Quick introduction to the problem: suppose that $F$ is a finite field of characteristic 2 (for purposes of this post $F = \mathbb{F}_2$ will suffice) and let $F[t]$ and $F(t)$ be its ring of polynomials (polynomials over a finite field are clearly finite enumerable objects) and field of rational functions respectively. Let $P_1, P_2, \ldots P_m$ be polynomials of variables $x_1, \ldots, x_n$ with coefficients in $F[t]$.

*Hilbert's tenth problem for $F(t)$* asks whether existence of $x_1, \ldots, x_n \in F(t)$ such that
$P_1 (x_1, x_2, \ldots, x_n) = P_2 (x_1, x_2, \ldots, x_n) = \ldots = P_m (x_1, x_2, \ldots, x_n) = 0$ is decidable (clearly, $n, m$ and $P_i$ are finite enumerable objects, so the problem is well-posed). Understandably, this problem
is often called *the diophantine problem for $F(t)$* and equations
$P_i (x_1, x_2, \ldots, x_n) = 0$ are called *diophantine*.

Paper in question supposedly proves that answer is indeed "undecidable" by reducing a undecidable problem in some first-order logic (that is not relevant to the question, I guess) to the diophantine problem over $F(t)$.

Well, preliminaries end here. I believe that lemma 2.1 (and its proof, of course) is incorrect. I will quote its statement here:

Let $x \in F(t)$. Then $x \in \{ t^{2^s} : s \geqslant 1 \} \Leftrightarrow \exists u, v, w, s \in F(t)$ such that: \begin{equation} x + t = u^2 + u \end{equation} \begin{equation} u = w^2 + t \end{equation} \begin{equation} x^{-1} + t^{-1} = v^2 + v \end{equation} \begin{equation} v = s^2 + t^{-1} \end{equation}

Well, following values of $u, v, x, s, w$ seem to be a counterexample to this lemma (clearly, $x$ is not a polynomial, let alone $t^{2^k}$ for some $k \geqslant 0$):

\begin{equation} u = \frac{t^1+t^5+t^6+t^7}{1+t^4+t^6} \end{equation} \begin{equation} v = \frac{1+t^1+t^2+t^6}{t^1+t^3+t^7} \end{equation} \begin{equation} x = \frac{t^2+t^6+t^{14}}{1+t^8+t^{12}} \end{equation} \begin{equation} s = \frac{1}{1+t^1+t^3} \end{equation} \begin{equation} w = \frac{t^3}{1+t^2+t^3} \end{equation}

Moreover, conditions $2$ and $4$ from the lemma look a bit fishy: $x + t = u^2 + u, u = w^2 + t \Leftrightarrow x + t = (w^2 + t)^2 + (w^2 + t) = (w^4 + t^2) + (w^2 + t) = w^4 + w^2 + t^2 + t \Leftrightarrow x = (w^2 + w + t)^2$ (here I use that $(a + b)^2 = a^2 + b^2$ in characteristic 2). In the same way $x^{-1} = (s^2 + s + t^{-1})^2$. Basically, they are rewritten in the form $x = y^2, y = (w^2 + w + t), y^{-1} = (s^2 + s + t^{-1})$ or $x = y^2, y + t = w^2 + w, y^{-1} + t^{-1} = s^2 + s$ (here I use that $a^2 = b^2 \Leftrightarrow a = b$ in $F(t)$). For $y$ we get just conditions $1$ and $3$ from the lemma. As it is said in paper itself (an explicit counterexample is given, on which the counterexample above is based) it is possible to choose $y \notin \{ t^{2^k} | k \geqslant 0 \}$ that it satisfies conditions $1$ and $3$. Then $x = y^2$ also does not lie in $\{t^{2^k} | k \geqslant 0 \}$.

And finally, an error in the proof seems to be in the following line (this requires actually reading the proof though): "if $q \not | ~n$ then there exists nonconstant polynomial $p$ such that $p | q, p | b, p \not | ~n$". This does not sound correct, because if, for example $b | n$ (I can't see why this can't happen), there is no such $p$. Moreover $q \not | ~n$ does not imply that a situation like $b = n, q | n^3$ can't happen ($q$ itself is not necessarily a prime).

You may wonder why I am asking this here. The reason is that I already talked about this issue with my scientific adviser and have written a letter to the author ($1,5$ months ago), who have not responded yet and I am not expecting response at all. So I am asking: are my doubts true? And what should I do about this situation?

to answer an email? $\endgroup$ – Francesco Polizzi Jun 27 '18 at 10:17