What is $\{u+u^p: u\in F\}$ in the finite field $F_{p^n}$?

Question

Let $F$ be a finite field of characteristic $p$ (including the case $p=2$). I know that $F^p:=\{u^p: u\in F\}=F$, because all finite fields are perfect. But what holds for the sets $\{u+u^p: u\in F \}$, $\{u+u^p+u^{p^2}: u\in F\}$ and so on, in general? I can only see that these are subgroups of $(F,+)$.

I would be very surprised if it were the case that $\{u+u^p+u^{p^2}+\ldots +u^{p^k}: u\in F \}=F$, $k\in \mathbb{N}$, but I don't see how to so that this is not the case in general.

Answer 1

Linear algebra to the rescue!

Let $K=\Bbb{F}_{p^n}$ be the finite field of cardinality $p^n$, $p=\mathrm{char} K$, a prime number. The first and most obvious observation is that $$T:K\to K, u\mapsto u+u^p$$ is a linear transformation over the prime field $\Bbb{F}_p$. This is because raising to $p$th power is additive (by Freshman's dream), and maps the elements of the prime field to themselves (by Little Fermat). As $T$ is a polynomial of degree $p$, it has at most $p$ zeros in the field $K$. We can thus conclude that $\mathrm{Ker}(T)$ has dimension zero or one. Hence, by rank-nullity, $\mathrm{Im}(T)$ is either all of $K$ or a subspace of codimension one. The next task is to determine, which case occurs when.

If $u\in K^*$ is in the kernel, then $u^p=-u$. In other words, the Frobenius automorphism $F:K\to K, x\mapsto x^p$, has $-1$ as an eigenvalue. Because $F^n$ is the identity on all of $K$, we arrive at the parity condition $$u=F^n(u)=(-1)^nu$$ implying that $(-1)^n=1$.

This means that for $T$ to have a non-trivial kernel, we must either have $1=-1$ (i.e. $p=2$) or $2\mid n$. It is easy to see that this necessary condition is also sufficient:

• If $p=2$, then $1_K\in\mathrm{Ker}(T)$.
• If $p>2$ and $2\mid n$, then the mapping $\Phi:x\mapsto x^{p-1}$ is a homomorphism from the cyclic group $\Bbb{F}_{p^2}^*\subseteq K^*$ of order $p^2-1$ to itself. Therefore the image $S$ of $\Phi$ consists of the roots of unity of order $(p^2-1)/(p-1)=p+1$. As $p+1$ is an even number, it follows that $-1\in S$. Therefore there exists an element $z\in\Bbb{F}_{p^2}$ such that $z^{p-1}=-1$. Consequently $z^p=-z$, and $z$ is a non-zero element in $\mathrm{Ker}(T)$. We can reach the same conclusion by using the fact that the minimal polynomial of the Frobenius automorphism of $K$ is $x^n-1$, hence the possible eigenvalue $\lambda=-1$ occurs whenever $2\mid n$.

Conclusion. If $p>2$ and $2\nmid n$, then $\mathrm{Im}(T)$ is all of $K$, and otherwise it is a codimension one subspace (over the prime field).

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Something more can be said about the image of $T$ in those cases where $T$ is not onto.

• When $p=2$ it follows easily that $\mathrm{Im}(T)$ is exactly the kernel of the absolute trace $tr:K\to\Bbb{F}_2$, $$tr(x)=\sum_{j=0}^{n-1}F^j(x)=\sum_{j=0}^{n-1}=x^{2^j}.$$ This is because obviously $tr(x)=tr(x^2)$ for all $x\in\Bbb{F}_{2^n}$
• When $p>2$, $2\mid n$, we can similarly describe the image of $T$ as the kernel of a linear combination of the powers of Frobenius, this time with alternating signs. See Daniel Schepler's comment under main.

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More generally, studying the image of $$ T_k:K\to K, u\mapsto u+u^p+u^{p^2}+\cdots+u^{p^k} $$ instead of $T=T_1$, we can do the following. First we think of $T_k$ as an $\Bbb{F}_p$-linear mapping from an algebraic closure $\overline{K}$ to itself. As an endotransformation of $\overline{K}$ we see that the kernel of $T_k$ consists of the trace zero elements of the finite field $L=\Bbb{F}_{p^{k+1}}\subset \overline{K}$. Therefore the kernel of the restriction of $T_k$ to the field $K$ must lie in the intersection $L\cap K$.

• If $L\cap K$ is a proper extension field of the prime field, then the kernel of $T_k$ is automatically non-trivial. This is because the kernel of the extended version of $T_k$ is a codimension one subspace of $L$. Observe that $L\cap K$ is non-trivial if and only if $\gcd(k+1,n)>1$.
• If $L\cap K=\Bbb{F}_p$ is the prime field, that is, if $\gcd(k+1,n)=1$, then the answer depends on whether the elements of the prime field belong to the kernel of the trace map of $L$. Because the Frobenius fixes the elements of the prime field, this happens if and only if $p\mid k+1$.

General conclusion. The mapping $T_k:K\to K$ is onto, unless either $p\mid k+1$ or $\gcd(k+1,n)>1$.

Observe that the earlier conclusion is just the special case $k=1$.

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With more general sums of powers of Frobenius the situation becomes more complicated. A general tool is to think of the algebraic closure $\overline{K}$ as an $\Bbb{F}_p[\tau]$-module (a simple polynomial ring, so among other things a Euclidean domain). We achieve this by letting the indeterminate $\tau$ act via Frobenius. So if $a(\tau)=\sum_i a_i\tau^i\in\Bbb{F}_p[\tau]$ and $x\in\overline{K}$ are arbitrary, we define $$ a(\tau)\cdot x=\sum_i a_ix^{p^i}. $$ The analogous question is then to fix a polynomial $a(\tau)$ and study the action of $a(\tau)$ on various finite subfields of $\overline{K}$. Above we got the transformation $T_k$ with the choice $a(\tau)=1+\tau+\cdots+\tau^k$. Anyway, if $a(\tau)$ has a non-zero constant term, then there exists an integer $m$ such that $a(\tau)\mid \tau^m-1$. It follows that all $a(\tau)$-torsion then lies in the field $E=\Bbb{F}_{p^m}$, and we can make deductions in certain cases as above. Of course, in the general case it is more difficult to identify the exact $a(\tau)$-torsion elements inside $E$.