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Let G be a group of order $2^n$ and suppose that $G=Gal(K/F)$ where $F \subseteq K$ is a Galois, separable, normal extension.

Then show that there exists an $\alpha \in K$ such that $\alpha^2 \in F$ but $\alpha \notin F$

I have no idea how to start this but I have some thoughts below.

Maybe if I suppose that $K(\alpha) = K$ is true for all $\alpha \in K\setminus F$: but I cant get that $F$ since $\alpha \notin F$, and it can't be larger than $K$ since both $1$ and $\alpha$ are in $K$.

Is it possible to be an intermediate field between $K$ and $F$ either, because there is no such field

$\alpha^2 \in K$, however, doesn't seem to be true for all $\alpha$.

For an arbitrary $\alpha \in F\setminus K$, you only know that $\alpha^2 = u\alpha+v$, so $\alpha^2-u\alpha = v \in K$.

But then $$\left(\alpha-\frac{u}{2}\right)^2 = \alpha^2-u\alpha+\frac{u^2}{4} = v+\frac{u^2}{4} \in K.$$

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    $\begingroup$ The claim is false for $n = 1$ and $F = \mathbb{F}_2$. $\endgroup$ – darij grinberg Apr 17 at 0:45
  • $\begingroup$ Actually @darijgrinberg's observation holds more generally. The claim is false for all $n$ when $\operatorname{char} F=2$. If $\alpha^2\in F,\alpha\notin F$, then $F(\alpha)/F$ is inseparable, and hence $K/F$ cannot be Galois. $\endgroup$ – Jyrki Lahtonen Apr 18 at 18:48
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First, you are mixing up $K$ and $F$. In the problem, $K$ is a field extension of $F$.

Next, the problem does not state that every $\alpha\in K\backslash F$ satisfies $\alpha^2\in F$, only that there exists one. Well, you've constructed it: $\alpha-u/2\notin F$, but as you've calculated, it squares to $u^2/4+v\in F$.

Finally, the problem is more general than when $[K:F]=2$. The hypothesis is that the Galois group $G$ has order $2^n$, which is equivalent to $[K:F]=2^n$. The group $G$ is a nilpotent $2$-group and has a subgroup $N$ of order $2$ with $N$ the Galois group of an intermediate extension $L/F$ with $[L:F]=2$. Now proceed as in your post.

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    $\begingroup$ As pointed out in the comments, you need to assume that the characteristic of the field is not $2$. That way you can divide by $2$. $\endgroup$ – David Hill Apr 17 at 1:27
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    $\begingroup$ BTW, the reason the statement is false for $F$ of characteristic $2$ is that the multiplicative group $K^\times$ is cyclic of odd order. An element $\alpha$ as in the problem will have even order, which is impossible. $\endgroup$ – David Hill Apr 17 at 1:47

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