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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.

The Count
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    For (i), I'm not sure you need to worry about the degree of the extension — fields are closed under multiplication. For (ii), how do you know that the minimal polynomial has that form? Do I have to have $\beta^2 \in F$? That seems like a strong statement. For the last part, it might help to view this $(q+1)$-th power operation as a homomorphism $K^* \to F^*$. You want this to be surjective. – Hoot Jul 04 '16 at 01:57
  • I think this is a duplicate of this older question. – Jyrki Lahtonen Jul 04 '16 at 04:56

2 Answers2

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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.

p Groups
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    Thanks very much for this response. I am not quite used to this level of language on the subject yet, but I'm sure it will be an important thing for me to revisit as I learn more. – The Count Jul 05 '16 at 01:07
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    In a word, $x^{q+1}$ is the field-theoretic norm of $x$ from $K$ down to $F$: the product of all ( both ! ) the $K$-conjugates of $x$. – Lubin Jul 14 '16 at 04:14
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Here is the first part:

Let $\beta\in K$. We can assume that $\beta\ne0$ because $0^{1+q}=0\in F$.

If $\beta\ne0$, then, by Lagrange's theorem, we have $(\beta^{q+1})^{q-1}=\beta^{q^2-1}=1$.

Now, the roots of $x^{q-1}-1$ in $K$ are exactly the non-zero elements of $F$.

Therefore, $\beta^{q+1} \in F$.

For the second part:

The map $\beta \mapsto \beta^{q+1}$ is a group homomorphism $K^\times \to F^\times$. (The codomain is right by the first part.)

Its kernel are the roots of $x^{q+1}=1$. Since $K^\times$ is cyclic and $q+1$ divides $q^2-1$, the kernel is a subgroup of order $q+1$. Therefore, the image has order $(q^2-1)/(q+1)=q-1$, which is the order of $F^\times$. Thus, the map is surjective.

lhf
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  • Thanks for this, Ihf. I am reading through it slowly, but I wanted to express my appreciation for the fast response right away! – The Count Jul 04 '16 at 02:24
  • This is great! I need to be sure I understand each statement exactly, but I get it. Thanks so much! One question: to justify the statement "Now, the roots of $x^{q−1}-1$ in $K$ are exactly the non-zero elements of $F$", I can just say "since $K$ is an extension over $F$, and hence $x^{q-1}-1=\prod\limits_{a\in F, a\neq 0}(x-a)$", right? – The Count Jul 04 '16 at 03:52