Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Suppose $K,L$ are finite fields with $|K|=p^n$ and the $L$ is a quadratic extension over $K$, i.e. $|L| = p^{2n}$. I am trying to show that for any element in the extension $\alpha\in L$ that $\alpha^{p^n+1} \in K$ and moreover that every element in $K$ is of the can be represented as $\alpha^{p^n+1}$ for $\alpha\in L$ . Further, I want to show that if an element in $\beta \in K$ is a generator for $K^*$, i.e. has order $p^n-1$, then there is a generator $\alpha\in L$, i.e. of order $(p^n)^2-1$, such that $\alpha^{p^n+1}=\beta$.

I tested this out with concrete examples with $p=2,3$ and $n=1,2$ but I couldn't really gain much insight from that. I am not sure if this needs to be broken into cases or something. I know that from the extension being quadratic every element $\alpha \in L$ satisfies some irreducible quadratic polynomial in $K$. That is $\alpha^2+b\alpha+c=0$ for some $b,c \in K$. I'm not sure if I need to specify whether $p$ must be odd, but the book does not seem to do so. More stuff I tried with the odd assumption was $\alpha^{p^n+1}=(\alpha^2)^{(p^n+1)/2}= (b\alpha+c)^{(p^n+1)/2}$ for some $b,c \in K$. This does not seem to lead anywhere useful though.

Also, even if I was assume the first part and moreover part, I am still confused by the "further part" . By the moreover part $\beta$ has some representation as $\alpha^{p^n+1}$. We know that $\beta^{p^n-1}=1$ thus $(\alpha^{p^n+1})^{p^n-1}=\alpha^{p^{2n}-1}=1$. Suppose though the order of $\alpha$ was some smaller number $d|p^{2n}-1$, i.e. $\alpha^d=1$. Then $\beta^d=(\alpha^{p^n+1})^d=(\alpha^d)^{p^n+1}=1$. But since $\beta$ is a generator of $K$ the $d$ must be a multiple of $p^n-1$ which does divide $p^{2n}-1$, with remainder $p^n+1$. So it seems that $\alpha$ is allowed to have order $p^n-1$. This would mean though that $\beta=\alpha^{p^n+1}=\alpha^{p^n-1}\alpha^2=\alpha^2$. I am not sure where to go from there though, not even sure if this is the right direction.

share|cite|improve this question
up vote 2 down vote accepted

Let $q = p^n$.

The roots of the polynomial $x^q - x$ are precisely the elements of $\mathbb{F}_q$. If $\alpha \in \mathbb{F}_{q^2}$, then $\alpha^{q^2} - \alpha = 0$, and

$$ \left(\alpha^{q+1}\right)^q = \alpha^{q^2 + q} = \alpha^{q^2} \cdot \alpha^q = \alpha \cdot \alpha^q = \alpha^{q+1} $$

Therefore $\alpha^{q+1} \in \mathbb{F}_q$.

In general, given a finite extension $L/K$ of fields, there is the concept of the norm (over $K$) of an element of $L$, which is the product of all of its conjugates, and is an element of $K$. If $K = \mathbb{F}_q$, the conjugates of any element are obtained by repeatedly raising it to the $q$-th power.

In this case, we have a quadratic extension, so

$$N(\alpha) = \alpha^q \cdot \alpha = \alpha^{q+1}$$

which gives another proof that $\alpha^{q+1} \in \mathbb{F}_q$.

To see that every element of $\mathbb{F}_q$ is a norm -- that is, a $(q+1)$-th power of an element in $\mathbb{F}_{q^2}$ -- we can appeal to polynomial equations again. Every $\alpha \in \mathbb{F}_{q^2}$ is a root of some equation

$$ f_\beta(x) = x^{q+1} - \beta = 0 $$

and each such equation has at most $q+1$ roots. $f_0(x)$ has a single root $0$ with multiplicity $q+1$. The remaining $q^2 - 1 = (q-1)(q+1)$ elements of $\mathbb{F}_{q^2}$ are distributed among the polynomials corresponding to remaining $q-1$ values for $\beta$, with each $f_\beta$ having at most $q+1$ of them. A counting argument thus shows that every $f_\beta(x)$ with $\beta \neq 0$ must have exactly $q+1$ distinct roots in $\mathbb{F}_{q^2}$.

We could (as the other answers have shown) argue by using the fact the unit group of $\mathbb{F}_{q^2}$ is cyclic.

