I'd rather not have the answer, because I feel like this should be a relatively easy question, and I'm just missing some key step, but could anyone give me a hint on showing that the norm (defined as $N(a)=\det(L_a)$) where $L_a$ is the linear transformation given by multiplication by $a$ is surjective in the case of a finite extension of a finite field?

I've been looking at the fact that $N(x)=x^n$ for any $x$ in the base field, where $n$ is the degree of the field extension. But this doesn't necessarily give me back every element in the base field, for instance $\mathbb{F}_3(\sqrt{2})$, where we have to apply norm to $\sqrt{2}$ to get $2$ back. Is there a basic fact about finite fields maybe that I'm missing, or something more clever regarding a field extension?



2 Answers 2


Two facts:

  • The norm is multiplicative: $N(ab) = N(a)N(b)$.
  • The nonzero elements of a finite field form a cyclic group (under multiplication).

So, if you can figure out what a generator of the multiplicative field of $\mathbb{F}_{p^k}$ maps to...

Added. Your comment below suggests you are a bit confused. So let me set things up a bit.

Given a field extension $F\subseteq K$ of finite degree, the norm from $K$ to $F$, $N_{K/F}\colon K\to F$ is the map that sends $N(a)$ to the determinant of $L_a\colon K\to K$, the linear transformation from $K$ to $K$ given by multiplication by $a$, considering $K$ as a vector space over $F$. The map is multiplicative, and always takes values in $F$. "Surjectivity" here would refer to surjectivity as a map $N_{K/F}\colon K\to F$.

You are considering $F=\mathbb{F}_{p^k}$, and $K=\mathbb{F}_{p^{kn}}$ for some positive integers $k$ and $n$ (remember that $\mathbb{F}_{p^a}$ is an extension of $\mathbb{F}_{p^b}$ if and only if $b|a$).

Since the multiplicative group of $K$ is cyclic, it is generated by some $a$; so the non-zero part of the image of $N$ is generated by $N(a)$, hence you only need to figure out what $N(a)$ is.

Because the powers of $a$ give all nonzero elements of $K$, then $K=\mathbb{F}_{p^k}(a)$; so $\{1,a,a^2,\ldots,a^{n-1}\}$ is a basis for $K$.

It is pretty easy to figure out what the matrix of $L_a$ is with respect to this basis (it will depend on the minimal polynomial of $a$, though). Then you want to argue that the determinant of this matrix is necessarily a generator of the multiplicative group of $\mathbb{F}_{p^k}$.

For example, with $\mathbb{F}_3(\sqrt{2})$, the multiplicative group is generated by $1+\alpha$, where $\alpha=\sqrt{2}$, since: $$\begin{align*} (1+\alpha)^2 &= 1+2\alpha+\alpha^2 = 3+2\alpha = 2\alpha;\\ 2\alpha(1+\alpha) &= 2\alpha+4 = 1+2\alpha;\\ (1+2\alpha)(1+\alpha) &= 2;\\ 2(1+\alpha) &= 2+2\alpha;\\ (2+2\alpha)(1+\alpha) &= \alpha;\\ \alpha(1+\alpha) &= 2+\alpha;\\ (2+\alpha)(1+\alpha) &= 1. \end{align*}$$ Note that $\{1,1+\alpha\}$ is a basis for $\mathbb{F}_3(\sqrt{2})$ over $\mathbb{F}_3$. If we let $a=1+\alpha$, then $L_a$ has matrix, relative to this basis, equal to $$\left(\begin{array}{cc} 0 & 1\\ 1 & 2 \end{array}\right)$$ (since $(1+\alpha)^2 = 2\alpha = 1 + 2(1+\alpha)$). So the determinant of this matrix is $-1 = 2$, hence $N(1+\alpha) = 2$, which happens to be a generator of $\mathbb{F}_3^{\times}$. That means that the image of $N$ consists of $0$ plus the subgroup of $\mathbb{F}_3^{\times}$ generated by $N(1+\alpha)=2$, which is all of $\mathbb{F}_3$.

Note that even though $K=\mathbb{F}_3(\sqrt{2})$, $\sqrt{2}$ does not generate the multiplicative group of nonzero elements of $K$; we needed to take a different element.

