Worked solution: Wiles's modularity proof via the Taylor–Wiles method (1994)

Step 3 of 11: Why the Frey curve is strange: semistability and a

2p

-th-power odd discriminant part

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In plain words

The Frey curve has roots $0,a^p,-b^p$ . Their pairwise differences include $a^p$ , $b^p$ , and $c^p$ , so the cubic root-discriminant is a $2p$ -th power. The elliptic-curve discriminant is instead $\Delta_E=16(abc)^{2p}$ : its odd-prime part has the special $2p$ -th-power form, while the factor $16$ must not be ignored.

\Delta(E_{a,b,c}) = 16(a^p - 0)^2(0 - (-b^p))^2(a^p - (-b^p))^2 = 16(abc)^{2p}, \quad N = \operatorname{rad}(abc)

Detailed analysis

For $y^2=(x-r_1)(x-r_2)(x-r_3)$ , the root-discriminant is $\operatorname{disc}=\prod_{i<j}(r_i-r_j)^2$ . With roots $0,a^p,-b^p$ and $a^p+b^p=c^p$ , this gives $\operatorname{disc}=(abc)^{2p}$ . The elliptic-curve discriminant includes the model factor: $\Delta_E=16(abc)^{2p}$ . Thus the odd-prime part is a $2p$ -th power, but the displayed elliptic discriminant is not literally one. For $q\nmid\Delta_E$ the reduction is smooth; the conductor in the Frey-curve argument is $N=\operatorname{rad}(abc)$ .

Terms in this step

Discriminant $\Delta$: For $y^2 = (x - r_1)(x - r_2)(x - r_3)$ , the quantity $\Delta = (r_1 - r_2)^2(r_1 - r_3)^2(r_2 - r_3)^2$ . It is nonzero if and only if the three roots are distinct, and its prime factors are the primes where the curve degenerates.
Semistable elliptic curve: An elliptic curve over $\mathbb{Q}$ that has only good reduction (smooth) or multiplicative reduction (a simple two-root collision, never a three-root collision) at every prime $q$ . Equivalently, its conductor $N$ is square-free (Wiles 1995, p. 448).
Conductor $N$: A positive integer encoding the primes where an elliptic curve has bad reduction and how severe the degeneration is. For a semistable curve, $N$ is simply the product of all primes of bad reduction.

Knowledge used in this step

Fermat's Last Theorem Arithmetic geometry Congruences and modular arithmetic