TITLE: Fermat's Last Theorem and Computability Theory
QUESTION [14 upvotes]: This question stems from the paper "Computably categorical fields via Fermat's Last Theorem," by Russell Miller and Hans Schoutens (available online at http://qcpages.qc.cuny.edu/~rmiller/Fermat.pdf). In this paper, they construct a field $F$ by starting with the field $\mathbb{Q}(x_0, x_1, x_2, . . .)$ of infinite transcendence degree over $\mathbb{Q}$, and then adjoining elements $y_i$ such that $(x_i, y_i)$ is a solution to the polynomial $X_i^{p_i}+Y_i^{p_i}=1$ for some odd prime $p_i$ (see paragraph 2, page 3); they then show that the resulting field has interesting computability-theoretic properties. In particular, they show that this field is computably categorical (i.e., any two computable presentations are computably isomorphic). I have only started reading this paper, but I have two questions, a simple one and a probably not-so-simple one:
Question 1: It is unclear to me exactly how much of Fermat's Last Theorem (FLT) is required for this paper, but certainly we need at least the existence of infinitely many primes $p$ such that $X^p+Y^p=1$ has no nontrivial rational solutions. How difficult is this fact to prove? (And, historically, when was it first known?)
Question 2: How much of FLT is actually required for the paper? I would be very interested if full FLT was required; although, as the authors point out, there has been at least one previous attempt made to prove the same result that apparently did not rely on FLT.
Thank you very much in advance,
Noah S

REPLY [16 votes]: The Fermat curves (and FLT) are used in the quoted paper to prove the following result:
Theorem 3.1: There exists a (recursive) infinite sequence $\mathcal{C} = \{C_i\}$ of curves $C_i/\mathbb{Q}$ such that:
(i) for all $_i$, one can effectively determine the set $C_i(\mathbb{Q})$ and
(ii) $\mathcal{C}$ is non-covering: for all $i \neq j$, there does not exist a finite $k$-morphism $C_i \rightarrow C_j$.
As the authors remark, any family of nonisomorphic curves of common genus $g \geq 2$ is noncovering, so it would suffice to exhibit such an infinite family of curves such that each one provably has no $\mathbb{Q}$-rational points.
This is easy, and we do so below for each $g \geq 2$:
Start with the (smooth projective model of the) curve given by the following equation.
$C: y^2 = - (x^{2g+2} + 1)$.  
This is a hyperelliptic curve of genus $g$ which is visibly without $\mathbb{R}$-points, so certainly $C(\mathbb{Q}) = \varnothing$.
Now, among many other constructions, one can perturb the coefficients of $C$ slightly to get infinitely other genus $g$ curves without $\mathbb{Q}$-rational points.  Indeed, write $P(x) = -x^{2g+2} - 1$ and for $a = (a_1,\ldots,a_{2g+2}) \in \mathbb{Q}^{2g+2}$, consider the polynomial 
$P_a(x) = P(x) + \sum_{i=0}^{2g+2} a_i x^i$.
There exists an effectively computable $\epsilon > 0$ such that if for all $i$, $|a_i|_{\infty} < \epsilon$, then $P_a(x)$ is negative for all real $x$, so the curve $C_a: y^2 = P_a(x)$ still has no $\mathbb{R}$-points and hence no $\mathbb{Q}$-points.  Here $|x|_{\infty}$ is the standard Archimedean norm.  
It is no problem to find infinitely many choices of $a$ leading to nonisomorphic curves $C_a$.  For instance, one can find choices of $a$ which give rise to curves with pairwise distinct moduli.  Alternately, by a standard weak approximation / Krasner's Lemma argument one can choose the coefficients so as to make the curve $C_a$ have / not have $\mathbb{Q}_p$-points for each prime $p$ in any finite set $S$ of rational primes.  More details for this can be supplied on request.
There are many other possible constructions.  For instance, in this paper I produced for each number field $K$ and each $g \geq 2$ a bielliptic curve $C_{/\mathbb{Q}}$ of genus $g$ which violates the Hasse Principle.  In fact by varying the elliptic curve that $C$ maps $2:1$ onto, one gets infinitely many such examples.  Note that here the curves have points over every completion but still no $\mathbb{Q}$-points.  In general, producing curves which do not have $\mathbb{Q}$-points "for local reasons" is much easier.
In my opinion emphasizing the connection to Fermat's Last Theorem here is a bit misleading.