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Using the primitive element theorem I got that $\mathbb{Q}(\sqrt{1+2i},\sqrt{1-2i}) = \mathbb{Q}(\sqrt{1+2i}+\sqrt{1-2i})$, and by noting that this is and normal extension of a field of characteristic zero and the irreducible polynomial of $\sqrt{1+2i} + \sqrt{1-2i}$ is $x^4-4x^2-16$, we get that $|Gal(\mathbb{Q}(\sqrt{1+2i},\sqrt{1-2i}) / \mathbb{Q})| = 4$.

However I don't get what other isomorphisms besides conjugation can be on $Gal(\mathbb{Q}(\sqrt{1+2i},\sqrt{1-2i}) / \mathbb{Q})$. I'm confused because $\mathbb{Q(\sqrt{1+2i},\sqrt{1-2i})} $ is the splitting field of $x^4-2x^2+5$, and all the roots of this polynomial are $\sqrt{1+2i},-\sqrt{1+2i}, \sqrt{1-2i},-\sqrt{1-2i}$ so any element of $Gal(\mathbb{Q}(\sqrt{1+2i},\sqrt{1-2i}) / \mathbb{Q})$ is a permutation of these roots but any isomorphism of this kind seems to be the conjugation so $|Gal(\mathbb{Q}(\sqrt{1+2i},\sqrt{1-2i}) / \mathbb{Q})| = 2$ which would be a contradiction. Am I doing something wrong or what isomorphisms am I missing?

Just a user
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H4z3
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  • $(1+2i)^{1/2}\mapsto-(1+2i)^{1/2}$ –  May 13 '24 at 00:53
  • But by taking a root $a+bi$ of $\sqrt{1+2i}$ this would map $a+bi$ to $-a-bi$ and this doesn't fix $Q$ – H4z3 May 13 '24 at 01:35
  • Oh ok thanks but then in this case $Gal(\mathbb{Q}(\sqrt{1+2i},\sqrt{1-2i}) / \mathbb{Q})$ would be $D_4$? – H4z3 May 13 '24 at 02:06
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    This old thread covers your question, I think. – Jyrki Lahtonen May 13 '24 at 02:40
  • The numbers $\sqrt{1+2i}$ and $\sqrt{1-2i}$ have the same minimal polynomial ($x^4-2x^2+5$), so it feels like you misapplied the mechanism in the usual proof of the primitive element theorem. I would think that $$\Bbb{Q}(\sqrt{1+2i}+\sqrt{1-2i})\subset\Bbb{Q}(\sqrt{1+2i},\sqrt{1-2i})$$ with a proper inclusion. – Jyrki Lahtonen May 13 '24 at 02:47
  • @JyrkiLahtonen Thanks I think you're right. I tried something different and I actually got $K = \mathbb{Q}(\sqrt{1+2i},\sqrt{1-2i}) = \mathbb{Q}(i,\sqrt{\frac{1+\sqrt{5}}{2}})$ (I got this by actually computing $a$ and $b$ for $a+bi = \sqrt{1+2i}$. But in this case you can see that $[K:\mathbb{Q}] = 8 = |Gal(K/ \mathbb{Q})|$. And in this case I'm even more confused because we're actually looking for a group of order $8$ and I don't see how can I get $8$ adecuate permutations of the original elements of the extension – H4z3 May 13 '24 at 05:11
  • Nevermind, I'm dumb – H4z3 May 13 '24 at 05:20

1 Answers1

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The primitive element theorem dictates there is an $c$ such that $\mathbb Q(a,b)=\mathbb Q(c)$, but $c=a+b$ may not work. For example, $$\mathbb Q(\sqrt 2, \sqrt 3) = \mathbb Q(\sqrt 2 + \sqrt 3, \sqrt 2 - \sqrt 3)\not=\mathbb Q(2\sqrt 2)=\mathbb Q(\sqrt 2)$$

Let $a=\sqrt{1+2i}, b=\sqrt{1-2i}$. In this case, there are some ambigurity of which square-roots we are taking, but at least we can find one pair such that $\bar a = b$, therefore $a+b$ is real, hence cannot generate $\mathbb Q(a,b)$.

"but any isomorphism of this kind seems to be the conjugation"

Do you mean "complex conjugation"? This is also unjustified. The splitting field of an irreducible polynomial of degree $4$ would have Galois group of order at least $4$ (and at most $4!=24$). The Galois action is transitive for the roots of an irreducible polynomial.

