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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.)

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    $\begingroup$ If you believe that, how do you account for $x = 1 + t + t^2z$ satisfying $Q(x)=0$, where $Q$ is the polynomial $Q(y) = y^3-(1+t)$? You can use Hensel's Lemma to solve for $z$. $\endgroup$ – Jason Starr Jul 7 '18 at 18:26
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    $\begingroup$ @JasonStarr You can multiply $Q$ by $y$. The resulting polynomial doesn't have to be irreducible. $\endgroup$ – MTyson Jul 7 '18 at 18:30
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    $\begingroup$ In fact it will never be irreducible, because it has to be divisible by $y$. $\endgroup$ – Kaban-5 Jul 7 '18 at 18:34
  • $\begingroup$ Now I understand. I remember something like this in one paper of Harm Derksen, but I may be misremembering. $\endgroup$ – Jason Starr Jul 7 '18 at 19:05
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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. 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 $q$.)

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    $\begingroup$ I am not sure but It seems that your argument works when $P$ is polynomial with coefficients from $F$, not $F[t]$. Do I miss something? $\endgroup$ – Kaban-5 Jul 7 '18 at 19:48
  • $\begingroup$ Good point! I just realized I was being sloppy about about that. The argument still works, I just need to be a little more careful because F[t] is not perfect. Will fix! $\endgroup$ – Alison Miller Jul 7 '18 at 19:53
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In fact for any integral domain $R$, if $x$ is any element that satisfies a polynomial equation of degree $n$ over $R$, and $S$ is any set of numbers of size greater than $n$, then $x$ satisfies a polynomial equation of the form $\sum_{i \in S} c_i x^i$ for $c_i \in R$ not all zero.

This is just because $\{x^i | i\in S\}$ cannot be linearly independent over the field of fractions of $R$, so some linear relation holds over the field of fractions, and we can clear denominators.

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