  • $\begingroup$ Right, so I was looking at that. Say $x$ is a generator of $\mathbb{F}_{p^k}^\times$, so $\mathbb{F}_{p^k}=\{0,1,x,\ldots x^{p^k-2}\}$, and by applying $N$ to everything we get $\{0,1,x^n,\ldots x^{n(p^k-2)}\}$, which isn't necessarily $\mathbb{F}_{p^k}$, UNLESS, it seems, $x$ maps to another generator. $\endgroup$ Apr 19, 2011 at 2:56
  • $\begingroup$ @JBeardz: Are you trying to show that the norm is surjective from $\mathbb{F}_{p^k}$ to itself? You do know that the norm depends on two fields (to figure out the determinant of $L_a$, you need to specify the ground field of the vector space). If you are considering the norm from $\mathbb{F}_{p^k}$ to itself, then you are looking at a 1-dimensional vector space, and the matrix of $L_a$ relative to the basis $\{1\}$ is just $(a)$, hence the norm of $a$ is $a$. (cont) $\endgroup$ Apr 19, 2011 at 2:59
  • $\begingroup$ @JBeardz: But that is not what you have: the fact that you specify that the norm of an element of $\mathbf{F}_p$ is its $n$th power when the extension is of degree $n$ means you are considering $\mathbb{F}_{p^k}$ as an $\mathbb{F}_p$-vector space. Then the norm always takes values in $\mathbb{F}_p$, not in $\mathbb{F}_{p^k}$. Surjectivity would refer to surjectivity as a map $N\colon\mathbb{F}_{p^k}\to\mathbb{F}_p$, not to $\mathbb{F}_{p^k}$. $\endgroup$ Apr 19, 2011 at 3:01
  • $\begingroup$ No, I'm not. I was hoping I could do it by just applying it to elements of $\mathbb{F}_{p^k}$ which seems to be fruitless, primarily because I had so little information about the elements of the extension. $\endgroup$ Apr 19, 2011 at 3:02
  • $\begingroup$ To clarify, I'm looking at a finite extension of $\mathbb{F}_{p^k}$, call it $K$, and showing that $N:K\to \mathbb{F}_{p^k}$ is surjective. But I think I see what you're saying. Look at a generator of the extension? $\endgroup$ Apr 19, 2011 at 3:03

Remark: I replied too soon and provided an answer for the trace of a field extension instead. I won't delete it because it's somewhat interesting, even though it's useless.

I have included but different calculation for the norm just for completion of my reply. $$ N\colon \mathbb F_{q^\ell}^*\to \mathbb F_{q^\ell}^* \colon x\mapsto xx^q x^{q^2}\cdots x^{q^{\ell-1}} = x^{\tfrac{q^\ell-1}{q-1}} $$ Notice that $N$ is a morphism of cyclic groups and that in fact $\mathrm{im} N \subseteq \mathbb F_q^*$. Now it's well known (or else easy to show) that if $$ \varphi \colon C\to C \colon x\mapsto x^m $$ is a morphism of cyclic groups, then $|\mathrm{im} \varphi| = |C|/(m,|C|)$. In this case $|\mathrm{im} N |= \frac{q^\ell-1}{(q^\ell-1,\frac{q^\ell-1}{q-1})} = q-1$ and indeed $N$ is onto $\mathbb F_q^*$.

Original reply: I'm not sure where exactly you are stuck, so I wouldn't know what to hint, but here is how I usually deal with such a problem.

Let $\mathbb F_{q^\ell}/\mathbb F_q$ be a field extension, with associated trace $$ T\colon \mathbb F_{q^\ell}\to \mathbb F_q \colon x\mapsto x+x^q + x^{q^2} + \dots + x^{q^{\ell-1}} $$ Notice that $T$ is a morphism of additive groups. (And note that indeed $T(x)^q = T(x)$ for all $x$ so it goes to $\mathbb F_q$ indeed).

Now first of all, all elements in $\ker T$ are solutions to an equation of degree $T^{\ell-1}$ therefore $|\ker T|\le q^{\ell-1}$. On the other hand $|\mathrm{im}\, N| = q^\ell/ |\ker N| \leq q$ therefore $|\ker T| \geq q^{\ell-1}$. Therefore $|\ker T|=q^{\ell-1}$ and $|\mathrm{im} T|=q$ and $T$ is a surjection.

-- Edit. I should clarify that the definition of "Trace" I use may seem different from yours; fortunately Arturo Magidin --as always-- provides an detailed explanation on that.

  • $\begingroup$ Sorry, I'm not familiar with that definition of norm, is that equivalent to the definition given in terms of determinant, or in terms of the product of different embeddings of an element raised to the inseparable degree, i.e. $N(x)=\left(\Pi_{\sigma}\sigma(x)\right)^{[K:L]_i}$? $\endgroup$ Apr 19, 2011 at 3:15
  • $\begingroup$ @JBeardz Actually I'm mistaken, it is $N(x) = \prod_{\sigma\in \mathrm{Gal}(K/F)}x^\sigma$. I'm sure Arturo Magidin's editted reply will help you further. (My reply holds for the Trace in fact...) $\endgroup$
    – Myself
    Apr 19, 2011 at 3:17
  • $\begingroup$ I thought that might be the trace. No worries, that's an important result as well! $\endgroup$ Apr 19, 2011 at 3:20
  • $\begingroup$ @JBeardz: Oops indeed. It's not wrong but a bit off topic, I'm afraid. $\endgroup$
    – Myself
    Apr 19, 2011 at 3:23

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