TITLE: Properties of a finite field extension of degree 2.
QUESTION [7 upvotes]: I am bad (but trying to improve!) at very basic number theory and algebra. I'm quite sure this question is easy, but I do not know what fundamentals I am missing. This is from Ireland & Rosen's "Classical Introduction to Modern Number Theory" and is question 10 of Chapter 7. I have copied it exactly.
Let $K\supset F$ be finite fields and $[K:F]=2$. For $\beta\in K$ show that $\beta^{1+q}\in F$ and moreover that every element in $F$ is of the form $\beta^{1+q}$ for some $\beta\in K$.
The question uses the context of the previous one, in which $|F|=q$.
What I've tried so far:
For the first part, it seems there are two cases: i) $\beta\in F$ or ii) $\beta\in K\backslash F$. In i) it is easy, since $\beta^q=\beta$ and the extension is of degree $2$, $\beta^{q+1}=\beta^2\in F$. In ii), I suppose the minimal polynomial of $\beta$ in $K[x]$ is $x^2-\beta^2=x^2-\beta^{q+1}$... but where to go from here? 
And I can't even begin to see what to do with the second part of the problem. Is any of this right so far? What do I do next if so? Thanks so much, ya'll. I'll go try to answer a question I can help with in the meantime.

REPLY [6 votes]: We have $x^q=x$ for all $x\in F$ and $x^{q^2}=x$ for all $x\in K$. Since $q$ is a power of $p$ and $p$ is characteristic of $K$, hence $\sigma\colon y\mapsto y^q$ is an automorphism of the field $K$. Note that $\sigma$ fixes all the points of $F$, and it can not fix any point of $K\setminus F$ (otherwise, the polynomial $X^q-X$ will have more than $q$ roots). 
Now, $\sigma(\beta^{q+1})=\beta^{q(q+1)}=\beta^{q^2}\beta^{q}=\beta\beta^q=\beta^{q+1}$, thus $\beta^{q+1}$ is a fixed point of $\sigma$, it must be inside $F$, thats what you expected.