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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

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Thanks to everyone for all the excellent responses. I wish I could accept multiple answers. –  Noah S Dec 23 '10 at 20:56
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4 Answers

up vote 8 down vote accepted

Actually, I am harder-pressed to say anything the supposedly simpler question. For Question 1, I would just add that we need the existence of an infinite computably enumerable set of primes satisfying FLT, and that this ought to suffice. (Since the set P of primes NOT satisfying FLT is obviously $\exists$-definable and hence c.e., another way to say this is that we need to show that P is not a simple set.) I would not have guessed that any odd prime was known to satisfy FLT, up until Wiles's proof, but I'm ready to be corrected on that.

For Question 2, I do think that one could come up with a proof avoiding FLT entirely. I can't do so myself; I'm in computability, and when I was thinking about computable categoricity for fields of infinite transcendence, I realized that we needed multivariable polynomials with known finite numbers of rational solutions, and I thought of the Fermat polynomials because I didn't know any other candidates. (Of course, this condition was not all that was needed, but clearly it's necessary.) I'm still a computability theorist, and I still don't know any other candidates, but field theorists have told me that they could come up with other such polynomials fairly readily. Whether those would satisfy the more difficult requirements (basically Theorem 3.1 in the paper) is not so clear, but I suspect that it can be done with other polynomials. Bjorn Poonen suggested at one point that the Fermat polynomials were actually a bad choice, because their symmetry creates an extra solution whenever one adds a single transcendental solution to the field.

As a related question: is there a computably stable field of infinite transcendence degree? A computable structure $\mathcal{A}$ is computably stable if, for every computable structure $\mathcal{B}$, every isomorphism from $\mathcal{A}$ onto $\mathcal{B}$ is computable. (Example: $\mathbb{Z}$ under the successor function.) A common way to build computably stable structures is to make them computably categorical and rigid, i.e. with no nontrivial automorphisms, so that the isomorphism from $\mathcal{A}$ onto any computable copy must be unique (by rigidity), hence computable (by categoricity). I would conjecture that it is possible to mimic the construction in the Fermat paper, with different polynomials, and to get a computably stable field of infinite transcendence degree, but I certainly don't know offhand what polynomials one might use.

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Russell, concerning your remark about odd primes known to satisfy FLT before Wiles, Euler had essentially treated the case of exponent 3 and Kummer gave conditions in the mid-19th century which were able to verify FLT for most odd primes below 100. –  KConrad Dec 23 '10 at 1:28
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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.

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From my reading of the cited paper, FLT is not really needed. The authors require a recursive sequence of curves $\langle C_i : i \in I \rangle$ for which the following holds:

(1) The function $i \mapsto |C_i({\mathbb Q})|$ is computable.

(2) There is a function (possibly required to be recursive) $B: \{ g \in \omega : g \geq 2 \} \to \omega$ so that for any $g \geq 2$ and any curve $X$ of genus $g$ over ${\mathbb Q}$ there are no nonconstant maps $C_i \to X$ or $X \to C_i$ for $i \geq B(g)$.

The Fermat curves conveniently satisfy these properties, but it is not so hard to build a sequence of curves over the rational numbers of growing genus for which local conditions preclude the existence of rational points and whose Jacobians are simple as abelian varieties (Edit: comment about prime genus removed; instead, one needs to argue that the generic curve does have a simple Jacobian ).

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I certainly agree with the overall assessment of the answer. The last sentence confuses me: what does the simplicity of the Jacobian have to do with the primality of the genus? –  Pete L. Clark Dec 23 '10 at 12:38
    
Good point --- primality of the genus has nothing to do with simplicity of the Jacobian. Obtaining curves of arbitrarily high genus with simple Jacobians and no rational points requires something a little more sophisticated than restricting properties of the genus. –  Thomas Scanlon Dec 23 '10 at 20:49
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Regarding question 1: I am pretty sure that only the full proof has given us FLT for infinitely many prime exponents. As soon as we know that there are infinitely many regular primes, we will have an alternative route to FLT for infinitely many prime exponents, but I am going to stick my neck out and say that that's unlikely to be easy. So at present, the answer is: to prove FLT for infinitely many exponents, you need the modularity theorem (at least for semistable elliptic curves).

Edit: to be clear, Kummer's proof that FLT holds for regular prime exponents is of course much simpler than the proof of the modularity theorem. Moreover, there is a probabilistic heuristic that suggests that more than half of the primes are in fact regular, and this agrees extremely well with numerical data. The "unlikely to be easy" comment referred to the missing tile of the puzzle: namely to turn the heuristic into a rigorous proof for infinitude of regular primes.

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Roger Heath-Brown (and Fouvry independently) proved FLT for infinitely many primes in 1985. See Heath-Brown's paper "The first case of Fermat's last theorem", Inventiones vol. 79. –  David Hansen Dec 23 '10 at 22:08
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@David I am aware of that paper. To prove the first case of FLT does not mean to prove it for that exponent. –  Alex B. Dec 23 '10 at 23:43
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