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Let be $K$ the finite splitting field of $f(x) (\in \Bbb Q[x])$ over the field, $\Bbb Q$(rational number set)

And say $E_H$ is a fixed field of $H\subset \operatorname{Gal}(K/Q) $.

Main Question) Find the fixed field $E_H$

(1) $f(x) = x^4 -2$, $H= \{ \sigma \}$ with $\sigma(\alpha) = -\alpha i $ , $ \alpha(i) = -i $, and $\alpha = 2^{1 \over 4}$

(2) $f(x) = x^8 +1$, $H= \{ \sigma_1, \sigma_7 ,\sigma_9, \sigma_{15 } \}$

with$\sigma_n (\omega) = \omega \to \omega^n $ for $\omega = e^{{2\pi i} \over 16} $ and $gcd(n,16)=1$

P.s.) I've solved the (1) by inefficient way that writing the element form like a method in attached image. So I use this method for solving (2) to find the fixed field for $H$.

But the process really complicated, so I can't find the fixed field.(C.f. the below of this post's image is my attempt)

Are there any simple method for finding the fixed field?

Thanks.

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  • $\begingroup$ see here for a related question $\endgroup$
    – vidyarthi
    Jul 25 '19 at 10:33
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Here’s a simpler way to approach this using Galois theory. I will carry out the computation in sketches for (1) and leave (2) for you to do. I use the notation $K^H$ instead of $E_H$.

  1. Determine the splitting field $K$ for $f$. Here, $K = ℚ(\sqrt[4] 2, \mathrm i)$, which is easy to see as $f$ splits in $K$ and $K$ clearly is generated by the roots of $f$ in $K$.
  2. Calculate the degree $[K : ℚ]$. Here, $[K : ℚ] = 8$, which we can deduce by noting that $ℚ(\sqrt[4] 2)$ is a subfield of $K$ not containing $\mathrm i$.
  3. Determine the structure of the Galois group $G = \operatorname{Gal} (K/ℚ)$. Here, Galois theory tells us that $\lvert G \rvert = [K : ℚ] = 8$, so there are really only two options up to isomorphism: The cyclic group $C_8$ or the dihedral group of the square $D_{2×4}$. If $G \cong C_8$, then, by Galois correspondence, there’d be only a total of four intermediate fields in $K/ℚ$, but $ℚ$, $ℚ(\sqrt 2)$, $ℚ(\sqrt[4] 2)$, $ℚ(i)$ and $K$ are five distinct intermediate fields, so we can conclude $G \cong D_{2×4}$.
  4. Determine the structure of the subgroup $H ⊆ G$. Here, it’s easy to check that $σ^2(α) = α$ and $σ^2(\mathrm i) = \mathrm i$ for $α = \sqrt[4] 2.$ Since $K = ℚ(\sqrt[4] 2, \mathrm i)$, $σ^2 = \mathrm{id}_K$, so $H = ⟨σ⟩$ is cyclic of order $2$.
  5. Determine the field $K^H$. Yet again by Galois theory, we now know $[K : K^H] = \lvert H \rvert = 2$, but $[K : ℚ] = \lvert G \rvert = 8$, so $[K^H : ℚ] = 4$. By trial and error, we quickly find that $$σ(α - α\mathrm i) = σ(α) - σ(α)σ(\mathrm i) = (-α\mathrm i) - (-α\mathrm i)(-\mathrm i) = α - α\mathrm i$$ for $α = \sqrt[4] 2$, so $α - α\mathrm i ∈ K^H$, but $α - α \mathrm i = \sqrt[4] 2(1 - \mathrm i)$ is a zero of $g = X^4 + 8$, which is irreducible over $ℚ$. Hence $[ℚ(α) : ℚ] = 4 = [K^H : ℚ]$ and, from $ℚ(α) ⊆ K$, we conclude $$K^H = ℚ(α - α\mathrm i) = ℚ(\sqrt[4] 2 (1 - \mathrm i)).$$

For (2), you need to know the Galois theory of cyclotomic extensions, that is just the bit that tells you what their Galois groups are. Then it should be easy sailing.

Edit. I made a mistake in the determination of $H$ earlier, which is now corrected. Now, determining the Galois group isn’t needed, so (3.) is superfluous. However, I’ll leave it because it is in general helpful for calculating the structure of $H$, which is in general helpful for calculating $K^H$.

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  • $\begingroup$ Thanks for answering. But the answer sheet said the answer is $Q(\alpha - \alpha i)$ not your answer Q($\alpha^2 $i). $\endgroup$ Jul 25 '19 at 10:53
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    $\begingroup$ @se-hyuckyang Er, yeah. Sorry. $σ^2(α) = α$, I did miscalculate this in my head. $\endgroup$
    – k.stm
    Jul 25 '19 at 12:57
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    $\begingroup$ @se-hyuckyang I’ve updated my answer accordingly now. Sorry for the mistake. Does this answer help you? $\endgroup$
    – k.stm
    Jul 25 '19 at 13:08
  • $\begingroup$ Thanks for your correction! :) But Are there any tricks that finding the fixed element like a $α - α\mathrm i$ in your perfect solution? Hmm.. Actually This was the key point of my question. It takes a long time finding the fixed element in my method.(I've tried to find $x$ by putting fixed element as the $x=\sum_{n=0} ^ {3} (\alpha ^n) + \sum_{n=0} ^ {3} (\alpha ^n) i$ $s.t$. $σ(x)=x$ ) $\endgroup$ Jul 25 '19 at 13:50
  • $\begingroup$ @se-hyuckyang Sorry for replying just now. The best approach in most elementary, exercise-level cases is to just guess it or find it by trial and error. That’s at least how I do it. If you want to proceed more structurally without writing down too much equations, you can try to compute the kernel of $σ - \mathrm{id}_K$ with basic linear algebra, provided that you know at least a $ℚ$-basis of $K$, which you do know as soon as you know a path of primitive extensions from $ℚ$ to $K$. Here, $\big((\sqrt[4] 2)^j \mathrm i^k\big)_{j,k}^{4×2}$ obviously gives a $ℚ$-basis for $K$. $\endgroup$
    – k.stm
    Jul 25 '19 at 21:16

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