a}>0$ and $b^2-a$ is a square in $\mathbb{Q}$. For example, if $a=5$ then taking $b=3$ works. Then consider the polynomial $p(x)=(x-\sqrt{b+\sqrt{a}})(x+\sqrt{b+\sqrt{a}})(x-\sqrt{b-\sqrt{a}})(x+\sqrt{b-\sqrt{a}})$. This is clearly irreducible over $\mathbb{Q}[x]$, and taking $N$ to be the splitting field for $p(x)$ we find that as a vector space over $\mathbb{Q}$ it is spanned by $\{1, \sqrt{b+\sqrt{a}}, \sqrt{b-\sqrt{a}}, \sqrt{a}\}$ which is due to our choice of $b$ so that $b^2-a$ is a square. Thus the extension is of degree $4$. Now it is clear that $\mathbb{Q}(\sqrt{a})$ is a subfield of $N$, so there should be an element of order $2$ the Galois group $G$ which fixes $\sqrt{a}$. Since the extension is degree $4$, the Galois group $G$ must be isomorphic to a group of degree $4$, that is either the Klein 4 group $V$ or the cyclic group $\mathbb{Z}_4$. Since $V$ has $3$ subgroups of order $2$, but $N$ has only one subfield of degree $2$, we conclude that $G\cong \mathbb{Z}_4$. Thus $G$ is generated by $\sigma$ such that $\sigma^2$ fixes $\sqrt{a}$. Then $\sigma$ is defined by its action on $\sqrt{b+\sqrt{a}}$ and $\sqrt{b-\sqrt{a}}$. Thus we must have $\sigma(\sqrt{b+\sqrt{a}})=\sqrt{b-\sqrt{a}}$ and $\sigma(\sqrt{b-\sqrt{a}})=-\sqrt{b+\sqrt{a}}$ (up to a minus sign). These two choices ensure that $\sigma$ is of order $4$, which is good because there are two generators for $\mathbb{Z}_4$.

Now we note that $\sqrt{a}=-b+(\sqrt{b+\sqrt{a}})^2$ and so $\sigma^2(\sqrt{a})=b+(\sigma^2(\sqrt{b+\sqrt{a}}))^2 = b+(-\sqrt{b+\sqrt{a}})^2=\sqrt{a}$ as required.

So I suppose the conclusion is that, if $a$ is a difference of squares in $\mathbb{Q}$ (since we needed that $b^2 - a = c^2$), then we can find a cyclic extension $N$ with $G\cong\mathbb{Z}_4$ such that $\mathbb{Q}(\sqrt{a})$ is a subfield of $N$. For example, $a=5=3^2-4^2$ and $a=7=4^2-3^2$ work.

Now I have two questions.

  1. Is everything I have done so far correct?
  2. Since my method was constructive, it is not clear that if $a$ isn't a difference of squares in $\mathbb{Q}$, that such an embedding is impossible. Is this the case? What else can we say about this problem?
Jay
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    I seem to remember that the criterion for $\sqrt a$ to lie in a cyclic quartic extension is that $a$ must be the sum of two squares. After all, every rational is the difference of two squares of rationals. – Angina Seng Sep 08 '20 at 11:04
  • Related 1, 2, 3. – Jyrki Lahtonen Sep 08 '20 at 11:14
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    What Angina Seng said: A lemma due to Albert: If $F$ is a field, $\sqrt{-1}\notin F$, and $E/F$ is a cyclic quartic extension, then the unique quadratic intermediate field is of the form $F(\sqrt{u^2+v^2})$, where $u,v\in F$ and (obviously) $u^2+v^2$ is a non-square of $F$. – Jyrki Lahtonen Sep 08 '20 at 11:31
  • @AnginaSeng After a quick google I found this link which agrees with what you are saying. After some thought, you are correct about what rationals all being the difference of two squares of rationals. But then where is my construction going wrong? What am I missing? – Jay Sep 08 '20 at 11:38
  • @JyrkiLahtonen If this is the case, then 7 should not work as I erroneously claimed in the post (since 7 is not the sum of squares). Can you help me see what is wrong with my construction? – Jay Sep 08 '20 at 11:51
  • Thinking about it. With $b=4$, $a=7$ you get the biquadratic $x^4-8x^2+9$. According to this result the Galois group should be the Klein four. Thinking... – Jyrki Lahtonen Sep 08 '20 at 11:54
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    $(\sqrt{4+\sqrt7}+\sqrt{4-\sqrt7})^2=8+2\cdot3=14$. Therefore $\sqrt{14}\in\Bbb{Q}(\sqrt{4+\sqrt7},\sqrt{4-\sqrt7})$. If the constant of the biquadratic is a square, this leads to another quadratic intermediate field. – Jyrki Lahtonen Sep 08 '20 at 11:57
  • @JyrkiLahtonen Ahh, I completely failed to consider that. If we look at 5 which is the sum of squares and perform the analogous calculation, we do not get an additional element of the extension. I wonder if my reasoning could have worked out with additional constraints to avoid such a situation. I'll think about it. Thank you for the help! – Jay Sep 08 '20 at 12:04
  • Alternatively: $(\sqrt{14}+\sqrt2)^2=4(4+\sqrt7)$, so $\sqrt{4+\sqrt7}\in\Bbb{Q}(\sqrt2,\sqrt{14}).$ – Jyrki Lahtonen Sep 08 '20 at 12:12
  • And also $$(\sqrt{3+\sqrt5}+\sqrt{3-\sqrt5})^2=10,$$ so $b=3$, $a=5$ won't give you a cyclic extenstion either. – Jyrki Lahtonen Sep 08 '20 at 12:19
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    I think this answer by Angina Seng may help you. – Jyrki Lahtonen Sep 08 '20 at 12:23

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