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Let irreducible polynomial $f(x)\in \mathbb{Q}[x]$ with $deg f =8$ and $f(\alpha) =0$.

$K$$(=SF(f/\mathbb{Q}))$ is the splitting field $f$ over $\mathbb{Q}$. Galois group $G(K/\mathbb{Q})$ is the Abelian group.

$H = \{\sigma_1, \sigma_2, \sigma_3, \sigma_4\}$ is the set which collects all the $\sigma_i(\alpha)= \alpha$ in $G(K/\mathbb{Q})$.

$\mathbb{Q} \leq F \leq K $ for the $F=\{x \in K \vert \sigma_1(x) = \sigma_2(x) = \sigma_3(x) = \sigma_4(x)\}$

Find the $[K : Q]$.

I got the answer $8$ because $K = \mathbb{Q}(\alpha)$ is the normal extension considering $G(K/\mathbb{Q})$ is Abelian. But the instructor claimed $32$ in his note. Here is the reason why he claimed like that.

$H=G(K/\mathbb{Q}(\alpha))$ and $F= K_{H}$(fixed field) $= \mathbb Q({\alpha})$ by the given condition.

$[K;\mathbb Q] = [K ; \mathbb Q({\alpha})] [\mathbb Q({\alpha}) ; \mathbb Q] = 4\cdot deg f = 32$

So my question is which one is right between him and I? Plus Taking the incorrect solution, Which point should be edited?

kechang lee
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1 Answers1

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You are right in the following sense. The problem is that the kind of an extension described in the exercise cannot exist – the requirements are contradictory. Therefore I find fault with the instructor's argument (assuming they composed the problem themself).

Recapping key steps in your argument:

  • It was assumed that $K/\Bbb{Q}$ is abelian. As you observed, this implies that $\Bbb{Q}(\alpha)/\Bbb{Q}$ is a normal extension because $Gal(K/\Bbb{Q}(\alpha)$ is automatically a normal subgroup of $Gal(K/\Bbb{Q})$.
  • As the irreducible polynomial $f(x)$ has a zero in the normal extension $\Bbb{Q}(\alpha)$, it follows that $f(x)$ splits completely over $\Bbb{Q}(\alpha)$. Therefore $\Bbb{Q}(\alpha)$ is the splitting field. In other words, $K=\Bbb{Q}(\alpha)$. Hence $[K:\Bbb{Q}]=8$.

Consequently $H=Gal(K/\Bbb{Q}(\alpha))$ is trivial, and cannot have those four elements.

For another approach consider the following. The Galois group $G=Gal(K/\Bbb{Q}(\alpha)$ can be thought of as a transitive subgroup of $S_8$, acting on the set of eight roots of $f(x)$ in $K$. Transitivity is a consequence of irreducibility of $f(x)$. But this implies that $|G|=8$. The argument has been given on our site many times, for example here. Reproducing it only to make this answer self-contained.

  • By the orbit-stabilizer theorem the stabilizer $H=\mathrm{Stab}_G(\alpha)$ has index eight in $G$.
  • If $\beta$ is some other root of $f(x)$, the stabilizer of $\beta$ is a conjugate of $H$ (a standard property of transitive group actions).
  • But $G$ is abelian, so $H$ is the only conjugate of itself. Therefore $\mathrm{Stab}_G(\beta)=H$ for every other root.
  • But the identity automorphism is the only one fixing all the roots of $f(x)$, so $H=\{1_G\}$.

An easy way of fixing this is to drop the assumption that $G$ should be abelian. In that case we can no longer conclude that $K=\Bbb{Q}(\alpha)$. The instructor's argument would then work (and you would have thought of something else).

I would expect the instructor to also recognize the problem, once you explain your argument. Possibly you can ask them to give an example of such a polynomial $f(x)$ :-).

If only I never fumbled like this when designing exercises! Happens easily, when you are in a hurry, and don't have the time to produce examples meeting all the criteria. The worst part is if you forget to make a note to yourself immediately, and reuse the exercise a year later. Been there :-)

Jyrki Lahtonen
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