# Invariant subspaces of Frobenius automorphism

Given finite field, is it true that any invariant subspace of its Frobenius automorphism is a subfield?

Firstly, I know that $F_{p^k}$ is a subfield of $F_{p^n}$ if and only if $k$ divides $n$.

For the second, it seems that we shall use that if finite field is of characteristic $p$ then any element satisfies $x^p=x$.

However, I can't proceed to the final answer from this remarks. Can anyone explain me is it truth or not and why?

• It is not true that every element of a field of characteristic $p$ satisfies $x^p = x$: only the elements of the minimal subfield do. – Vik78 Mar 27 '17 at 16:54
• Characteristic pp means that $p=0$ i.e. $\underbrace{a+\ldots+a}_p = pa = 0$. So that $(a+b)^p =\sum_{n=0}^p {p \choose n} a^n b^{p-n} =a^p +b^p$ and $(a+b)^{p^k} = \sum_{n=0}^p {p^k \choose n} a^n b^{p^k-n} = a^{p^k} +b^{p^k}$ – reuns Mar 27 '17 at 21:19

$\newcommand{\F}{\mathbb{F}}$I understand the question to be

Suppose the $\F_{p}$-subspace $V$ of the finite field $\F_{p^{n}}$ is invariant under the Frobenius automorphism. Is $V$ a subfield of $\F_{p^{n}}$?

The answer is negative.

Let $E = \F_{p^{6}}$.

Then $U = \F_{p^{2}}$, $W = \F_{p^{3}}$ are subfields of $E$, and thus $\F_{p}$-subspaces of $E$.

Then $V = U + W$ is an $\F_{p}$-subspace of $E$, which has dimension $2 + 3 - 1 = 4$ over $\F_{p}$, and thus order $p^{4}$, so that it is not a subfield of $E$.

However $V$ is invariant under the Frobenius automorphism, as $U$ and $W$ are.

Andreas Caranti's answer gives a nice example. Another one that comes to mind right away is the kernel of the trace map. The trace is the function $$tr:\Bbb{F}_{p^n}\to\Bbb{F}_p, tr(x)=x+x^p+x^{p^2}+\cdots+x^{p^{n-1}}.$$ We have $tr(x+y)=tr(x)+tr(y)$, so the kernel of the trace is a subspace.

Because we also have $tr(x^p)=tr(x)$, the kernel is stable under Frobenius. Hence $\operatorname{Ker}(tr)$ is an $(n-1)$-dimensional subspace of $\Bbb{F}_{p^n}$ stable under Frobenius. [Edit: Thanks, Andreas!] This is not a subfield when $n>2$ because then $n-1\nmid n$. [/Edit]

You can also use images of linear combinations of (iterated) powers of the Frobenius to generate more examples. There will be overlap in images and kernels though. Possibly the best known example of such is the additive version of Hilbert's Satz 90: $$\operatorname{Ker}(tr)=\operatorname{Im}(\phi),$$ where $\phi=F-id$, i.e. $\phi(x)=x^p-x$ for all $x\in\Bbb{F}_{p^n}$.

• Oh, +1, sure, that's more natural than mine, and allows for a smaller example. (Nitpick: take $n > 2$.) – Andreas Caranti Mar 29 '17 at 8:34

It seems that I found a proof. Denote our finite field of characteristic $p$ by $F_{p^n}$. Then set of all elements invariant under $x \mapsto x^p$ contains prime subfield $F_p$ and roots of the polynomial $t^p-t$, but $\vert F_p \vert = p$, so the invariant subset is exactly the prime subfield $F_p$ and the answer is yes.

Moreover, multiplication group $F_{p^n}^*$ is cyclic of order $p^n-1$ and by Lagrange theorem any its element has order $k \in \mathbb{N}$, where $k$ divides $p^n-1$.

Is everything correct?

• No. Your proof does not make sense. You should start with a fixed subspace $V$ and show under the hypothesis that $\sigma(V) = V$ that $V$ is a field, i.e that it is stable by multiplication and inverse. What you said is true but it only show that $F_p$ is fixed by $\sigma$, which is weaker that your exercise. – user171326 Mar 27 '17 at 21:09
• @N.H. But for $a \in \mathbb{F}_p,\sigma(a) = a$. Thus, if $\sigma(V) = V$ then $\sigma(V \setminus \mathbb{F}_p) =V \setminus \mathbb{F}_p$ and hence $V \setminus \mathbb{F}_p$ is not a field. So the answer is no ? – reuns Mar 27 '17 at 21:16
• $V \backslash \mathbb F_p$ is not a subspace. – user171326 Mar 27 '17 at 21:18
• @N.H. Why that ? We are seeing $\mathbb{F}_{p^k}$ as a $k$-dimensional $\mathbb{F}_{p}$-vector space so $\mathbb{F}_{p^k} \setminus \mathbb{F}_{p}$ is a subspace – reuns Mar 27 '17 at 21:21
• $0$ is not in your "subspace". – user171326 Mar 27 '17 at 21:22