# Show that there exists an $\alpha \in K$ such that $\alpha^2 \in F$ but $\alpha \notin F$

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

• The claim is false for $n = 1$ and $F = \mathbb{F}_2$. – darij grinberg Apr 17 at 0:45
• 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. – Jyrki Lahtonen Apr 18 at 18:48

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.
• 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 $2$. – David Hill Apr 17 at 1:27
• 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. – David Hill Apr 17 at 1:47