Now we figure out the Galois group. So $\mathbb Q(a,b)$ contains $a^2$ hence $i$, as well as $ab=\sqrt{5}$. Therefore it contains a subextension $\mathbb Q(i, \sqrt 5)$ of degree $4$. It's also easy to show that $\mathbb Q(a,b)$ is also the splitting field of $p(x)=x^4-4x^2-16$ by checking $\pm a\pm b$ are all roots of the polynomial. We have $p(x)$ is irreducible, because otherwise, its Galois group is of degree $2$ or $3$ or $3!$, but none of them has a factor $4$. But since $p(x)$ has a pair of real roots, while $i\in\mathbb Q(a,b)$, we know that the splitting field is not simply $\mathbb Q$ joined by a single root, but more roots are needed.

Let $\alpha = \sqrt{2+2\sqrt 5}$ be a real root of $p(x)$. Now we claim the splitting field of $p(x)$ is just $\mathbb Q(\alpha, i)$ whose degree is $[\mathbb Q(\alpha, i):\mathbb Q]=[\mathbb Q(\alpha)(i):\mathbb Q(\alpha)][\mathbb Q(\alpha):\mathbb Q] = 8$. This is easy since the other roots are $-\alpha$ and $\beta = i\sqrt{2\sqrt 5 - 2}, -\beta$ which satisfies $\alpha\beta = 4i$. Therefore $\beta\in\mathbb Q(\alpha, i)$. Now it's pretty clear that $\operatorname{Gal}(F/\mathbb Q)$ where $F:=\mathbb Q(a,b)$ is generated by $\operatorname{Gal}(F/\mathbb Q(i))$ and the complex conjugation.

$\operatorname{Gal}(F/\mathbb Q(i))$ has order $4$ and each element is completely determined by its action on $\alpha$. Therefore there exists $\sigma\in \operatorname{Gal}(F/\mathbb Q(i))$ such that $\sigma(\alpha)=-\beta$, now we have $\sigma(-\alpha)=\beta, \sigma(\beta)=\sigma(4i/\alpha)=4i/\sigma(\alpha)=4i/(-\beta)=-\alpha$, $\sigma(-\beta)=-\sigma(\beta)=\alpha$. In similar ways, we can figure out all four elements in $\operatorname{Gal}(F/\mathbb Q(i))$, and if we label $\alpha, -\alpha, \beta, -\beta$ as $1,2,3,4$ respectively

$$\operatorname{Gal}(F/\mathbb Q(i)) = \{(1), (12)(34), (13)(24), (14)(23)\}$$

Note that the complex conjugation is just $(34)$. Therefore

$$\operatorname{Gal}(F/\mathbb Q) = \langle (1), (12), (34), (13)(24)\rangle \simeq D_4$$

The permutaitons form $D_4$ because it's precisely the symmetric group of the square:

$$\require{AMScd}\begin{CD} 1 @= 3\\ @| @|\\ 4 @= 2 \end{CD}$$

Just a user
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  • I don't get how are you concluding that $Gal(F/\mathbb{Q})$ has order 4 once you proved that an intermediate field ($\mathbb{Q}(i)$) also has galois group of order 4. I'm confused because $\mathbb{Q}(i) \subseteq F$ is a separable and normal extension so $[F: \mathbb{Q}(i)] = |Gal(F/ \mathbb{Q}(i)) | = 4$, however $[F:\mathbb{Q}] = [F: \mathbb{Q}(i)][\mathbb{Q}(i):\mathbb{Q}]$ and this last expresion is equal to 4 (because $\frac{5}{\sqrt{1+2i}} = \sqrt{1-2i}$ so $\sqrt{1-2i} \in \mathbb{Q}(\sqrt{1+2i})$ – H4z3 May 13 '24 at 06:22
  • @H4z3 $\operatorname{Gal}(F/\mathbb Q)$ has order $8$ as proved. If you mean I have only listed four elements at the end, I carefully used $\langle \cdots \rangle$, not ${\cdots}$, those four are just the generators of $D_4$. – Just a user May 13 '24 at 06:36
  • But why does it has $8$ elements if $\sqrt{1-2i} \in \mathbb{Q}(\sqrt{1+2i})$ and the irreducible polynomial of $\sqrt{1+2i}$ has degree $4$? – H4z3 May 13 '24 at 06:47
  • @H4z3 It's not true that $\sqrt{1-2i}\in\mathbb Q(\sqrt{1+2i})$. If you follow the argument, you know this cannot be right as the splitting field indeed has degree $8$. Essentially what we can show is that $\sqrt{1-2i}\in\mathbb Q(\sqrt{1+2i}, i)$, but $i$ is not in $\mathbb Q(\sqrt{1+2i})$. And $\sqrt{1+2i} \cdot \sqrt{1-2i}=\sqrt 5$, not $5$ as you claimed, so your "proof" doesn't work. – Just a user May 13 '24 at 06:51