Having the idea of the norm, it's easy to generalize to arbitrary finite extensions of $\mathbb{F}_q$: the norm of an element of $\mathbb{F}_{q^n}$ is

$$ N(\alpha) = \alpha \cdot \alpha^q \cdot \ldots \cdot \alpha^{q^{n-1}} = \alpha^{(q^n-1) / (q-1)}$$

Of course, this again could have been discovered by using the fact the unit group is cyclic.

(For extensions $L/K$ of arbitrary fields, it's not always true that every element of $K$ is a norm)

share|cite|improve this answer
I'm still having trouble showing this last part that if we given a generator $\beta\in K$ then we can find a generator $\alpha\in L$ such that $\beta=\alpha^{q+1}$ – Steven-Owen Dec 1 '12 at 17:17


The multiplicative group of a finite field is always cyclic. Thus $L^\times$ is cyclic of order $p^{2n}-1=(p^{n}+1)(p^n-1)$ and $K^\times$ is reobtained back as the only cyclic subgroup of order $p^n-1$.

By writing $L^\times=\langle\gamma\rangle$ and writing every element as a power of $\gamma$ all the answers to your questions should be straightforward.

share|cite|improve this answer

Work in progress

The litmus test for deciding whether an element $\beta$ in an extension field $L = \mathbb F_{q^m}$ belongs to the base field $K = \mathbb F_q$ is to check whether $\beta^q = \beta$ or not. If $\beta^q = \beta$, then $\beta \in K$; else $\beta \notin K$. In your case, $m = 2$. Given any $\alpha \in L$, write $\alpha^{q+1} = \beta$, and note that $$\beta^q = \left(\alpha^{q+1}\right)^q = \alpha^{q^2+q} = \alpha^{q^2}\cdot\alpha^q = \alpha\cdot\alpha^q = \beta$$ where we have used the litmus test for membership in $\mathbb F_{q^2}$: $\alpha^{q^2} = \alpha$ for all $\alpha \in \mathbb F_{q^2}$. Thus $\alpha^{q+1} \in \mathbb F_q$ for all $\alpha \in \mathbb F_{q^2}$.

Turning to the matter of the orders of the elements, let $\alpha \in L$ denote a generator of $L^*$ so that $\tt{ord}(\alpha) = q^2-1$. then $$\tt{ord}(\alpha^{q+1}) = \frac{\tt{ord}(\alpha)}{\gcd(\tt{ord}(\alpha),q+1)} = \frac{q^2-1}{\gcd(q^2-1,q+1)} = q-1$$ and so $\beta=\alpha^{q+1}$ is a generator of $K^*$ whenever $\alpha$ is a generator of $L^*$. Now, any generator $\gamma$ of $K^*$ can be expressed as $\beta^k$ where $\gcd(k,q-1) = 1$. Then, $$\left(\alpha^k\right)^{q+1} = \alpha^{k(q+1)} = \left(\alpha^{q+1}\right)^k = \beta^k = \gamma.$$ Furthermore, we can use the fact that $\tt{ord}(\alpha) = q^2-1$ to deduce that for any integer $i$, $$\left(\alpha^{k+i(q-1)}\right)^{q+1} = \alpha^{k(q+1)+i(q^2-1)} = \alpha^{k(q+1)}\cdot\alpha^{i(q^2-1)} = \alpha^{k(q+1)} = \beta^k = \gamma.$$ Thus, the $q+1$ elements $\alpha^k, \alpha^{k+(q-1)}, \alpha^{k+2(q-1)}, \ldots, \alpha^{k+q(q-1)}$ are the $q+1$ roots of $x^{q+1} - \gamma \in \mathbb F_q[x]$. At least one of these elements must be a generator of $L^*$. Work in progress

share|cite|improve this answer
This last part shows the converse of what I was trying to show, which is much easier. I want that if we given a generator $\beta\in K$ then we can find a generator $\alpha\in L$ such that $\beta=\alpha^{q+1}$ – Steven-Owen Dec 1 '12 at 17:16

I keep your notations K, L, q etc. We have a cyclic extension L/K, and from the theory of finite fields, Gal(L/K) is generated by the Frobenius automorphism defined by Fr(x) = $x^q$ for any x in L. The norm map N from L* to K* is defined by taking the product of conjugates, i.e. N(x) = x. Fr(x) = $x^(q+1)$. Since $(Fr)^2$ = Id, it is obvious that Fr(N(x)) = N(x), hence N(x) belongs to K, as desired. Let us determine the image of the norm. Since N is the same as raising to the (q+1)-power in the cyclic group L*, Ker N has order (q+1), hence Im N has order $(q^2 - 1)/q + 1$ = q - 1 = order of K* , i.e. the norm is surjective, as desired.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.