1
$\begingroup$

This question is an exact duplicate of:

Let $a = \sqrt{2+i}$ and $K$ is the splitting field of minimal polynomial of $a$ over $\mathbb{Q}$. Prove that $Gal(K/\mathbb{Q})$ is $D_4$.

I find the minimal polynomial of $a$ is $p(x)=x^4-4x^2+5$ and its roots are $\sqrt{2+i},-\sqrt{2+i},\sqrt{2-i},-\sqrt{2-i}$. Let $b=\sqrt{2-i}$. So the splitting field of $p$ is $K=\mathbb{Q}(a,b)$. Also by Eisenstein's criterion to $p(x-1)=x^4 - 4 x^3 + 2 x^2 + 4 x + 2$ with the prime 2 we can conclude, $p$ is irreducible over $\mathbb{Q}$. Thus $[\mathbb{Q}(a):\mathbb{Q}]=4$. Also $b\not\in\mathbb{Q}(a)$ (since if it were then $\sqrt{2+i}\cdot\sqrt{2-i}=\sqrt{5}$ would also be in $\mathbb{Q}(a)$), and minimal polynomial of $b$ over $\mathbb{Q}(a)$ is $x^2-2+i$. Thus $[\mathbb{Q}(a,b):\mathbb{Q}]=[\mathbb{Q}(a,b):\mathbb{Q}(a)]\cdot[\mathbb{Q}(a):\mathbb{Q}]=2\cdot 4=8.$ Now since $K$ is the splitting field over $\mathbb{Q}$ of a separable polynomial, $K/\mathbb{Q}$ is Galois. Hence $|Gal(K/\mathbb{Q})|=8$. Hence the 8 automorphisms are $$a\to \begin{cases}a\\-a\\b\\-b\end{cases}\quad\text{and}\quad b \to \begin{cases}b\\-b\end{cases}.$$ But there is no automorphism of order 4. Then how can the Galois group be $D_4$?

Can somebody correct me what I am missing?

$\endgroup$

marked as duplicate by Jyrki Lahtonen Mar 31 '18 at 15:56

This question was marked as an exact duplicate of an existing question.

  • 1
    $\begingroup$ See the same question. The Galois group is not abelian, has order $8$, and the automorphisms do not satisfy the quaternion relation, so it must be $D_4$. In the duplicate, the automorphism called $\sigma$ has order $4$. $\endgroup$ – Dietrich Burde Mar 31 '18 at 13:03
  • $\begingroup$ @DietrichBurde If $\sigma$ is given by $a\to -b, \quad b\to -b$, how it become automorphism of order 4? $\endgroup$ – pie Mar 31 '18 at 13:42
  • 1
    $\begingroup$ How does the rational root theorem make a four degree polynomial irreducible? At most you could tell it doesn't have linear factors... You also mention that the minimal pol. of $\;b=\sqrt{2-i}\;$ over $\;\Bbb Q(\sqrt{2+i})\;$ is $\;x^2-2+i\;$, which would imply that $\;2-i\in\Bbb Q(\sqrt{2+i})\;$ ...are you sure of all this? $\endgroup$ – DonAntonio Mar 31 '18 at 13:52
  • 1
    $\begingroup$ @DonAntonio since $$\sqrt {2+i}\in\mathbb{Q}(\sqrt {2+i})\implies (\sqrt {2+i})^2\in\mathbb{Q}(\sqrt {2+i})\implies 2+i\in\mathbb{Q}(\sqrt {2+i})\implies i\in\mathbb{Q}(\sqrt {2+i})\implies 2-i\in\mathbb{Q}(\sqrt {2+i})$$. Is this wrong? $\endgroup$ – pie Mar 31 '18 at 14:01
  • 1
    $\begingroup$ Observe that what you call the 8 automorphisms have a fishy part: one of the mixes you show there is $\;\begin{cases}a\mapsto b\\b\mapsto b\end{cases}\;$ , for example...but this is impossible as an automorphism is injective...This must me checked and made accurate. $\endgroup$ – DonAntonio Mar 31 '18 at 14:04
1
$\begingroup$

