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My question is related to the fact that Frey's curve:

$$y^{2} = x(x-a^{p})(x+b^{p})$$

Could not be constructed if Fermat's last theorem holds true. I mean because Fermat's last theorem implies that there is no solution for $a$, $b$, $c$, and prime number $p > 2$ such that:

$$a^{p}+b^{p}=c^{p}$$

As a result the Frey's curve which is semi-stable elliptic curve could be not constructed because according to Ribet's theorem this curve is not modular but Taniyama-Shimura-Weil conjecture implies that each elliptic curve should be modular. The thing I can't understand that is when we say Frey's curve could not be constructed in what sense it could not be constructed? I mean the equation $y^{2} = x(x-a^{p})(x+b^{p})$ contains only $a^{p}$ and $b^{p}$ but it's not completely symmetric to contain all $a$, $b$, $c$ in a symmetric way. I appreciate if someone could explain this to me.

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    $\begingroup$ If I recall correctly, the entire point of the Frey curve is that the pairwise distances between the three places it crosses the $x$-axis are $a^p, b^p$ and $a^p+b^p$. If $a^p+b^p=c^p$, then all three distances (and therefore the discriminant) are $p$th powers, and somehow such an elliptic curve cannot have an associated modular form. It's been a while since I last heard a lecture on the subject, though, so I'm somewhat uncertain on the details. $\endgroup$
    – Arthur
    Sep 9, 2018 at 18:46
  • $\begingroup$ You mean its roots are $a^{p}$, $b^{p}$, and $c^{p}$? I don’t think so... $\endgroup$
    – GGG
    Sep 9, 2018 at 18:49
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    $\begingroup$ No, the roots are $0, a^p$ and $-b^p$, obviously. That much you can read off directly from $y^2=x(x-a^p)(x+b^p)$. Now, what are the pairwise distances between those three roots? $\endgroup$
    – Arthur
    Sep 9, 2018 at 18:50
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    $\begingroup$ You can certainly write down the curve $y^2 = x(x - a^p)(x + b^p)$. But this curve is not called a Frey curve unless there is a $c$ such that $a^p + b^p = c^p$ (and $p \ge 3$), and since there is never such a $c$, there aren't any Frey curves. $\endgroup$ Sep 9, 2018 at 18:51
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    $\begingroup$ No, the construction of the curve itself is really as straight-forward as this. Proving that the curve cannot exist is where the work lies. $\endgroup$
    – Arthur
    Sep 9, 2018 at 19:00

1 Answer 1

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In 1969 Hellegouarch performed the elliptic curves $F (a, b)$, which were later named after Gerhard Frey and were constructed from the solutions of the Fermat equation, he reported on the curves in an article in Acta Arithmetica. I have found the article but it is not in English.

Let me give you a over-simplified version taken form the lecture-series "Riemann Hypothesis and its Applications" (Lecture-29) by Prof. Manindra Agrawal, Department of Computer Science and Engineering, IIT Kanpur (won the 2006 Fulkerson Prize, and the 2006 Gödel Prize for the AKS primality test).

First Step

Let $F:= y^2=x^3+ax^2+bx+c$, then discriminant $\Delta_F=(ab)^2+4a^3c-4b^3+18abc-27c^3.$

If $c=0$, then $y^2=x^3+ax^2+bx$, $\Delta_F=(ab)^2-4b^3.$

Now the Frey's curve is , $y^2=x(x-A)(x+B)=(x^2-Ax)(x+B)=x^3-Ax^2+Bx^2-ABx.$

Let, $a=(B-A), b=-AB,$ then $\Delta= \{(B-A)(-AB)\}^2-4(-AB)^3= \{(B-A)^2-4(-AB)\}(-AB)^2=(A^2+B^2-2AB+4Ab)(AB)^2=(A+B)^2(AB)^2.$

Now, let, $A=a^q, B=b^q,$ then, $\Delta_F=(a^q+b^q)^2(ab)^{2q},$

if $a^q+b^q=c^q$ is true for $q>2$ (this is Fermat's last theorem),

then the discriminant $\Delta_F=(c^q)^2(ab)^{2q}=(abc)^{2q}.$

Second Step

If $\Delta_F$ is $l^{th}$ power of an integer, then it has a point of order $l$. If $a^q+b^q=c^q$ is true for $q>2$, then the Frey curve $F$ has a point a point of order $2q$, since it's discriminant $\Delta_F$ is $2q^{th}$ power of an integer $(abc)$.

Third Step

But if $F$ is modular, then, it does not have a point of order $ > C$ where $C$ is a constant.

Fourth Step

It has been proved that every elliptic curve is modular (Taniyama–Shimura conjecture, Andrew Wiles proved the Taniyama–Shimura conjecture for semistable elliptic curves, which was enough to imply Frey curve is modular).

Final Step

Since Frey curve $F$ is modular, then it can not have a point of order $ \geq C$, but the Frey curve $F$'s discriminant $\Delta_F$ is $2q^{th}$ power of an integer $(abc)$, so $q$ can not be $> \frac{C}{2}$, does there could not be solution $a^q+b^q=c^q$ for $q> \frac{C}{2}$, i.e. there are finite $q$ for which Fermat's Last Theorem is true.

Frey's curve can not be constructed since by the definition of the Frey's Curve, it must have a point of order $2q$ but at the same time it is an elliptic curve, thus modular, so, the Frey's Curve can not have a point of order $ \geq C$, so after certain value of $q$, Frey's curve can not be constructed.

Of-course I have missed many intricate details for the sake of accessibility.

In a bit more technical terms: The group of $p$-torsion points $F[p]$ on $F$ (which is a two-dimensional vector space over the field $\mathbb F_p$ of $p$-elements),

equipped with an action of the Galois group of $\overline{\mathbb Q}$ over $\mathbb Q$) has very special properties,

in algebraic number theory terms, it is very close to being unramified, specifically, but more technically, it is unramified except possibly at $2$ and $p$, and at $p$ the ramification is very mild, it is finite flat.

Now the Shimura-Taniyama conjecture, which is what Wiles (together with Taylor) proved, shows that $F$, and so $F[p]$, arises from a weight two modular form. Ribet's earlier results on Serre's epsilon conjecture imply that this modular form must actually be of level $2$. (This is where we use the above information about the ramification.) But there are no non-zero cuspforms of weight $2$ and level $2$, and we get a contradiction.

Although it is much harder (in that the only way we know to rule out the existence of $F[p]$ is by the - quite difficult- Shimura-Taniyama conjecture, or else by related more recent results such as Khare and Wintenberger's work on Serre's conjecture), one can think of the non-existence of $F[p]$ as being analogous to Minkowski's theorem in algebraic number theory, which says that an everywhere-unramified extension of $\mathbb Q$ cannot exist (See this answer).

For detail proof, read following books:

1. Fermat's Last Theorem: Basic Tools by Takeshi Saito (Publication Year: 2013),

2. Fermat's Last Theorem: The Proof by Takeshi Saito (Publication Year: 2014).

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    $\begingroup$ Your answer is not up to the level of expectation, 3rd and 4th step is unclear, however I am awarding the bounty, not because you deserve it, because, otherwise it will be wasted. $\endgroup$
    – Michael
    Oct 22, 2020 at 8:59

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