Is $x+x^{2^t}$ a cube in the finite fields with characteristic 2?

Let $m,t$ be two odd positive integers with $m>3$. Let $x\in \mathbb{F}_{2^{2m}}$. I conjecture that the following assertion holds due to some numerical experiments. However, I only can prove this for the case of $t=1$.

If $x^{2^m+1}=1$ or $x^{2^m+1}=x+x^{2^m}$, then $x+x^{2^t}$ is a cube in $\mathbb{F}_{2^{2m}}$.

Here, an element $c$ is called a cube in $\mathbb{F}_{2^{2m}}$, if $c=a^3$ for some $a \in \mathbb{F}_{2^{2m}}$.

Solution 1:

Let $k = \mathbf{F}_{2^m}$ be the finite field of order $q = 2^m$. Let $l = \mathbf{F}_{2^n}$ be the finite field of order $q^2$.

Since $m$ is an odd, it follows that $q \equiv 2 \bmod 3$, and $q^2 \equiv 1 \bmod 3$. In particular, $|k^{\times}| = q-1$ has order prime to $3$, and $|l^{\times}| = q^2 - 1$ has order divisible by $3$. Since the non-zero elements of finite fields form a cyclic group, it follows that any element of $k^{\times}$ thought of as an element of $l^{\times}$ will always be a cube.

The Galois group $\mathrm{Gal}(l/k)$ of order $2$ is generated by the Frobenius map $\sigma = \mathrm{Frob}_{q}$ such that

$$\sigma(z) = z^q$$

for all $z \in l$.

Let me prove the claim in the case that $N=1$ and $t$ is odd.

Suppose that $N = x^{q+1} = 1$. The same equation holds for all the conjugates of $x$ under $\mathrm{Gal}(l/\mathbf{F}_2)$, and hence it also holds for $y$ where $y^2 = x$ (so $\mathrm{Frob}_2 y = x$). We deduce that $\sigma(x) = x^q = x^{-1}$ and also $\sigma(y) = y^q = y^{-1}$.

Now consider the element

$$z = x + x^{2^t} = y^2 + y^{2^{t+1}} = y^{2^t + 1} (y^{1-2^t} + y^{2^t - 1}).$$

If $t$ is odd, then $2^t + 1$ is divisible by $3$, and thus $y^{2^t+1}$ is a cube. Hence to show that $z$ is a cube it suffices to show that

$$y^{1-2^t} + y^{2^t - 1}$$

is a cube. But it is invariant under $\sigma$, so it lies in $k$, and hence it is a cube from the discussion above.

On the other hand, the result doesn't appear to be true in general when $N = 1$ and $t$ is even. Suppose that $m=5$, so $q=32$. Suppose that $x$ is a root of

$$x^{10} + x^9 + x^5 + x + 1 = 0,$$

which is irreducible over $\mathbf{F}_2$, then $x^33 = 1$. On the other hand,

$$(x+x^4)^{341} = x^9 + x^6 + x^5 + x^2 + 1,$$

where the latter is a cube root of unity. So $(x+x^4)$ is not a cube in this case. In fact, it has order $1023$ and is a generator of $l^{\times}$.

Solution 2:

@user994373 Thanks a lot for your help. With your help, I can prove this quenstion. The proof is as follows.

For the case that $x^{2^m+1}=1$, the expert @user994373 had shown the assertion. To show the assertion for the case of $x^{2^m+1}=x+x^{2^m}$, we first show that the following asssertion holds.

$\bf{Assertion}:~$ if $x+x^{2^m}=1$, then $$x+x^{2^t}$$ is a cube.

In fact, if $x+x^{2^m}=1$, then from the fact that $x^2+(x+x^{2^m})x=x^{2^m+1}$, we have $x+x^2=x^{2^m+1}\in \mathbb{F}_{2^m}$, which implies that $x+x^{2^t}=(x+x^2)+\cdots+(x+x^2)^{2^{t-1}}\in \mathbb{F}_{2^m}$ is a cube.

Now, assume that $x^{2^m+1}=x+x^{2^m}$, and $x\neq 0$. Then we have

$$\frac{1}{x}+(\frac{1}{x})^{2^m}=1,$$

which implies that $\frac{1}{x}+(\frac{1}{x})^{2^t}=\frac{x+x^{2^t}}{x^{2^t+1}}$ is a cube by the above assertion. However, $x^{2^t+1}$ is a cube in $\mathbb{F}_{2^{2m}}$, since $t$ is odd. Hence $x+x^{2^t}$ is a cube.