# If $K/F$ is a quadratic field extension, any generator of $F^*$ is a $|F|+1$ power of some generator of $K^*$? (From Ireland/Rosen)

I was pondering this question yesterday, but couldn't complete it.

Suppose $F\subset K$ are finite fields, with $[K:F]=2$, and $|F|=q$. If $\alpha\in F$ has order $q-1$, then why does there exist some $\beta\in K$ with order $q^2-1$ such that $\beta^{q+1}=\alpha$?

I know $K^*$ is cyclic with order $q^2-1$, I let $\gamma$ be a generator. Then $F^*$ is the unique cyclic subgroup of $K^*$ or order $q-1$, so $F^*=\langle\gamma^{(q^2-1)/(q-1)}\rangle=\langle\gamma^{q+1}\rangle$. I let $\delta=\gamma^{q+1}$. So if $\alpha\in F$ has order $q-1$, it must be a generator of $F^*$, and thus has form $\alpha=\delta^a$, where $(a,q-1)=1$. If $\beta$ has order $q^2-1$, then it is a generator for $K^*$, and thus has form $\gamma^b$ where $(b,q^2-1)=1$.

Then $\beta^{q+1}=\alpha$ is equivalent to $\gamma^{(q+1)b}=\gamma^{(q+1)a}$, or $\delta^b=\delta^a$. This is asking for $b$ and $a$ such that $b\equiv a\pmod{q-1}$, where $(a,q-1)=1$ and $(b,q^2-1)=1$. But I can't tell if given any $a$ such that $(a,q-1)=1$, if there exists $b$ satisfying the above properties. Thanks.

Source: Chapter 7, #11.

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One can also use the Chinese remainder theorem. Clearly $2$ is the only nontrivial common divisor for $q-1$ and $q+1$. Decompose $q-1$ and $q+1$ into prime factors :

$$q-1=2^j u_1^{e_1} u_2^{e_2} u_3^{e_3} \ldots u_r^{e_r},$$

$$q+1=2^{j'} v_1^{f_1} v_2^{f_2} v_3^{f_3} \ldots v_s^{f_s}$$

then there is a $b$ with

$$b \equiv a \ ({\sf mod} \ 2^{{\sf max}(j,j’)}), b \equiv a \ ({\sf mod} \ u_i^{e_i}), \ b \equiv 1 \ ({\sf mod} \ v_i^{f_i})$$

This $b$ will do.

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Thanks for completing my original approach! At least it was salvageable. –  Noomi Holloway Jan 25 '13 at 7:19

We know that $K^\ast$ is generated by some $\gamma$ where $\gamma$ has order $q^2 - 1$. Now if we consider $\alpha \in F$ as an element of the bigger field $K$, then $\alpha$ must be equal to some power of $\gamma$, say $\alpha = \gamma^k$ for some $k$ in the range $1 \leq k \leq q^2 - 2$. Now because $\alpha$ has order $q - 1$ this means that

$$\gamma^{k(q-1)}= 1$$

and hence $q^2 - 1| k(q-1)$. Thus $q+1|k$ which shows that there is $\beta$ such that $\beta^{q+1} = \alpha$.

Note: This does not show that $\beta$ has order $q^2 - 1$.

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Ah, thank you BenjaLim. That is much more straightforward. –  Noomi Holloway Jan 25 '13 at 7:17
@NoomiHolloway With these sorts of questions it boils down to looking it the right way, once you've done that it is not so bad. –  user38268 Jan 25 '13 at 7:18
Come to think of it, this just proves $\alpha=\beta^{q+1}$, but I don't think it proves that $\beta$ itself has order $q^2-1$? –  Noomi Holloway Jan 26 '13 at 23:29
@NoomiHolloway You are right, at the moment I have only been able to prove that the order of $\beta$ has to be more than $q + 1$. Perhaps you should accept Ewan's answer then. –  user38268 Jan 27 '13 at 4:32
You can add one more step to make it work. Let $k = r(q+1)$. Show that 1. $gcd (r,q-1) = 1$. 2. By adding some multiples of $q-1$ to $r$ if necessary, we can assume that $gcd(r,q+1) = 1$ too, i.e. $gcd(r,q^2-1) = 1$. –  user27126 Jan 27 '13 at 8:33