We put as before $\;a=\sqrt{2+i}\;,\;\;b=\sqrt{2-i}$ . I think the basic automorphisms are ("copying" the embedding $\;D_4\hookrightarrow S_4$ ):

$$\begin{cases}\sigma:\;\;a\mapsto -b\;,\;\;b\mapsto a\;\\{}\\\tau:\;\;a\mapsto a\;,\;\;b\mapsto -b\;\end{cases}$$

Observe then

$$\begin{cases}\sigma^2(a)=\sigma(-b)=-a\;,\;\;\sigma^2(b)=\sigma(a)=-b\;,\;\\{}\\ \sigma^3(a)=\sigma(-a)=b\;,\;\;\sigma^3(b)=\sigma(-b)=-a\;,\\{}\\ \sigma^4(a)=\sigma(b)=a\;,\;\;\sigma^4(b)=\sigma(-a)=b\;.\end{cases}\;\;\;\;\;\;\;\;\;\;\implies \text{ord}\,\sigma=4$$

$${}$$

$$\tau^2(a)=a\;,\;\;\tau^2(b)=b\;\;\;\;\implies \text{ord}\,\tau=2$$

And also

$$\begin{cases}\tau\sigma\tau(a)=\tau\sigma(a)=\tau(-b)=b=\sigma^3(a)\\{}\\ \tau\sigma\tau(b)=\tau\sigma(-b)=\tau(-a)=-a=\sigma^3(b)\end{cases}\;\;\;\;\;\;\;\;\;\;\implies \tau\sigma\tau=\sigma^3$$

and we thus got

$$\text{Gal}\,\left(K/\Bbb Q\right)=\left\{\;\sigma,\,\tau\;/\;\sigma^4=\tau^2=1\;,\;\;\tau\sigma\tau=\sigma^3\;\right\}\cong D_4$$

$\endgroup$
  • $\begingroup$ $\Bbb Q(\sqrt{2+i},\,\sqrt{2-i})\not=\Bbb Q(\sqrt{2+i},\,i)$ since $\sqrt{5}\in \Bbb Q(\sqrt{2+i},\,\sqrt{2-i})$ but it does not belong to $\Bbb Q(\sqrt{2+i},\,i)$. Also a theorem in the Patrick Morandi's book says that all the automorphisms of the Galois group can be determined by the action on a generating set. But $i$ is not in the generating set. Then why are you considering $i$ to define the automorphisms? Also, any automorphism permutes the roots of its minimal polynomial, then why in definition of $\sigma$, $b\to a$ (since $a$ is not a root of the minimal polynomial of $b$)? $\endgroup$ – pie Mar 31 '18 at 16:34
  • $\begingroup$ @pie I need to check the $\;i\;$ thing, but the reason is pretty clear from what I did: to get an easy (yet perhaps erroneous...need to check) involution in the Galois group. Anyway, even without that I thin it can be achieved. I shall check this later. $\endgroup$ – DonAntonio Mar 31 '18 at 16:42
  • $\begingroup$ Sorry, I can't edit my answer now since every time I click on "edit" I get the original post all with gibberish and I can't see clearly what's going on (This bug has hit this site several times in the past years...). It will have to wait. $\endgroup$ – DonAntonio Mar 31 '18 at 16:45
  • 1
    $\begingroup$ @pie The bug disappeared and I edited my answer. No need to get into the game that annoying $\;i\;$ .\ $\endgroup$ – DonAntonio Mar 31 '18 at 16:49
  • 1
    $\begingroup$ Sorry, yes you are right. That's what I was missing. Thank you. $\endgroup$ – pie Mar 31 '18 at 17:05

Not the answer you're looking for? Browse other questions tagged or ask your own question.