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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 rational root theorem, $p$ is irreducible over $\mathbb{Q}$. Thus $[\mathbb{Q}(a):\mathbb{Q}]=4$. Also $b\not\in\mathbb{Q}(a)$, 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 $$a\to \begin{cases}a\\-a\\b\\-b\end{cases}\quad\text{and}\quad b \to \begin{cases}b\\-b\end{cases}.$$ From this how to conclude $Gal(K/\mathbb{Q})$ is isomorphic to $D_4$?

Thanks

  • @ChrisCuster Why? The automorphisms are not the same –  Mar 30 '18 at 19:57
  • How would Eisenstein apply to show your polynomial is irreducible? – sharding4 Mar 30 '18 at 19:59
  • @sharding4 Sorry by rational root theorem it can be concluded. –  Mar 30 '18 at 20:02
  • @sharding4 sorry. I undid it... –  Mar 30 '18 at 20:09
  • Yikes. Actually you were better off with Eisenstein. It does apply to $p(x-1)$ with the prime $2$. – sharding4 Mar 30 '18 at 20:13
  • Also how do you know $b \not \in \Bbb{Q}[a]$ There's something to be shown there I think. If you want to continue down the path your on, you need to show that there is an autmorphism, say $\sigma$, of order 4 and an automorphism, say $\tau$ of order 2 with $\tau \sigma \tau = \sigma ^{-1}$ Otherwise you could just appeal to a big theorem once you know the degree of the splitting field over $\Bbb{Q}$ is 8 and the minimum polynomial has degree 4. $D_4$ is the only group possible. – sharding4 Mar 30 '18 at 20:42
  • @sharding4 $b \not \in \mathbb{Q(a)}$, otherwise $|Gal(K/\mathbb{Q})|=4$, so it will not be $D_4$. Also then among the 8 automorphisms none of them is of order 4. Then how can the Galois group be $D_4$? Is the question wrong or my logic is wrong? –  Mar 30 '18 at 20:59
  • You're trying to prove that $Gal(K/\Bbb{Q})\simeq D_4$, so you can't assume a contradiction if the order of the galois group were 4. If your Galois group is $D_4$, it necessarily contains an automorphism of order 4. You must have a problem with your automorphisms. – sharding4 Mar 30 '18 at 22:28
  • @amWhy Can you please explain to me why this is a duplicate? –  Mar 31 '18 at 05:09
  • @DietrichBurde Can you please explain to me why this is a duplicate? –  Mar 31 '18 at 05:11
  • @EthanBolker Can you please explain to me why this is a duplicate? –  Mar 31 '18 at 05:11
  • @XanderHenderson Can you please explain to me why this is a duplicate? –  Mar 31 '18 at 05:11

2 Answers2

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This follows from Corollary $4.5$ in K. Conrad's notes. The minimal polynomial is $x^4-4x^2+5$, so its Galois group is $D_4$ of order $8$.

Dietrich Burde
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Find the minimal polynomial of $a$. \begin{align*} a &= \sqrt{2+i} \\ a^2 &= 2+i \\ a^2- 2 &= i \\ a^4 - 4a^2 +4&= -1 \\ \end{align*} So $a$ satisfies the polynomial equation $f(x)=x^4-4x^2+5$. You should check that this polynomial is irreducible, say, by applying Eisenstein's criterion to $(x-1)^4-4(x-1)^2 +5 = x^4 - 4 x^3 + 2 x^2 + 4 x + 2 $.

By inspection $b=\sqrt{2-i}$ is another root of $f$. $b$ does not belong to $\Bbb{Q}[\sqrt{2+i}]$, because if it did, $\sqrt{2+i}\cdot\sqrt{2-i}=\sqrt{5}$ would belong to $\Bbb{Q}[i]$ as $a^2$ and $b^2$ do which it does not. So if we call $K$ the splitting field of $f$ we have $[K:\Bbb{Q}]=8$ since it's generated by $a=\sqrt{2+i}$ and $\sqrt{5}$.

For automorphisms of $K/\Bbb{Q}$ we have the obvious choice of complex conjugation, say $\tau: i \rightarrow -i$. If we define $\sigma$ by \begin{align*} \sigma &: \sqrt{2+i} \rightarrow -\sqrt{2-i} \\ \sigma &: -\sqrt{2+i} \rightarrow \sqrt{2-i} \\ \sigma &: \sqrt{2-i} \rightarrow \sqrt{2+i} \\ \sigma &: -\sqrt{2-i} \rightarrow -\sqrt{2+i} \\ \end{align*}

$\sigma$ has order 4 and $\tau \sigma \tau = \sigma ^{-1}$, so the group generated by $\sigma$ and $\tau$ is isomorphic to $D_4$

sharding4
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  • First of all your automorphisms are not stated correctly. Also why $\sigma$ is of order 4? Also $K=\mathbb{Q}(a,b)$, from where do you get $i$ as a generating element of $K$? –  Mar 31 '18 at 05:06
  • @readon You would need to apply $\sigma$ four times for it to act like the identity on all elements, so it has order four. Start with $\sigma^2:\sqrt{2+i}\rightarrow -\sqrt{2+i}$.
    I don't think I ever said that $i$ generated the splitting field, but only that there is an automorphism of the splitting $i\rightarrow -i$     – sharding4 Mar 31 '18 at 14:15
  • But the automorphisms of the Galois group is given by : any generating element of the extension field to any of the roots of its minimal polynomial. –  Mar 31 '18 at 14:34
  • @readon If you don't like defining $\tau$ that way take it to be $\tau: \sqrt{2+i} \rightarrow \sqrt{2-i}$ – sharding4 Mar 31 '18 at 14:40
  • Can you please explicitly tell me what are the 8 automorphisms? I am really confused. –  Mar 31 '18 at 14:50
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    @readon Take $\sigma$ as above. Take $\tau$ as $\tau: \sqrt{2+i} \rightarrow \sqrt{2-i}$ and $\tau: \sqrt{2-i} \rightarrow \sqrt{2+i}$. ($\tau$ has order 2, and with the roots are described explicitly as complex numbers, $\tau$ would be complex conjugation). The automorphisms are then what you should expect them to be for $D_4$ $1, \sigma, \sigma^2, \sigma^3, \tau, \sigma \tau , \sigma^2 \tau , \sigma^3 \tau$. Their action is determined by composition. What are you confused about? – sharding4 Mar 31 '18 at 16:09