Let $F = \mathbb{F}\_2$ be the field with two elements. I will denote the rings of polynomials and formal
power series over $F$ as $F[t]$ and $F[[t]]$ respectively. Suppose that $x \in F[[t]]$ is algebraic
over $F[t]$ (there exists a non-zero polynomial $P$ with coefficients in $F[t]$ such that $P(x) = 0$).
Is it true that there exists a non-zero polynomial $Q(y) = \sum\limits\_{k = 0}^m q\_k y^{2^k}$ with coefficients $q\_k$ 
from $F[t]$ such that $Q(x) = 0$ (the difference with is that all non-zero coefficients in $Q$ correspond to the powers of two)? 


(I encountered this statement being used without proof in literature, so it most probably is true.)



Algebraic power series over $\mathbb{F}_2$ as roots of polynomials of special form
Getting rid of the power series, your question boils down to showing that any polynomial $P(x)$ divides some polynomial $Q(x)$ which has the form $Q(x) = \sum\_{i} c\_i x^{2^i}$. Polynomials $Q$ which have this property are known as [additive polynomials](https://en.wikipedia.org/wiki/Additive_polynomial "additive polynomials"). They have some nice alternative descriptions: for instance, if $K$ is a field of characteristic $2$ then a polynomial $Q(x) \in K(x)$ with distinct roots is additive if and only if the set of roots of $Q(x)$ in $\overline{K}$ form a $\mathbb{F\_2}$-vector space. (This is a nice exercise, or see *Basic Structures of Function Field Arithmetic* by Goss for a proof.) 


Using this description, it's easy to construct $Q(x)$. Let $K = \mathbb{F\_2}(t)$ and let $V$ be the $\mathbb{F\_2}$-subspace of $\overline{K}$ generated by the roots of $P$. We set $Q(x) = (\prod\_{\alpha \in V} (x-\alpha))^{2^N}$ for $N$ chosen to be large enough so that $P(x)$ divides $Q(x)$. Then $Q(x) \in K(x)$ because the finite vector space $V$ is stable under the action of $\mathrm{Gal}(\overline{K}/K)$, and $Q(x)$ is additive by the fact of the previous paragraph + Frobenius


(Note that everything here generalizes in the obvious way if you replace $\mathbb{F}\_2$ by $\mathbb{F}\_q$ for any prime power $q$.)


